Induced Modules over Group Algebras (North-Holland Mathematics Studies)

INDUCED MODULES OVER GROUP ALGEBRAS

NORTH-HOLLAND MATHEMATICS STUDIES 161 (Continuation of the Notas de Matematica)

Editor: Leopoldo NACHBIN Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro, Brazil and University of Rochester New York, U.S.A.

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

INDUCED MODULES OVER GROUP ALGEBRAS

Gregory KARPILOVSKY Department of Mathematics California State University Chico, CA, U.S.A.

1990

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

ELSEVIER SCIENCE PUBLISHERS B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands Distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 655 Avenue of the Americas New York, N.Y. 10010, U.S.A.

Library of Congress Cataloging-in-Publication D a t a K a r p i l o v s k y . G r e g o r y , 1940Induced modules over group algebras Gregory Karpilovsky. p. cm. -- ( N o r t h - H o l l a n d m a t h e m a t i c s s t u d i e s , 161) Includes bibliographical references. ISBN 0-444-88414-9 ( U . S . ) 1 G r o u p a l g e b r a s . 2. M o d u l e s ( A l g e b r a ) I. T i t l e . 1 1 . S e r i e s . P A 1 7 1 . K 3 7 8 1990 512 .24--dC20 @9-49199 f

CIP

ISBN: 0 444 88414 9 ELSEVIER SCIENCE PUBLISHERS B.V., 1990 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science Publishers B.V./Physical Sciences and Engineering Division, P.0.Box 103, 1000 AC Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center lnc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Printed in the Netherlands

To my wife Helen, who typed it all

This Page Intentionally Left Blank

vii

In 1898 Frobenius discovered a construction which, in present terminology, associates with every module of a subgroup the induced module of a group. This construction proved t o be of fundamental importance and is one of the basic tools in the entire theory of group representations. The present monograph is designed for research mathematicia,ns and advanced graduate students and gives a picture of the general theory of induced modules as it exists at present. Much of the material presented has heretofore been available only in research articles. Our approach is not intended to be encyclopedic, but each topic is considered in sufficient depth that the reader may obtain a c1ea.r idea of the major results in the area. A word about notation. As is customary, Theorem 5.4.2 denotes the second result of Section 4 of Chapter 5; however, for simplicity, all references to this result within Chapter 5 itself are designated as Theorem 4.2. The following is a brief description of the content of the book. After establishing algebraic preliminaries (Chapter l),the general facts about induced modules are provided (Chapter 2). Following presentation of some formal properties of induced modules, we examine in detail their annihilators and record a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. In Chapter 3 we provide a detailed information on the process of induction from normal subgroups. Among other results, we discover conditions under which induced modules are totally indecomposa.ble or absolutely indecomposable. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to establish a criterion for induced modules to be homogeneous. The chapter ends with the discussion of the Loewy length of induced modules.

viii

Preface

Let H be a subgroup of G, let F be a field and let V be an F H module. It is natural to investigate circumstances under which the induced module V G has projective indecomposable summands. An important contribution to this problem was marked by the appearance of the paper of Robinson (1989) to which Chapter 4 is devoted. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G (if there are any) have relatively large dimension. The aim of Chapter 5 is twofold: first to present some basic results of Green theory with refinements and extensions achieved in scope of recent developments, and second to provide a number of applications. The chapter ends with a result of Alperin (1986) which relates the Green correspondence with normal subgroups. In Chapter 6 we investigate in detail circumsta.nces under which restriction and induction of irreducible modules are completely reducible modules. As one of the applications, we show that if H is a subgroup of G and F is an arbitrary field of characteristic p > 0, then the following conditions are equivalent: (i) V Gis completely reducible for any irreducible FH-module V and MH is completely reducible for any irreducible FG-module M . (ii) There exists a normal suhgroup N of G such that N H and N has p'-index in G. Chapter 7 provides a detailed account of a distinguished class of induced modules, namely permutation modules. These modules hold much information about the p-modular representations of G, as well as information about the fusion of p-subgroups of G. The main results presented are due to Alperin (1988), Brouk and Robinson (1986), Robinson (1988), Dress (1975), Saksonov (1971) and Brouk (1985). Chapter 8, the final chapter, is based exclusively on an important work of Weiss (1988). A number of applications to the isomorphism problem for group rings is also presented. I would like to express my gratitude to my wonderful wife for the tremendous help and constant encouragement which she has given me in the preparation of this book (as well as in the previous 11 books).

ix

Contents Preface Chapter 1. 1. 2. 3. 4.

5. 6.

Preliminaries Notation and terminology Tensor products Artinian, noetherian and completely reducible modules The radical of modules and rings The Wedderburn- Artin theorem Group algebras and group representations

Chapter 2. 1. 2. 3. 4. 5. 6. 7. 8.

General properties of induced modules . Induced modules, representations and characters Formal properties of induced modules Annihilators of induced modules Clifford’s theorem Dual and contragredient modules Induction, restriction, and outer tensor products Mackey theorems and their applications Counting induced modules and characters 9. Relative trace maps 10. Induction and relative projectivity 11. Unique decompositions 12. Projective covers

Chapter 3. 1. 2. 3.

4. 5. 6. 7. 8. 9.

Induction from normal subgroups Complete noetherian local rings Reduction to G-invariant modules Group-graded algebras and crossed products The endomorphism ring of induced modules Relations between the decomposition of V G and E n d ~ cV( c ) Twisted group algebras over fields Total and absolute indecomposability of induced modules Crossed products over prime rings Homogeneity of induced modules

1 1 4 8 19 29 37

45 45 54 66 74 79 87 89 97 103 107 120 123 131 132 139 141 153 159 167 180 191 208

Contents

X

Frobenius and symmetric algebras Symmetric crossed products EndFc( V') is symmetric Graded modules Induction from irreducible modules and their projective covers Inflated modules over twisted group algebras Induction of absolutely irreducible modules Applications The Loewy length of induced modules

210 220 230 232

Chapter 4. 1. 2. 3.

Projective summands of induced modules The Reynolds ideal Projective summands Applications

273 273 280 295

Chapter 5. 1. 2. 3.

Green theory Vertices and sources The Green correspondence The endomorphism ring of the Green correspondents The Green correspondence and Brauer's induction theorem The Green correspondence and the Brauer lift The Green correspondence and normal subgroups

299 299 309

Chapter 6. 1. 2. 3. 4. 5.

Simple induction and restriction pairs Blocks of algebras Defect groups of blocks Blocks and vertices Simple induction and restriction pairs Complete reducibility of induced modules

34 1 34 1 350 358 365 373

Chapter 7. 1. 2. 3. 4. 5. 6. 7.

Permutation modules Preliminary results Hecke algebras Fusion and permutation modules Complete reducibility of ( 1 ~ ) ~ Induction from Sylow p-subgroups Loewy series for transitive permutation modules Characterizations of p-perrnutaion modules

383 383 392 403 409 416 426 433

10. 11. 12. 13. 14. 15. 16. 17. 18.

4. 5. 6.

238 242 256 26 1 266

318 329 334 337

xi

Contents 8. 9. 10.

Chapter 8. 1. 2. 3. 4. Bibliography Notation Index

The Brauer morphism Scott modules The Brauer morphism and p-permutation modules

438 448 45 1

Permutation lattices Generalized permutation lattices Permutation lattices and normal subgroups Some bimodule isomorphisms Applications

459 459 473 484 489

499 511 516

1

Chapter 1 Preliminaries Our aim here is to review some basic ring-theoretic results and to fix conventions and notations for the rest of the book. Since we presuppose a familiarity with various elementary ring-theoretic terms, only a brief description of them is presented. Many readers may wish to glance briefly at the contents of this chapter, referring back to the relevant sections when they are needed later.

1.

Notation and terminology

All rings in this book are associative with 1 # 0 and subrings of a ring R are assumed to have the same identity element as R. Each ring homomorphism will be assumed to preserve identity elements. Let R be a ring. An element x in R is called nilpotent if xn = 0 for some integer n 2 1. An ideal J in R is nil if every element of J is nilpotent, while J is nilpotent if there is an integer n 2 1 such that J" = 0, where J" is the product of J with itself n times. An element e in R is an idempotent if e2 = e. Two idempotents u,v of R are orthogonal, if uv = vu = 0. A nonzero idempotent is primitive if it cannot be written as a sum of two nonzero orthogonal idempotents. We denote by Z ( R )the centre of R. An idempotent e is called centrally primitive if e is a primitive idempotent of Z ( R ) . The ring of all n x n matrices over R is denoted M,(R). Let R be a ring. From this we can construct a new ring R", called the opposite ring of R. Both the underlying set and the additive structure

Preliminaries

2

of R" are just those of R. But the multiplication, denoted by o is given bY z ~ =y y z for all z,y E R It is easy to verify that R" is a ring with these operations. Clearly, Z ( R ) = Z(R")and R is commutative if and only if R = R". Assume that A is a ring, R a commutative ring, and p : R -+ Z ( A ) a ring homomorphism. The resulting system ( A ,R, p ) is called an Ralgebra. It will be convenient to suppress the p and speak of A as an R-algebra or as an algebra over R. Thus A is an R-algebra (with respect to some p ) if and only if there is an ideal I of R with R / I isomorphic to a subring of Z ( R ) . Setting r a = p ( r ) a , r E R, a E A, it follows that A is an R-algebra such that r(zy) = z(ry) = ( w ) y

for all

r E R,x,y E A

Conversely, if A is an R-module for which the above equalities hold, then the map R + Z ( A ) , r H r 1 is a ring homomorphism and hence A is an R-algebra. By a homomorphism of R-algebras, we understand a ring homomorphism which is also a homomorphism of R-modules. Let (&), i E I , be a family of rings and let R be the direct product set &rRj. We can define addition and multiplication on R by the rules

-

(G) t (Yi) = ( X i

t Y;)

(Z;>(Yi)

= (Wi)

(z;,y; E R;) . An immediate verification shows that R is a ring ; we shall refer to R as the direct product of the f a m i l y (Ri),i E I . We now record some conventions concerning modules. By an Rmodule, we always understand a unital R-module; the symbols RM or M R will be used to underline the fact that M is a left or right Rmodule, respectively. We usually consider left R-modules, and, in this case, speak simply about R-modules. A nonzero R-module V is indecomposable if 0 and V are the only direct summands of V . A nonzero R-module V is irreducible if 0 and V are the only submodules of V . Let V and W be R-modules. We denote by H o r n ~ ( VW , ) the additive abelian group of all R-homomorphisms V + W . The same notation will be used in case V and W are right R-modules. In case V = W ,

1. Notation and terminology

3

we write E n d ~ ( vinstead ) of Horn~(V, V ) . Note that E n d ~ ( vis) a

ring under the multiplication

Let V be an R-module. A submodule W of V is called maximal if W is a proper submodule (i.e.W # V ) and W is not contained in any proper submodule of V . An R-module V is said to have an R-basis (vi), i E I, if there exist elements o; E V such that each o E V can be written as a finite sum = Crp; with uniquely determined coefficients r; E R. Such modules V are called free R-modules. Given a pair of rings R, S, we say that V is an ( R ,S ) - b i m o d u f e if V is a left R-module and a right S-module, with the actions of R and S on V commuting : ( r v ) s = ?-(‘us)

(7.

E R, v E

v,s E S )

Unless explicitly stated otherwise, all groups are assumed to be multiplicative. The multiplicative group of a field F is denoted by F*. If S is a subset of G, < S > will denote the subgroup of G generated by S, while cG(s) and NG(s) is the centralizer and normalizer of S in G, respectively. As usual, Z(G) denotes the centre of G. Let H be a subgroup of a group G. A subset of G containing just one element from each left coset zH is called a left transversal for H in

G, and right transversals are defined correspondingly. Let g be an element of a finite group G and let p be a prime. We say that g is a p e l e m e n t if the order of g is a power of p ; we say that g is a p’-element (or g is p-regular) if its order is prime to p. Each g E G can be written in a unique way g = gpgpt, where g p is a pelement, gpt is a p’-element, and gP and gpt commute. The elements g, and gpt are called the p and p’-parts of g , respectively. A conjugacy class C of G is called p r e g u f a r (respectively, p-singular) if the order of the elements of C is not divisible by p (respectively, divisible by p ) . Let S be an arbitrary set. An action of G on S is a homomorphism

Preliminaries

4

8 of G into the permutation group of S. When the action is understood, we shall suppress the symbol 8 and instead write s H g s for O(g),g E G, s E S. The stabilizer G(s) of s E S is defined by G(s) = { g E G, 's = s} and the orbit of Gs of s E S is defined by G~ = {'slg

E G}

We say that G acts transitively if Gs = S for some (and hence all) s E S. One readily verifies that G(s) is a subgroup of G and that IGsl = (G : G(s))

Throughout the book, S, and A, denote symmetric and alternating groups of degree n, respectively, while Z n is a cyclic group of order n. 2.

Tensor products

Let R be a ring. Given a right R-module V and a left R-module W, the 23-module V @ R W , called the tensor product of V and W , is defined as follows. Let F be a free 23-module with V x W as a basis. Each element of F can be uniquely written in the form

c

Zij(Vi, Wj)

(Zjj

E

z,

vj

E

v,wj E W )

with finitely many zij distinct from 0. Let T be the subgroup of F generated by all elements of the form-

(21,

w1+ w2)- (v,w1) - (21, w2)

(v,r w ) - ( ' U f , 20) with v,v; E V and w,wi E W . Then V @R W is defined as the factor group FIT. The image of (v,w) under the natural homomorphism F + F / T is denoted by v 8 w. Observe that the Z-module V @ R W consists of all finite sums C vi @ w;with vi E V,wi E W . Assume that

2. Tensor products

5

V is an ( S ,R)-bimodule. Then V

@R

W can be regarded as a (left)

S-module by putting s(v@w) = s v @ w

( W E

v , w E w,s E S )

In particular, if V and W are modules over a commutative ring R, then V @ R W is an R-module. Let R be a ring, V,V' two right R-modules, W,W' two left Rmodules and 4 : V + V',$ : W -+ W' two R-linear maps.Then the map

$@$:V@RW+VI@RW' defined by

(4 0$)(W

€3

= +)

@ $(w)

is a 23-linear map. Moreover, if V,V' are ( S ,R)-bimodules and 4 is also S-linear, then $ 8 $ is S-linear. We next record a number of standard properties of tensor products (see Bourbaki (1974), (1972)).

2.1. Proposition. Let V,V',V" be right R-modules, let W,W',W'' be left R-modules and let

be exact sequences. (i) The sequences

w 3' V ~ wR'5 vN@R w o v@Rw'3v @ R w 'Yv 8 R w"+ o

v'@R

4

of 23 -homomorphisms are exact. (ii) The homomorphism g 0t ; V € 3 W ~ -+ V" @ R W" is surjective and its kernel is equal to I m ( f @ 1 ) + I m ( 1 0s ) . In particular, if V' and W' are submodules of V and W , respectively, and f and s are inclusion maps, then the map

(v/v')@R (w/w') ( v@ R w ) / ( I m ( f@ 1 ) + I m ( 1 €3 s))

(v

+ V ' )@ (wt W')

(W

€3 w)t (h(f 6 1)

+ I m p @ s))

Preliminaries

6

is a Z-isomorphism. 2.2. Proposition. Let ( K ) , i E I , be a family of right R-modules and (Wj),j E J a family of left R-modules. Then the map

is a Z-isomorphism. 2.3. Proposition. Let V be an ( S ,R)-bimodule. Then

as S-modules. 2.4. Proposition. If V and W are too free modules over a commutative ring R and (v;),(wj) are R-bases of V and W , respectively, then (v;8 wj) is an R-basis of V @ R W .

2.5. Proposition. Let R be a subring of a ring S and let V be a free left R-module with el,. . . , en as an R-basis. Then S @R V is a free left S-module with 1 8 el,. .. ,1 @ en as an S-basis. 2.6. Proposition. Let V be a right R-module, W a left R-module, V' a submodule of V and W' a submodule of W . If V' is a direct summand of V and W' is a direct summand of W , then the canonical

homomorphism

v' @ R w' v @ R w 4

is injective and the image of V' @R N .

Z-module M

@R

W' is a direct summand of the

Let R be a ring, V a right R-module and W a left R-module. The module V is said to be flat i f for every injective homomorphism f : W' t W of left R-modules, the homomorphism

7

2. Tensor products

is injective. By a projective module, we understand a direct summand of a free module.

2.7. Proposition.Every projective right R-module is pat. Let V be a right R-module and let A be a left ideal of R. The homomorphism V @R A -i V which sends v 8 a to va is called canonical: its image is the subgroup V A consisting of all finite sums C v ; u ; with v; E V,a; E A. The induced homomorphism V 8~A -iV A is called canonical. The canonical map V @ R R + V is an isomorphism. Note that if V is an ( S ,R)-bimodule, then all canonical maps are homomorphisms of S-modules. 2.8. Proposition. Let V be a right R-module. Then the following properties are equivalent: (i) V is flat. (ii) For any finitely generated submodule W' of a left R-module W , the canonical homomorphism l v @ i : V @ R W' + V @ R W (i is the inclusion map) is injective. d (iii) For every exact sequence of left R-modules W' + W + W" the sequence

*

is exact. (iv) For every finitely generated left ideal A of R, the canonical map V @ R A + V A is an isomorphism. Let V be a flat right R-module and let W' be a submodule of a left R-module W . Then the canonical injection V @ R W' + V @ , RW allows us to identify V @ R W' with its image in V @ R W . In the future, we shall always use this identification. 2.9. Proposition. Let V be a flut right R-module. (i) I f f : X + Y is a homomorphism of left R-modules, then

Preliminaries

8

and

Im(1v 8 f ) = v

OR

(Imf)

(ii) If W',W" are two submodules of a left R-module W , then

3.

Artinian, noetherian a n d completely reducible modules

Throughout this section, R denotes an arbitrary ring. Let V be an R-module. We say that V is artinian (respectively, noetherian) if every descending (respectively, ascending) chain of submodules of V terminates. The ring R itself is called artinian (respectively, noetherian) if the left regular R-module R R is artinian (respectively, noetherian). We call V finitely generated if V = Cy=lRvi for some wl, . . . $0, in V . The module V is said to be finitely cogenerated if for every family{ li E I}of submodules of V with niErK= 0, there exists a finite subset J of I such that

3.1. Proposition.For any R-module V , the following conditions are equivalent: (i) V is artinian. (ii) Every nonempty set of submodules of V has a minimal element. (iii) Every factor-module of V is finitely cogenerated. Proof.(i)=+ (ii): Assume that there is a nonempty set S of submodules of V such that S does not have a minimal element. Then, for any given W E S, the set {W' E SlW' c W } is nonempty. Owing to the axiom of choice, there is a function W L-+ W' with W 3 W' for any W in S. By choosing any W in S, we therefore obtain an infinite descending

chain

w 3 W ' 3 W " > ....

of submodules of V. (ii)+ (iii): It suffices to show that if W is a submodule of V and

3. Artinian, noetherian and completely reducible modules

{Eli E I} is

nj&

a family of submodules of V with n;& = W for some finite subset J of I . Let us put

9

= W , then

S = {nkEKvklICI is finite } By assumption, S has a minimal element, say njE& with J C_ I and J finite. It is obvious that W = njEJv,as required. (iii)+ (i): Consider a descending chain & _> & 2 . .. 2 V , 2 . . . of submodules of V . If W = nn>lVn, then by hypothesis V/W is finitely cogenerated. Thus W = V, for some n 2 1. But then V,+j = V, for all i 2 1, as desired. W 3.2. Corollary,Let V irreducible submodule.

#0

be an artinian module. Then V has an

Proof.Apply Proposition 3.1. for the nonempty set of all nonzero submodules of V . W Turning to noetherian modules, we next record

3.3. Proposition. For any R-module V , the following conditions are equivalent: (i) V is noetherian. (ii) Every nonempty set of submodules of V has a maximal element. (iii) Every submodule of V is finitely generated. Proof.The proof of this result is dual to that of Proposition 3.1. and therefore will be omitted. Proposition. If V is a finitely generated R-module, then every proper submodule of V is contained in a maximal submodule ( in particular, if in addition V # 0 , then V has a maximal submodule ). 3.4.

Proof. By hypothesis, V = Ci=lRv; for some v1,. . . ,v, in V . Denote by W a proper submodule of V and by S the set of all proper submodules of V which contain W . Then W E S and if (Wj) is a chain in S , then U;W; is a submodule. If V = U;W;, then vj E W,, for

10

Preliminaries

some nj, 1 5 j 5 r. Let W, be the largest of the modules W,, , . . . ,W,,. Then W contains vl,,. .,u, and hence V = W,, a contradiction. Thus S is inductive and therefore, by Zorn's lemma, it has a maximal element. This gives us a maximal submodule of V containing W , as desired. W

Let 0 --f U --f V -, W --f 0 be an exact sequence ofR-modules. Then V is artinian (noetherian) i f and only i f both U and W are artinian (noetherian). 3.5. Proposition.

Proof. Suppose that V is artinian. Then, because U and W are isomorphic to a submodule and factor module of V , respectively, it is obvious from the definition that both U and W are artinian. Conversely, assume that U and W are artinian. To show that V is artinian, we may assume that U 2 V and W = V/U. Consider a descending chain

v, 2 vz 2 ... 2 v, 2 ...

of submodules of V . Since V/U is artinian, there is an integer rn such that (i = 1,2, . . .) = vm+; u

v, + u

+

Because U is artinian, there is an integer n 2 rn such that

Taking into account modularity and that V, 2 V,+i, we have for each i = 1,2, ...

V, = V , n (v, U ) = V , n (Vn+l

+

=

+U )

K+i + (V, n U )= Vn+i + (K+;n U )

= Vn+i

Hence V is artinian. The proof of the noetherian case is dual. W 3.6. Corollary. Let V = @y=lq.Then V is artinian (noetherian) if and only if each is artinian (noetherian).

P r o o f . The case n = 2 follows from Proposition 3.5. The general case is a straightforward induction on n. '

3. Artinian, noetherian and completely reducible modules

11

3.7. Corollary. Let R be a ring. Then R is artinian (noetherian) if and only if every finitely genemted R-module is artinian (noetherian). Proof. Assume that R is artinian. If V is a finitely generated R-module, then V is a factor module of a free R-module W of finite rank. By Corollary 3.6, W is artinian, hence so is V , by Proposition 3.5. Conversely, if every finitely generated R-module is artinian, then so is R, since R is generated by 1. The proof of the noetherian case is similar . 3.8. Corollary. Let A be an algebra over a commutative ring R such that A is a finitely generated R-module. If R is artinian (noetherian), then A is artinian (noetherian).

Proof. This is a direct consequence of Corollary 3.7. Let V be a nonzero module. A finite chain of n -t 1 submodules of

V

v = vo 3 v, 3 . . . 3 v, = 0

is called a composition series of length n for V provided that each K-I/V, is irreducible, 1 5 i 5 n , i.e. provided each term in the chain is maximal in its predecessor. The irreducible modules

are called the composition factors of the series. 3.9. Proposition. A nonzero module V has a composition series if and only if V is both artinian and noetherian.

Proof. Suppose that V is both artinian and noetherian. Because any factor module of V is artinian, it follows from Corollary 3.2 that there is an ascending chain

0

c v1 c vz c .

of submodules of V where each term in the chain is maximal in its

successor. But V is noetherian, hence the chain terminates at V after

Preliminaries

12

a finitely many terms. Hence

V has a composition series .

Conversely, assume that V has a composition series. We argue by induction on n, where n is the minimum length of all such series. If n = 1, then V is irreducible and there is nothing to prove. Otherwise, if = vo 3 6 3 . . . 3 =0

v

v,

is a composition series of minimal length for V ,then V/V1is irreducible and V1 has a composition series of length n - 1. Now apply Proposition

3.5. Suppose that and

v = vo 3 6 3 ... 3 v, = 0 v = wo 3 w,3 ... 3 w, = 0

are two composition series for V. We say that these two series are equivalent in case n = rn and there is a permutation r~ of { 1,2,. . . ,n} such that vi/K+l Wo(i)/WC+)+l ( 2 = 1,. . ., n )

V V

3.10. Proposition. (The Jordan-Holder theorem). If a module has a composition series, then every pair of composition series for are equivalent.

Proof. We argue by induction on c ( V ) , where c ( V ) is the minimum length of composition series for V. The case c ( V ) = 1 being obvious, we may assume that c(V)= n > 1 and that any module with a composition series of smaller length has all of its composition series equivalent. Choose a composition series

(1)

v = v o 3 VI 3 . . . 3 v, = 0

of minimal length and fix any other composition series for

(2)

v = wo 3 w13 ... 3 w, = 0

V:

3. Artinian, noetherian and completely reducible modules

13

If & = W1, then by the induction hypothesis and in view of c ( K ) 5 n - 1, (1) and (2) are equivalent. We may thus assume that V1 # W1, in which case V = V, -t Wl since V, is a maximal submodule of V. It follows that

V, n W1 is maximal in both Wl 3.9, V1 n IV1 has a composition series whence

and V1. Owing to Proposition

and hence

are composition series for V, and W l . Since c(Vl) < n, every two composition series are equivalent. Thus the two series

and

v = VO 2 v1 2 xo 3 . .. 3 XI, = 0

are equivalent. It follows that k < n - 1 and c(W1) < n , proving that every two composition series for Wl are equivalent. Hence the two series = 3 3 ... 3 W, = 0

v wo w,

and

v = w o > w l 2x03 ... > x k = o

are equivalent. Because, by (3), V/Vl Z Wl/XO and V/W1 &/XO, the series (1) and (2) are equivalent, thus completing the proof. Assume that V is both artinian and noetherian. Then, by Propositions 3.9 and 3.10, for such a module V we can define its composition len,gth c(V) unambiguously by c(V) = 0 if V = 0, and c(V) = n if V has a composition series of length n. We next examine direct decompositions of modules.

Preliminaries

14

Let R be any ring and let V # 0 be an Rmodule which is either artinian or noetherian. Then V is a finite direct sum of indecomposable submodules. 3.11. Proposition.

For any nonzero R-module M that is not a finite direct Proof. sum of indecomposable submodules, choose a proper decomposition M = M' @ N' where M' is not a finite direct sum of indecomposable submodules. Assume by way of contradiction that V cannot be written as a finite direct sum of indecomposable submodules. Then

v = V' @ w',V' = v"@ W",.. . is a sequence of proper decompositions. Hence there exist infinite chains

W' c W' @ W" c . . . and V 3 V' 3 V'' 3 . . . proving that V is neither artinian nor noetherian. w

3.12. Proposition. Let R R = & @ . . . @ Vn with K # 0 for i = 1,2,. . . ,n, and write 1 = el - . en with e; E K. Then { e l , . . . e,} is a set of pairwise orthogonal idempotents in R and V , = R e ; , 1 5 i 5 n. Conversely, if { e l , . . . , e n } is a set of pairwise orthogonal idempotents

+ +

in R, then

n

R(Ce;) = @y=lRei i=l

Proof. Given r E R , we have r = r . 1 = 1 . r = re1 Hence, in particular,

+ . - + re,.

which shows that e;ej = &jei for all i , j . Moreover, Rei V , and RR = Re1 @ - . @ Re,, which implies that = R e ; for all i. Conversely, suppose that { e l , . . . ,e n } is a set of pairwise orthogonal idempotents and put e = CZ1e;. Then e2 = e and ee; = e;e = e; for R e ; . If E r i e ; = 0 , then multiplication on the all i. Hence R e = C:=t=, right by ej gives r j e j = 0 for all j. Thus R e = @y=lRe;, as asserted.

3. Artinian, noetherian and completely reducible modules

15

3.13. Corollary. Assume that R is artinian or noetherian. Then there exists a set { e l , . . . ,en} of pairwise orthogonal primitive idempotents of R with el - - . en = 1 . Moreover

+ +

where each Re; is an indecomposable R-module. Proof.

Apply Propositions 3.11 and 3.12.

We shall refer to the R-modules R e l , . . ., Re, in Corollary 3.13 as the principal indecomposable R-modules. An R-module V is said to be completely reducible if every submodule of V is a direct summand of V .

3.14. Proposition. Let V be a completely reducible R-module. (i) Every submodule ofV is isomorphic to a homomorphic image of V and every homomorphic image ofV is isomorphic to a submodule of

V. (ii) Every submodule and every homomorphic image ofV is comple t ely reducible, (iii) If V # 0 , then V contains an irreducible submodule. Proof. (i) If W is a submodule of V , then V = W @ W’ for some submodule W’ of V . Then W E V/W’ and V/W W‘, as required. (ii) Let W be a submodule of V and let U = V / W . By (i), it suffices to show that U is completely reducible. Assume that U1 = & / W is a submodule of U . We may choose V2 with V = Vl @ V2. Then

U = Ul@

(vz + W ) / W

and therefore U is completely reducible. (iii) Fix a nonzero 2, in V , By Zorn’s lemma, there is a submodule W of V maximal with respect to the property that TI 4 W . Write V = W @ W’ for some submodule PV’ of V . If W’ is not irreducible, then W‘ = W1@W2 for some nonzero submodules Wl, W2of V . Because

(W @ Wl) n (W @I W 2 )= W

16

Preliminaries

it follows that w $ W @ Wi for i = 1 or i = 2. But this contradicts the maximality of W . Thus W’ is irreducible, as desired.

We are now ready to provide two important characterizations of completely reducible modules.

3.15. Proposition. For any nonzero R-module V , the following conditions are eqivalent: (i) V is completely reducible. (ii) V is a direct sum of irreducible submodules. (iii) V is a sum of irreducible submodules.

Proof. (i)Ij(ii) : Consider the collection of sets of irreducible submodules of V whose sum is direct. By Proposition 3.14 (iii), it is nonempty and, by Zorn’s lemma, there is a maximal element, say {K}, in this collection. Let W = and let V = W @ W’. If W’ # 0, then by Proposition 3.14 (ii), (iii), W‘ contains an irreducible submodule V‘. Hence W V’ = V‘ @ (@V,),contrary to the maximality of {K}. Accordingly, W’ = 0 and V = W , as required, (ii)+ (iii) : Obvious. (iii)+ (i) : Let W be a submodule of V . Owing to Zorn’s lemma, there is a submodule W’ of V maximal with the property that WnW’ = 0. Hence we must have W W’ = W @ W’. We are thus left to show that V = W e W’. Assume by way of contradiction that W $ W’ # V and let v E V be such that w $ W @ W‘. By hypothesis, w = v1 . . . w, where v; E V , and V , is an irreducible submodule of V , 1 5 i 5 n. Accordingly, v j 6 W @ W’ for some j and thus

+

+

+ +

V, n (W $ W’) # V, However,V, is irreducible, so

V, n (W @ W‘) = 0 and hence

W 03 W’+ v, =

w @ W‘$ v,

which shows that W n (W’ @ V,) = 0. Since this contradicts the maximality of W’, the result follows. m

3. Artinian, noetherian and completely reducible modules

17

3.16. Corollary. Let V # 0 be a completely reducible R-module. Then the following conditions are equivalent: (i) V is artinian. (ii) V is noetherian. (iii) V is a finite direct sum of irreducibze submodules. Proof. By Corollary 3.6, if V = @LIK then V is artinian (noetherian), if and only if each V , is artinian (noetherian). Now apply Proposition 3.15 (ii). 3.17. Corollary. If R R is completely reducible, then every Rmodule is completely reducible. Proof. Let V be an R-module and let v E V . Then the map R R --$ Rv, r H r v is a surjective homomorphism of R-modules. Therefore, by Proposition 3.14 (ii), Rv is completely reducible. Because V = CvEVRv,the result follows by applying Proposition 3.15 (iii).

3.18 Lemma. Let V = @K where each Vi is irreducible and let W be an irreducible submodule of V . Then W @I$ where j ranges over those i for which W E V,. Proof. Given a nonzero w E W , we may write w = Cv;with v; E L$ and with finitely many v; # 0. If v; # 0, then the map r w I-+ rvi,r E R, is a nonzero homomorphism from W = Rw to Vl = Rv;.Since W,K are irreducible, W V , as desired.

Let V be a completely reducible R-module. We say that V is homogeneous if it can be written as a sum of isomorphic irreducible submodules. The sum of all irreducible submodules of V which are isomorphic to a given irreducible submodule of V is called a homogeneous component of V .

Let V = @iEzV,, where each V , is irreducible 3.19. Proposition. j E J , are all representatives of the and let J C I be such that the isomorphism classes of V,, i E I . For each j E J , denote b y Wj the sum

K,

18

Preliminaries

of all F with r/: E 6 . Then (i) The Wj, j E J , are all homogeneous components of V . (ii) V = @ j E ~ W j .

Proof. (i) Apply Lemma 3.18. (ii) This follows from the definition of Wj. Let V be any R-module. A submodule W of V is said to be fully invariant in V if it is mapped into itself by all R-endomorphisms of V .

3.20. Proposition. Let V # 0 be a completely reducible Rmodule. A submodule W of V is fully invariant if and only i f W is a sum of certain homogeneous components of V .

Proof. Keeping the notation of Proposition 3.19, fix f E EndR(V). If V , 2 V, then either f ( X ) = 0 or f(V,) Z V , %' vj, since V;: is irreducible. Thus f(Wj) C Wj, so each Wj (and hence the sum of certain Wj) is fully invariant. Conversely, assume that W is fully invariant. It suffices to show that if P is an irreducible submodule of W and Q is an irreducible submodule of V such that Q Z P , then Q C W. If Q = P , this is clear; otherwise P n Q = 0, so P -t Q = P @ Q and thus V = P @ Q @ U for a submodule U of V . Let f : P -+ Q be the given isomorphism and define an endomorphism g of V by

Then W _> g(W) 2 g ( P ) = f ( P )= Q, as asserted. 1 We close by recording the following useful observation. 3.21. Corollary. Let V # 0 be a completely reducible R-module and write V as the direct sum of its homogeneous components: V = @ i ~ W jThen . En&(V) % EndR(Wj)

n

jEJ

4. The radical of modules and rings

19

Proof.

It is clear that any family (fj) with f j E E n d ~ ( W j ) determines a unique endomorphism f of V . The map

is obviously an injective ring homomorphism. Because each Wj is fully invariant (Proposition 3.20), any endomorphism f of V maps each Wj into itself. Setting f, = flWj,j E J , it follows that ( f j ) H f, as desired. 4.

The radical of modules and rings

Throughtout, R denotes an arbitrary ring. Given an R-module V , the radical J ( V ) of V is defined to be the intersection of all maximal submodules of V . If V contains no maximal submodules, then by definition J ( V ) = V . Observe that if V # 0 is finitely generated, then J ( V ) # V by virtue of Proposition 3.4. The Jacobson radical J ( R )of R is defined by

J ( R )= J(RR) Hence, by definition, J ( R ) is the intersection of all maximal left ideals of R. We say that R is semisimple if J ( R ) = 0. 4.1 Proposition. For any nonzero R-module V , the following conditions are equivalent: (i) V is a finite direct sum of irreducible modules. (ii) V is artinian and com.pletely reducible. (iii) V is artinian and J ( V ) = 0.

Proof.

(i)+ (ii): This follows from Corollary 3.6 and Proposition

3.15. (ii)+ (iii): We may clearly assume that V # 0, in which case V is a direct sum of irreducible submodules, by Proposition 5.15. Therefore, by the definition of J ( V ) ,J ( V ) = 0 (iii)+ (i): Owing to Proposition 3.1, V is finitely cogenerated. Since J ( V ) = 0 it follows that f~r=,,K= 0 for some maximal submodules V,,. . . ,V, of V . Thus V is isomorphic to a submodule of (V/K).

ny=,

Preliminaries

20

Hence, by Proposition 3.14 (ii), V is completely reducible. Now apply Corollary 3.16 (iii). 4.2 Corollary.

For any ring R, the following conditions are equiv-

alent:

(i) RR is a finite direct sum of irreducible modules. (ii) RR is completely reducible. (iii) R is semisimple artinian. (iv) Every R-module is completely reducible. Proof. (i)+ (ii): Follows from Proposition 3.15. (ii)+ (iii): Because 1 lies in the sum of finitely many irreducible submodules, it follows that R R is a finite direct sum of irreducible modules. Now apply Proposition 4.1 for V = R R. (iii)+ (iv): By Proposition 4.1 (applied for V = R R),RR is completely reducible. Now apply Corollary 3.17. (iv)+ (i): By assumption, RR is completely reducible. Thus R is semisimple artinian, by the implication (ii)+ (iii). The desired conclusion follows from Proposition 4.1 applied for V = R R. We now proceed to examine J ( V ) in detail.

4.3. Lemma. (i) Let f : W + V be a homomorphism of Rmodules.Then f ( J ( W ) )& J ( V ) with equality i f f is surjective and Ke rf C J ( W ) . (ii) If W is a submodule of an R-module, then J ( W ) C J ( V ) and

J ( V / W )2 ( J ( V )t W ) / W . (iii) If W is a submodule of an R-module V with W 2 J ( V ) , then J ( V / W )= J ( V ) / W . (iv) IfV is an R-module, then J ( V / J ( V ) )= 0 and J ( V ) W for any submodule W ofV for which J ( V / W )= 0 (v) If W is a submodule of an R-module V , then W = J ( V ) if and only ifW J ( V ) and J ( V / W )= 0 . Proof. (i) If V has no maximal submodules, then J ( V )= V and therefore f ( J ( W ) )C J ( V ) .Now assume that M is a maximal submodule of V . Then the map f * : W --+ VIM given by f*(w) = f ( w ) M

+

4. The radical of modules and rings

21

is an R-homomorphism. Since K e r f * is a maximal submodule of W,J ( W ) C K e r f * and so f ( J ( W ) )C M . Hence f ( J ( W ) ) J ( V ) . Suppose that f is surjective and K e r f C J ( W ) . If W has no maximal submodules, then so does V , in which case

f ( J ( W ) )= f ( W )=

v = J(V)

We may thus assume that the set {MiliE I } of all maximal submodules of W is nonempty. By assumption, K e r f Mi for all i E I . If X and Y are the lattices of submodules of W containing K e r f and submodules of V , respectively, then the map U H f ( U ) , U E X , is an isomorphism of X onto Y. Therefore { f ( M i ) l i E I } is the set of all maximal submodules of V and

as asserted.

(ii) and (iii). This follows from (i) applied to the cases where f is the inclusion map and f : V --f V/W is the natural homomorphism. (iv) This follows from (iii) and (ii). (4 Apply (iv>* 4.4.Corollary.

Let R be an arbitrary ring and let I be an ideal of

R. (i) I = J ( R ) if and only if I J ( R ) and J ( R / I ) = 0 (ii) If1 C J ( R ) , then J ( R / I ) = J ( R ) / I . Proof. This is a direct consequence of Lemma 4.3 (v), (iii) and the obvious fact that the radical of the R-module R / I is the same as the Jacobson radical of the ring R / I .

A submodule W of an R-module V is called superfluous if for every submodule X of V , W + X = V implies X = V . 4.5. Proposition. Let V be an R-module. Then J ( V ) is the su'm of all superfluous submodules in V . Futhermore, if V is finitely generated, then J ( V ) is the unique largest superfluous submodule of V .

Preliminaries

22

Proof. Suppose that W is a superfluous submodule of V . To prove that W C J ( V ) ,we may assume that V has a maximal submodule M . If W M , then M + W = V and so M = V , a contradiction. Hence every superfluous submodule of V is contained in J ( V ) . On the other hand, let z E V . If W is a submodule of V with Rz + W = V , then either W = V or there is a maximal submodule M of V such that W & M and z $! M . If z E J ( V ) ,then the latter cannot occur. Thus z E J ( V )implies that Rz is a superfluous submodule of V , proving the first assertion. Suppose that V is finitely generated. By the foregoing, it suffices to show that J ( V )is a superfluous submodule of V . To this end, assume that V = W J ( V ) for a submodule W of V . Owing to lemma 4.3 (ii), J ( V / W )= V / W . Since V/W is finitely generated it follows from Proposition 3.4 that V/W = 0. Hence V = W as required.

+

4.6 Proposition. Let V be an R-module generated b y v l , . . . ,v, and let v E V . Then v E J ( V ) if and only if for all r; E R, 1 5 i 5 n, the elements v; r;v, 1 5 i 5 n, generate V .

+

Proof. Let v E J ( V )and let W be the submodule of V generated by the elements vi + riv, 1 5 i 5 n. Then W + J ( V ) = V so that W = V , by Proposition 4.5. Conversely, suppose that v 6 J ( V ) , Then there is a maximal subW , so for each module W of V such that v $! W . Hence V = Rv i there exists r; E R such that vi E -r;w W , whence v; r;v E W . Thus the elements v; + r;v, 1 5 i 5 n , do not generate V .

+

+

+

Let V be an R-module. The annihilator of V , written ann(V),is defined by

ann(V)= { T

E

RlrV = 0)

It is clear that ann(V) is an ideal and that V may be viewed as an R/ann(V)-module. We say that V is faithful if ann(V)= 0. An ideal I of R is said to be primitive if the ring R / I has a faithful irreducible module. Obviously, 1 is primitive if and only if I is the annihilator of an irreducible R-module.

4. The radical of modules and rings

23

Lemma. (i) An R-module V is irreducible if and only if V E R / X for some maximal left ideal X of R. (ii) Every maximal left ideal of R contains a primitive ideal and every primitive ideal is the intersection of the maximal left ideals containing it. (iii) J ( R ) is the intersection of the annihilators of irreducible Rmodules. In particular, J ( R ) is an ideal of R. 4.7.

Proof. (i) It is obvious that R / X is irreducible. Conversely, suppose that V is irreducible and let 0 # u € V . Then Ru is a nonzero submodule of V and so V = Rv. The map R -+ V,r H ru is therefore a surjective R-homomorphism. Thus R / X E V for some left ideal X of R. Since V is irreducible, X is maximal. (ii) Let X be a maximal left ideal of R. By (i), R / X is an irreducible R-module. Hence the annihilator of R / X is a primitive ideal contained in X . Let I be a primitive ideal in R and let V be an irreducible R-module whose annihilator is I . Given a nonzero IE: in V , put V, = { r E Rlrz = 0). Then V, is a left ideal in R. Since V is irreducible , V = Ra: E R/V, and so V, is a maximal left ideal of R. Because I is the intersection of all V, with 0 # II: E V , (ii) is proved. (iii) Apply (i) and (ii). Let V be an R-module. (i) J(R)V C J ( V ) with equality i f R / J ( R )is artinian. (ii) Assume that R/ J ( R)is artinian. Then V is completely reducible if and only if J(R)V = 0. 4.8. Proposition.

(i) If V has no maximal submodules, then J ( V ) = V and so J(R)V C J ( V ) . Let M be a maximal submodule of V . Then VIM is an irreducible R-module. Because J ( R ) annihilates VIM (Lemma 4.7 (iii)), it follows that J(R)V & M . Thus J(R)V & J ( V ) . Suppose that R / J ( R ) is artinian. Owing to Lemma 4.3 (v), it suffices to verify that J ( V / J ( R ) V )= 0. Because J ( R / J ( R ) )= 0 (Corollary 4.4 (i) and Lemma 4.7 (iii)) and R / J ( R )is artinian, every R / J (R)-module is completely reducible (Corollary 4.2). In particular, the R/J(R)-module V / J ( R ) Vis completely reducible. Hence V / J ( R ) V is a completely reducible R-module and therefore J ( V / J ( R ) V )= 0. Proof.

Preliminaries

24

(ii) If V is completely reducible, then J(R)V = 0 by Lemma 4.7 (iii). Conversely, assume that J(R)V = 0. Then V can be regarded as a module over the semisimple artinian ring R/J(R). Hence, by Corollary 4.2 V is a completely reducible R / J ( R)-module. Thus V is a completely reducible R-module. 4.9. Proposition.

(Nakayama’s lemma.) Let W be a submodule If W J(R)V = V , then W = V .

of a finitely generated R-module V .

+

Proof. Owing to Proposition 4.8 (i), J(R)V C_ J ( V ) and hence W + J ( V ) = V . But, by Proposition 4.5, J ( V ) is a superfluous submodule of V , hence the result. W The rest of this section will be devoted to ring-theoretic properties of J ( R ) . An element x of R is called a left (respectively, right) unit if there exists y E R such that yx = 1 (respectively, xy = 1). By a unit, we understand an element x in R which is both a left and a right unit. Hence x E R is a unit if and only if there exists y E R (denoted by x-l) such that xy = yx = 1 4.10. Proposition. Let x be an element of a ring R. Then x E J ( R ) if and only if for all r E R, 1 - rx is a left unit. In particular

J(R ) contains no nonzero idempotents. Proof. We first note that 1 - T X is a left unit if and only if R( 1 rx) = R. Therefore the first assertion follows from Proposition 4.6 applied to V = R R. Assume that e is an idempotent in J ( R ) . Then, by the above, x(l - e) = 1 for some x E R. Thus e = 1 e = x(1 - e)e = 0, as we wished to show. W 4.11. Proposition. For any ring R, J ( R ) is the unique largest ideal I of R such that 1 - rx is a unit of R for all r E R, x E I .

Proof. Owing to Proposition 4.10, it suffices to show that 1 - x is a unit for all 2 E J ( R ) . Since, by Proposition 4.10, 1 - x is a left unit, we have y ( 1 - x) = 1 for some y E R. Therefore z = 1 - y = -ya:

4. The radical of modules and rings

25

is in J ( R ) and so 1 = y'( 1 - z ) = y'y for some y' E R. Thus y is a unit and 1 - x = y-* is also a unit.

4.12 Corollary. For any ring R, J(R")= J ( R ) and, in particular, J ( R ) is the intersection of all maximal right ideals of R. Proof.

Apply Proposition 4.11.

4.13. Corollary. Let I be a left or right nil ideal of R. Then

I

C J(R).

Proof. Suppose that I is a left nil ideal of R and let x E I . Then, for any r E R,rx E I and so (rx)" = 0 for some n 2 1. Invoking the identity

(1 - y ) ( l

+ y + . . + y"-') *

= (1

+ y + - + yn-')(l * *

- Y) = 1 - y n

we see that 1 - r x is a unit. Thus by Proposition 4.10, I C J ( R ) . If I is a right nil ideal, then the same argument applied to R" yields the

result. W 4.14. Proposition.

For a n y ring R, the following conditions

are equivqlent: (i) R is artinian. (ii) R is noetherian, R / J ( R ) is artinian and J ( R ) is nilpotent. Proof. (i)+ (ii): Put J = J R and consider the chain J 2 J 2 2 . . .. By hypothesis, J k = J"' = . - . for some k 2 1. Setting I = J k we have I 2 = I . Assume by way of contradiction that I # 0. Then there exists a left ideal X in R with I X # 0, for example X = R. Let M be a minimal element in the set of all such X . Then I ( J M ) = ( I J ) M = I M # 0 and, becuase J M C M , we have J M = M . By Proposition 4.9, we are thus left to show that A4 is finitely generated. By assumption, there exists xlinM with I x # 0 and hence I ( R x ) # 0. Since R x C_ M , w e have M = R x , proving that J ( R ) is nilpotent. Since R is artinian, so is R / J ( R )by Proposition 3.5. To prove that R is noetherian, we argue by induction on the nilpotency index n of

Preliminaries

26

J ( R ) . If n = 1, then there is nothing to prove. Consider the exact sequence 0 + J"-1 + R + R/J"-' + 0 Since J ( R / J " - ' ) = J/Jn-' (Corollary 4.4 (ii)), it follows from the induction hypothesis that R/J"-' is noetherian. Hence, by Proposition 3.5, we need only show that Jn-' is noetherian. But J"-' is artinian (as a submodule of an artinian module RR) and J"-' is completely reducible, by Proposition 4.8 (ii). Hence, by Corollary 3.16, J"-l is noetherian, as required. (ii)+(i): We argue by induction on the nilpotency index 7~ of J(R). If n = 1 , there is nothing to prove. By induction hypothesis, R/J(R)"-l is artinian. On the other hand, J(R)"-' is noetherian (as a submodule of a noetherian module RR). Since R / J ( R )is artinian and J(R)" = 0, it follows from Corollary 3.16 and Proposition 4.8 (ii) that J(R)"-' is artinian. Thus, by Proposition 3.5, R is artinian. W 4.15. Corollary. Let R be an artinian ring. Then every finitely generated R-module is both artinian and noetherian. Proof. Apply Proposition 4.14 and Corollary 3.7. W

We next record the following useful observation. 4.16. Proposition,

Let (Ri),i E I , be a family of rings. Then

Proof. An element (r;) E n i E I R is ; a left unit if and only if ri, is a left unit of R; for all i E I . Now apply Proposition 4.10. H

In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Sez as left S-modules if and only if elS 2 e2S as right S-modules (see Jacobson (1956)).

s

4.17. Proposition. Let I be a nil ideal of a ring S and, for each E S , let S be the image of s in S = S / I . Then

4. The radical of modules and rings

27

(i) Each idempotent c E S can be lifted to an idempotent e E S , that

e = E. (ii) If e and f are idempotents of S , then e S E fS as right S modules .if and only if & Z $9 as right S-modules, (equivalently, S e E S f as left S-modules aj and only if SZE Sf as left S-modules). (iii) If I = el + €2 + . . . + en is a decomposition of I into orthogonal idempotents in S, then there exist orthogonal idempotents e l , e2,. . . ,en E is

S such that 1 =el+e2+-**+en, Moreover,

~j

e; = E;

is primitive if and only if so is ei.

Proof. (i) Fix an idempotent 6 E S and choose u E S with ii = E . Then u - u2 E I and therefore (u- u2)rn= 0 for some m 2 1. We have

Observe that on the right each term after the first m is divisible by (1 - u ) ~ while , the first rn terms are divisible by urn. Hence if e denotes the sum of the first m terms, then 1 = e + (1 - e)rng,where g is a polynomial in u. Now u ( 1 - u) E I, so

that is = ii = E . Since e ( l - e ) = e ( l - u)"g = 0 , e is a required i dempo tent. (ii) Let e , f be idempotents in S. We first show that e S E f S as right S-modules if and only if there exists a,b E R such that a = e a f , b = fbe,ab=e,

and

ba=

f

(1)

First suppose that 0 : e S ---f fS is an isomorphism that maps e to b and a to f. Then f b = b and O ( e ) e = O ( e ) ,hence be = b, so f b e = b, and similarly eaf = a. Hence e = ( @ - l @ ) ( e )= 0 - ' ( b ) = 0 - ' ( f b ) = O - ' ( f ) b = ab, and similarly f = ba. Conversely, if a and b satisfy (l), then x H bx is a homomorphism from e S to fS with inverse y H ay. Now assume that e S E f S and let a , b E S satisfy (1). Then 2 , b

Preliminaries

28

satisfy the same conditions for the isomorphism ES E fS, so &' E Now assume that d E fS and a, b E S are such that a

G

e a f , b 3 fbe,ab

e , ba

f

fs.

(modl)

and put a1 = e a f , bl = f b e . Then albl = e - z , where z E e l e . Because z is nilpotent, 1 - z has an inverse 1 - z' and so z z' = zz' = z'z. Setting 2" = ez'e, we have z zN = zz" = z"z and it follows that a l b l ( e - z") = e. Now put a2 = a l , b2 = bl(e - z"), then

+

+

a2 3

a , b2

b ( m o d l ) , a2b2 = e

Next write b2a2 = f -y, then y E flf.Because (ha2)' = bzeaz = bza2, we have

f-y=(f

-Y)2=f2-fy-yf+y2=

f -2y+y2

Hence y2 = y and because y is nilpotent, we find that y = 0 and thus bzaz = f. Since a2bz = e and b2a2 = f imply a2 = eu2 f,b2 = f bze, the required assertion follows. (iii) Let e , f be idempotents in S such that ef = f E = 0. We first claim that there exists an idempotent g such that g = f and eg = ge = 0 . To verify the claim, we note that f e is nilpotent and so 1 - f e is a unit. Put

h = (1 - f e ) - ' f ( 1 - f e ) Then h2 = h, he = 0 and

h = f. Now put

g = (1 - e ) h . Then

proving the claim. We now demonstrate that if e l , . . . ,en are idempotents such that Eiej = 0 for all i # j , then there exist idempotents e: such that = ~i and e:es = 0 for all i # j. For n = 1 there is nothing to prove, so we may assume that n > 1 and use induction on n. In other words, we may assume that e;ej = 0 for i # j, i , j > 1. Put e = e2 + e n , then e is an idempotent such that Fli? = EFl = 0. Applying the claim, there exists an idempotent el, such that e: = Cl and eel, = e i e = 0. It follows

4

+

5. The Wedderburn-Artin theorem

29

that e:, e 2 , . . . ,e, are pairwise orthogonal idempotents. By (i) and the above, there exist pairwise orthogonal idempotents e l , e2,. . . ,en such that E; = c i , 1 5 i 5 n. Thus for u = el - - - e, we have U = i, so 1 - u is a nilpotent idempotent, and therefore u = 1. To prove the final assertion, suppose that e E S is a primitive idempotent. Because e # 0, we have e # 0. If 6 = ~t: y for some orthogonal idempotents 2,y E 3, then there exist orthogonal idempotents el, e2 E S with C1 = 1t:,t?2 = y and hence with E = el e2. By (ii), (el e 2 ) S , which is possible only in the case el = 0 or e2 = 0. eS Thus either 5 = 0 or y = 0 and so iii is primitive. Conversely, assume that e is primitive. Then E # 0 and so e # 0. If e = u v, where u , v are orthogonal idempotents, then ?I = U 5 , ii5 = V i i = 0 and so either U = 0 or V = 0. Thus either u = 0 or v = 0, proving that e is primitive. This completes the proof.

+ +

+

+

+

5.

+

+

The Wedderburn-Artin theorem

In this section we provide the ring-theoretic structure of semisimple artinian rings. Throughout, R denotes an arbitrary ring and, for any interger n 2 l , M , ( R ) is the ring of all n x n matrices over R. Let i,j E (1,. . . ,n } and let e;j be the matrix with (i,j)-th entry 1 and 0 elsewhere. The elements eij E M,(R) are called matriz units. As is customary, we shall identify R with its image in M,(R) consisting of all scalar matrices diag(r, T , . . . ,T ) , r E R. As a preliminary to our first result, let us observe that the matrix units satisfy the following properties : ( i ) e;jeks = 0 if j # k and e i j e k , = ei, if j = k. (ii) 1 = ell . . - enn. (iii) The centralizer of { e ; j } in M,(R) is R. (iv) R Z ellM,(R)ell. It turns out that (i)) (ii), and (iii) determine M,(R) up to isomorphism. Namely, we have the following result.

+ +

5.1. Proposition. Let S be any ring containing elements v,j, 1 5 i ,j 5 n , satisfying (i) and (ii) and let R be the centralizer of {vij} in

Preliminaries

30

S . Then the map (aij) H C i , j aijvij

is an isomorphism of rings and R-modules. Furthermore, R Z v l l S q l .

Proof. Note that M,(R) is a free R-module freely generated by the matrix units eij. Hence the map ejj H v;j extends to a homomorphism of R-modules, which clearly coincides with $. Applying (i), (ii), and (iii), we immediately derive that 1c, is also a ring homomorphism. Suppose that Ci,ja;jv;j = 0 and fix k,s E { 1,.. . ,n}. Then, for all t E { I , . ..,n}, 0 = v t k ( x aijvij)vst = aksvtt i,j

and so 0=

c

aksvtt

= a k s ( z vtt) = ak3

t

t

Thus (ajj) = 0 and therefore $ is injective. To show that 1c, is also surjective, fix s E S and for each i,j put

Then, for all

v,t,

we have

and aijvrt

=

VkiSVjkvrt

= VTjSVjt

k

Hence a;j commutes with all vTt and therefore the foregoing, +((Ujj))

=

c i,j

ajjv;j =

c

aij

E R. Moreover, by

v1;svjj

i,j

= (cv;j)s(xvjj) = s

31

5. The Wedderburn-Artin theorem

proving that 1c, is surjective. Finally, taking into account that

the result follows. For any integer n 2 1 and any (left) R-module V , we write V" for the n-th direct power of V . 5.2. Proposition.

Let V be an R-module and let n

Proof. Fix z,j E { 1,. . . ,n } and define

f&, where

vj

f;j

2 1. Then

E E n d ~ ( v "by )

. . . ,W") = (0,. . . ,v j , o , . . . , O )

is in the i-th place. Then clearly n

fijfks

= ajkfis

and

cfi; = i=l

1

+

If E EndR(V") and $ ( v l , . . . ,on) = ($l ( V l ) , * * * ,+n(vn)), then $ obviously centralizes all f i j if and only if t,bl = $2 = . . . = +". Thus the centralizer of {fij) in EndR(V")is identifiable with EndR(V). The desired conclusion is therefore a consequence of Proposition 5.1. Let V be an R-module and let n 2 1. Then the R-module V" can also be viewed as an M,(R)-module in the following natural way. We visualize the elements of V" as column vectors and, for each A E M,(R) and x E V " , define Az as the matrix multiplication. We now are ready to record the following result. 5.3. Proposition. (i) The map W H W" is an isomorphism from the lattice of submodules of V onto the lattice of M,( R)-submodules of V " . (ii) EndM,,(R)( V " ) EndR(V ).

Preliminaries

32

(iii)The map V H V" induces a bijective correspondence between the isomorphism classes of R ans M,(R)-modules. The inverse of this correspondence is given b y W H ell W . f W" obviously preserves orProof. (i) The correspondence W +

der. Thus we need only show that it has an inverse which is also order-preserving . Consider the projection ?rl : V" + V on the first factor and, for any submodule X of V", put g ( X ) = ?rl(X). Then the correspondence X H g ( X ) is order-preserving and clearly ( g f ) ( W = ) W . From the action of M,(R) on V", we see that X can be written in the form X = V;.. for some submodule 6 of V. Thus ( f g ) ( X )= f(6)= V;2 = X , proving (i). (ii) Put S = M,(R) and choose f E Ends(V"). Then, by the nature of action of S on V", f has the same projections, say X j , on all factors. Conversely, any given 1c, E EndR(V) determines an element of Ends(V") whose projections on all factors are equal to +. Thus the map f H X j provides a desired isomorphism. (iii) It suffices to show that ellVn 2 V and (ellW)" W , where V and W are R and M,(R)-modules, respectively. By the definition of

ell 7

ellVn = V x 0 x . . . x 0 Z V

Observe also that

Becuase eklw = 0 if and only if ellw = 0, 1 5 Ic

5 n, w E W , the map

is well defined and is at least an R-isomorphism of W onto (ellW)". Consequently, we need only show that

Since the latter is a consequence of the action of M,(R) on (ellW)", the result is verified.

5. The Wedderburn-Artin theorem

33

For any ideal I of R, put

M,(I) = {(a,j)E M,(R)la;j E I

for all i , j E {1,..., n } }

5.4. Proposition. (i) The map I H M,(I) is a bijection between the sets of ideals of R and M,(R). In particular, R is simple if and only i f so is M,(R). (ii) Mn(R)/Ma(I) Mn(R/I). (iii) If R = I1 @ . . . @ I, is a two-sided decomposition of R, then

M,(R) = Mn(Ii)@ . . . @ Mn(Is) is a two-sided decomposition of M,(R). Furthermore, the ideal I; is indecomposable if and only if so is M,(I;).

Proof. (i) It is clear that M,(I) is an ideal of M,(R) and that the given map is injective. Suppose that J is an ideal of R and let I C_ R consist of all entries of elements in J. Then I is an ideal of R such that

J = M,(I). (ii) The natural homomorphism R + R / I induces a surjective homomorphism M,(R) + M,(R/I) whose kernel is M n ( I ) . (iii) This follows from the facts that Z(M,(R)) = Z(R) and that, for any r E R, M,(R)r = M,(Rr). H 5.5. Proposition.

For any ring R and any positive interger n,

J(Mn(R)) = M"(J(R)) Proof. Let V be an R-module and let V" be an M,(R)-module as in Proposition 5.3. Then ann(V") = M,(ann(V)),by the definition of V". Now apply Proposition 5.3 and Lemma 4.7 (iii). H

Let R be a finite direct product of f i l l matrix 5.6. Corollary. rings over division rings. Then R is semisimple artinian. Proof. That R is semisimple follows from Propositions 4.16 and 5.5. To prove that R is artinian, we may assume that R = M,(D)for some n 2 1 and some division ring D. Now apply Proposition 5.3.

Preliminaries

34

Our next aim is to establish the converse of the above result. As a preliminary, we record the following classical observation.

.

(Schur's lemma). Let V be an irreducible module. 5.7. Lemma Then E n d R ( V ) is a division ring. Proof. Let f : V + V be a nonzero R-homomorphism. Since f(V) is a nonzero submodule of V ,f ( V ) = V . Because f # 0, K e r f # V and so K e r f = 0. Thus f is an isomorphism, as required. Let V be an R-module and let e be an idempotent of R. Then HomR(Re, V ) E eV as additive groups. Similarly, i f V is a right R-module, then H o m R ( e R , V ) V e . 5.8. Lemma.

Proof. If f E H o m R ( R e , V ) , then e f ( e ) = f ( e 2 ) = f ( e ) E eV. Therefore the map f H f ( e ) is a homomorphism from HornR(Re, V ) to e V . Conversely, if v E e V , then the map g , : z e H zv is an Rhomomorphism from R e to V . Because the map H gv is an inverse of f H f ( e ) , the first isomorphism follows. The second isomorphism is proved by a similar argument. W 5.9. Lemma. Let e be an idempotent of a ring R. Then

EndR(Re) E (eRe)" and E n d R ( e R ) E eRe

as rings

In particular,

End&R)

E R"

and E n d R ( R R )

R

Proof. Applying Lemma 5.8 for V = R e , it follows that the map f H f ( e ) is an isomorphism of the additive group of E n d R ( R e ) onto the additive group of eRe. Given f , g E E n d R ( R e ) , write f ( e ) = erle and g ( e ) = er2e for some r1, r2 E R. Then ( f g ) ( e )= f(er2e) = erzerle = (erze)(erle)= g ( e ) f ( e ) proving that f H f ( e ) reverses multiplication. Since e is the identity element of the ring eRe, the above map preserves identity elements.

5. The Wedderburn-Artin theorem

35

This establishes the first isomorphism. The second isomorphism follows from Lemma 5.8 applied for V = eR. W We have now come to the demonstration of the following classical result. 5.10 Theorem.

(Wedderburn-Artin). Let R be a semisimple ar-

tinian ring. (i) There exist only finitely many, say Vi, . . . , V , of nonisomorphic irreducible R-modules and V , 2 Re; for some primitive idempotent e; o f R , 15 i 5 r . (ii) RR =" nl& @ . . .@ n,V, for some positive integers n;, 1 5 i 5 r , where n;T/: is a direct sum ofn; copies of K. (iii) R 2 niz1Mn,(D;),where D;= EndR(x)" S e;Re; is a division ring, 1 5 i 5 r . (iv) The integers n; and r are unique and each D; is determined up to isomorphism.

Proof. (i) and (ii): Owing to Corollary 4.2, RR is a finite direct sum of irreducible modules. Therefore, by Proposition 3.12, there exist idempotents e l , . . . ,en of R such that R R = Rel$. . .$Re, and each Re; is irreducible (in particular, e; is a primitive idempotent of R). We may assume that Rel, .. . , Re,, 1 5 r 5 n , are all nonisomorphic modules of the set { R e l , .. . ,Re,}. If V is any irreducible R-module, then by Lemma 4.7 (i) and Proposition 3.10, V 2 Re; for some i E {1,2,. . . ,r } . This establishes (i) and (ii). (iii) Put Di = EndR(V,)",1 5 i 5 r . Then we have

R"

EndR(RR)

(by Lemma 5.9)

r

!2

nEnd~(n;K)

(by Corollary 3.21)

i=l

n r

Mni(D;o)

(by Proposition 5.2)

i=l

and thus r

Preliminaries

36

(e;Re;)" we have D; Z e;Rei, Because, by Lemma 5.9, EndR(K) proving (iii). (iv) Assume that R z Mk,(Dj), where each Di is a division ring. Then there exist central idempotents u1,. . .,us of R such that

ng=,

Mkj(Di)is a simple ring. Hence Ruj is a direct sum of kj and Ruj copies of an irreducible R-module and Rui, Ruj, i # j , have no common composition factors. Thus r = s and, by renumbering the u j , we have I C3. = n,, 1 5 j 5 r, by Proposition 3.10. Put A = Mkj(Di),D = Di and let V be an irreducible A-module. E n d o ( o D ) . Hence, by Then, by Proposition 5.3(ii), EndA(V) Lemma 5.9, EndA(V) 2 Do. On the other hand, V % V,, hence

This proves that

Di

Dj and the result follows. W

5.11. Corollary. For any ring R, the following conditions are equivalent : (i) R is semisimple artinian. (ii) R is a finite direct product of full matrix rings over division rings. (iii) RR is completely reducible. (iv) Every R-module is completely reducible.

Proof. This is a direct consequence of Corollaries 4.2 and 5.6 together with Theorem 5.10. We close by proving the following classical result. 5.12. Theorem. (Wedderburn.) Let A be a finite-dimensional algebra over a field F and let I be an ideal o f A which has an F-basis consisting of nilpotent elements. Then I is a nilpotent ideal of A .

Proof. We may harmlessly assume that F is algebraically closed. Moreover, because (I J ( A ) ) / J ( A )is an ideal of A / J ( A ) having a

+

6. Group algebras and group representations

37

basis consisting of nilpotent elements, we may assume that J ( A ) = 0. Hence, by Theorem 5.10,

. . ,n,. Let

be the projection of I into M n i ( F ) 1, 5 i 5 r . It obviously suffices to show that 7 r i ( l ) = 0 for all i E ( 1 , . . . , r } . Since 7 r i ( I ) is an ideal of M n i ( F ) ,either 7 r ; ( l ) = 0 or 7 r ; ( l ) = M n i ( F ) . In the latter case, it follows that M n i ( F )has a basis consisting of nilpotent elements. But then the trace of each matrix in M n i ( F )would be zero, a contradiction. Thus 7r;(I) = 0 and the result is established. for some positive integers

6.

121,.

7ri

Group algebras and group representations

Let R be a commutative ring and let G be an arbitrary group. The group algebra RG of G over R is defined to be the free R-module on

the elements of G , with multiplication induced by that in G. More explicitly, RG consists of all formal linear combinations C xg - g, xg E R,g E G, with finitely many xg # 0 subject to for all g E G (i) C xg - g = C yg - g if and only if zg = yg (4 c x g ‘ 9 C Y g Y = C(xg Yg) ‘9 (iii) (C xg . g ) ( C yh - h ) = C zt . t , where t = Cgh=t zg- y h for all r E R (iv) r(C xg - g ) = C ( r x g ). g One easily verifies that these operations define RG as an associative R-algebra with 1 = 1~ l ~where , 1~ and 1~ are identity elements of R and G, respectively. Applying the injective homomorphisms

+

+

we shall in future identify R and G with their images in RG. With

these identifications, the formal sums and products become ordinary sums and products. For this reason, we drop the dot in xg - g. We shall also adopt the convention that RG 2 R H means an isomorphism of R-algebras.

Preliminaries

38

Let x = C x g g E RG. Then the support of x , Supp(x), is defined by

SUPP(4 = ( 9 E Glx,

# 0)

It is obvious that Supp(x)is a finite subset of G that is empty if and only if x = 0. For convenience of reference, we next record some elementary properties of group algebras. 6.1. Proposition.

Let A be an R-algebra and let

f : G -+ U ( A ) be a homomorphism of G into the unit group U ( A ) o f A . Then the map f* : RG -+ A defined b y

is a homomorphism of R-algebras. In particular, i f f is injective and A is R-free with f(G) as a basis, then RG Z A.

Proof. Because RG is R-free freely generated by G , f* is a homomorphism of R-modules. Let

be two elements of RG. Then

as required.

Let R be a commutative ring and let G be a group. Given a nonzero R-module V , by a representation of G on V we understand a homomorphism p : G + AutR(V)

6. Group algebras and group representations

39

We say that p is faithful if Kerp = 1. By Proposition 6.1, to each representation p of G on V there corresponds an RG-module structure on V given by (v E

V,zgE R,g E G)

Conversely, any nonzero RG-module V determines a representation p of G on V given by p(g)v = gv for all g E G,v E V . Thus there is a bijection between the class of all nonzero RG-modules and the class of all representations of G on R-modules. We say that the representation p of G is irreducible (indecomposable, completely reducible) if the corresponding RG-module is irreducible (indecomposable, completely reducible). Let V be an R-module which is R-free of finite rank n. Then

A u ~ R ( V ) GLn(R) where GLn(R) is the unit group of M,(R). Thus to each representation of G on V , there is a matrix representation p* : G + GL,(R). In particular, if R = F is a field, then all finitely generated F-modules are free of finite rank. In this case the study of FG-modules, representations of G on F-modules, and matrix representations of G over F are essentially equivalent. For the rest of this section, all vector spaces are assumed to be finite-dimentional over a ground field and G is a finite group. Let V be a vector space over a field F having a basis v l , . . . ,v, of n elements, and let f E EndF(V). With respect to this basis, f can be represented as an n x n matrix ( a ; j ) over F . By the trace of f , written tr f, we mean n

t r f = tr(a;j)= Caii i=l

If a different basis for V is chosen, then the two different matrices for f are similar, and hence have the same trace. Thus f H t r f is a well defined F-valued function on EndF(V). The map

Preliminaries

40

g) = t r f . trg. Now suppose that V is an FG-module. Then V determines a homomorphism 1c, : F G + EndF(V) of F-algebras. By the character of V, written x v , we understand the map xv : F G + F given by x v ( x ) = tr+(x) for all x E FG. The map x : G --f F defined by x ( g ) = xv(g) is called the character of G a$orded by V. In view of the equality X V E "99) = %XV(9) = xgx(9) is obviously F-linear, but need not satisfy the equality t r ( f

c

c

the characters x v and x determine each other. Suppose that p : G + GL(V) is a representation of G. By the character of p, (or character of G) we understand the map x : G + F given by x ( g ) = trp(g) for all g E G. It is clear from the foregoing discussion that : (i) We may treat the terms character of G and character of G afforded by an FG-module as interchangeable. (ii) The character of an FG-module V is completely determined by the character of G afforded by V. Let V and W be FG-modules. Define the sum xv xw of the characters of V and W , respectively , by the rule

+

+

Then obviously xV@w = xv xw and therefore the sum of two characters is again a character. Note that if charF = p > 0, then the constant function 0 is the character of the direct sum of p copies of any FG-module V . On the other hand, if charF = 0, then identifying Z with its image in F , we obtain

In this case, we say that x(1) is the degree of x. Let p : G 3 GL(V) be a representation of G and let x be the character of p. Then x is a class function; that is its value is cwstant on any conjugacy class of G. Indeed, for all x , y E G, we have

41

6. Group algebras and group representations

If C is a conjugacy class of G, then we put x(C)= ~ Suppose that the map

pi

( z with ) z E

C.

: G + GL(K),i = 1,2,is a representation of G. Then

P*

@'2

:G

+

GL(V1 @F

VZ)

defined by ( P I @ P 2 ) ( 9 ) = /h(g) @ P 2 ( 9 )

is a.gain a representation of G. We shall refer to pi 8 p2 as the (inner) tensor product of pi and p 2 . Note that if X I , x 2 are the characters of p1, p2, respectively, then the product xlx2defined by

is the character of pi @ p2.

We next record the following classical result. 6.2. Proposition. (Maschke's theorem). Let G be a finite group and let F be a field of characteristic p 2 0 . Then F G is semisimple i f and only i f p does not divide the order of G.

Proof. If p > 0 divides [GI, then z = CSECgsatisfies z E Z(FG) and z2 = IGlz = 0. Hence F G z E J ( F G ) # 0. Conversely, assume that p does not divide [GI, and let W be a submodule of an FG-module V . We may write V = W@W'for some F-subspace W'. Let 8 : V -+ W be the projection map. Define 1c, : V + V by

Because for all v E V and y E G,

1c, is an FG-homomorphism. Now assume that v E V . Then, for any z E G,z-'v E W , so O(x-lv) = z-lv. Accordingly, z8z-l~= v and $(v) = v. Setting

Preliminaries

42

W" = Ker$, it follows that W'' is an FG-submodule of V such that W" n W = 0. Finally, let v E V. Then, by the above, v - $(v) E W", so 21 = $(v) (v - $(v)) E W W'. Thus V = W @ W' and the result follows. rn

+

+

Suppose that A is a finite-dimensional algebra over a field F. Let V be a vector space over F, let E be a field extension of F , and form the tensor products AE=E@FA

and

VE=E@FV

If V is an A-module, then the vector space VE becomes an AE-module under a module action

An A-module V is said to be absolutely irreducible if, for every field extension E/F, VE is an irreducible AE-module. We say that F is a splitting field for A if every irreducible A-module is absolutely irreducible. We say that F is a splitting field for G if F is a splitting field for FG. We close by quoting certain standard facts pertaining to group representations ( see Curtis and Reiner (1962) Dornhoff (1971), Huppert and Blackburn (1982)).

.

6.3. Proposition Let F be an arbitrary field and let ~ 1 , ... ,X , be the characters of G aforded b y all nonisomorphic irreducible FGmodules. Then the characters xl,.. . ,xT are linearly independent (as functions from G to F . )

6.4.Proposition. Let F be a splitting field for G of characteristic 0, and let X I , . . . ,xT be all irreducible characters of G over F. For each i E (1, ...,r } , put

ei = Xi(i)/IGl

c

x;(g-l)g

g€G

Then e l , . . . ,e, is a complete system of centrally primitive idempotents of F G .

6. Group algebras and group representations

43

6.5. Proposition. Let F be a field of characteristic 0, and let V and W be arbitrary FG-modules. Then V 2 W i f and only i f xv = X W .

6.6. Proposition. (Orthogonality relations). Let xv and xw be the characters afforded b y the irreducible FG-modules V and W . Then (i) If F then CgECxv(9)XW(9-1)= 0, (ii) If F is a splitting field for F G and charF does not divide /GI, then

v w,

c xv(s)xv(g-l)

= IGI

g€G

(iii) If F is a splitting field for FG with charF not dividing \GI, CI, . . . ,C, conjugacy classes of G and X I , . . . ,xT irreducible characters of G over F , then

where gj E Cj.

In what follows if p = 0, then by definition all elements of G are p'-element s. Let F be an arbitrary field of characteristic p 2 0, let n be the least common multiple of the orders of the p'-elements in G and let E be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers r , taken modulo n, for which E I+ E' defines an automorphism of F ( E )over F . Two $-elements a, b E G are called F-conjugate if z-lbz = a' for some z E G and some r E I,. Thus an F-conjugacy class is a union of ordinary conjugacy classes. 6.7. Proposition. (Witt-Berman theorem). Let F be an arbitrary field of characteristic p >_ 0 and let G be a finite group. Then the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G.

We close by quoting some standard results on absolutely irreducible modules (see Huppert and Blackburn (1982)).

Preliminaries

44

Let A be a finite-dimensional algebra over a field F , and let V be an irreducible A-module. Then the following conditions are equivalent: (i) V is absolutely irreducible. (ii) If E is the algebraic closure of F , then VE is an irreducible AEmodule. (iii) EndA(V) = F , i.e. each A-endomorphism o f V is a left multiplication b y an element of F . 6.8. Proposition.

6.9. Proposition. Let A be a finite-dimensional algebra over a field F . (i) F is a splitting field for A if and only if

for some r 2 1 and some n; 2 1, 1 5 i 5 r . (ii) If F is a splitting field for A, then for any field extension E I F , E is a splitting field for A E . 6.10. Proposition. (Burnside). Let V be an absolutely irreducible A-module, where A is a finite-dimensional algebra over a field F . Then the homomorphism A EndF(V), a H $a, where $a(v) = av, a E A, v E V , is surjective.

45

Chapter 2 General properties of induced modules In this chapter we record a number of basic results concerning induced modules. After presenting some formal properties of these modules, we examine in detail their annihilators and provide a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. The chapter ends with a short discussion of projective covers of induced modules.

1.

Induced modules, representations and characters

Throughout this section, R denotes a commutative ring and G a group. Let H be a subgroup of G. Then the group algebra RH can be identified with the subalgebra of RG consisting of all R-linear combinations of the elements h E H . If V is an RG-module, then we shall denote by VH th RH-module obtained by the restriction of algebra; thus as an R-module, VH equals V , but only action of RH is defined on VH. This process will be called restriction and it permits us to go from any RG-module V to a uniquely determined RH-module V'. There is a dual process of induction. Let V be any RH-module. Since we may consider RH as a subalgebra of RG, we can define an RG-module structure on the tensor product

General properties of induced modules

46

This is the induced module and we denote it by V G . 1.1. Lemma. Let H be a subgroup of G. If T is a left (right) transversal for H in G, then RG is a free right (left) RH-module with T as a basis.

Proof. Suppose that T is a left transversal for H in G. Then, for any t E T , t(RH) is the R-linear span of the coset t H . Therefore, for any tl,. . . ,in E T ,t l ( R H ) -.. t,(RH) is the R-linear span of UY="=,t;H.As is apparent from the definition of RG, if X and Y are disjoint subsets of G, their R-linear spans meet at 0. Thus

+ +

proving that RG is a free right RH-module with basis T . A similar argument proves the case where T is a right transversal. 1.2. Corollary. Let H be a subgroup ofG, let T be a left transversal for H in G and let V be an RH-module. (i) VG = @V (direct sum of R-modules) (ii) V G= (l@V)$(CtET,tgH t@V) (direct sum of RH-modules) (iii) The map V -+ t @ V, v H t @ v is an R-isomorphism, while the map V + 1 8 V , v H 1 @I v is an RH-isomorphism. Proof.

(i) By Lemma 1.1, RG = etETt(RH),hence

as required.

(ii) By (i), it suffices to prove that CtET,tgHt@V is an RH-submodule . h E H and t E T - H. Then ht = t'h' for some h' E H of ( V G ) ~Fix and t' E T - H. Since h(t @ V) = ht @ V = t'h' @ V = t' 8 V, (ii) is established. (iii) The map V -+ R H @wV = 1 @ V, v H 1 @ v is obviously an RH-isomorphism. If t @ v = 0 then 18v = 0, hence v = 0. This shows that V -+ t @ V , v H t @ v is an injective R-homomorphism. Since the above map is obviously surjective, the result follows.

1. Induced modules, representations and characters

47

Let V be an RG-module. By the kernel of V , written I<erV, we understand the kernel of the representation of G afforded by V , i.e. I<erV = { g E Glgv = v for all v E V } 1.3. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

Proof. Let T be a left transversal for H in G. Then obviously

By Corollary 1.2(i), g E Ii'erVG if and only if g ( t @ v) = t 8 v

for all t E T ,v E V

If g t = t'h with t' E T , h E H , then g ( t @ v) = t ' @ hv. Hence, by Corollary 1.2(i), g ( t 8 v) = t 8 v if and only if g t = t h with hv = v. Thus g E I<er(VG)if and only if g E ntETtKerVt-l,as required. The process of induction is closely related to the concept of imprimitivity of a module. We say that an RG-module V is imprimitive if V can be written as a direct sum

of R-modules r/: with 111 a permutation of the set

> 1 such that for each g E G, V;:

g V , is {r/:li E I } . The latter is equivalent to the assumption that G acts on the set { K J iE I } with each g E G sending V, to g v . We refer to { K l i E I } as a system of imprimitivity for V . The following result ties together imprimitive and induced modules. H

1.4. Proposition. (i) Let H be a proper subgroup of G, let V be an RH-module and let T be a left transversal for H in G . Then { t 8 Vlt E T } is a system of imprimitivity for V G and the action of G on it is transitive. (ii) Let V be an imprimitive RG-module with {r/:liE I } as a system

General properties of induced modules

48

of imprimitivity and let G act transitively on

{xli E I ) . If H is the

stabilizer of any W E { K l i E I } , then W is an RH-module such that VS'G

Proof. (i) This follows from Corollary 1.2(i) and the fact that the left multiplication by g E G induces a transitive action of G on the set

{ t 63 Vlt E T } . ( 2 ) Let T be a left transversal for H in G. Then { K l i E I } = { t W l t E T } and V = etETtW (direct sum of R-modules). Hence, by Corollary l.2(i), the map t 63 W -+ t W , t 63 w H t w induces an Risomorphism W G V . Given g E G, t E T , we may write gt = t'h for some t' E T , h E H . Hence f ( g ( t 8 w)) = f ( t ' h @ W ) = f (t' 63 h w ) = t'hw = gtw = g ( t w ) = g f ( t 8 w ) as required.

Let V be an RH-module and let g E G. We define gV to be the R(gHg-')-module whose underlying R-module is V and on which the elements z E gN9-l act according to the rule 2

* 2, = (g-1zg)v

1.5. Lemma. Let H be a subgroup o f G , let V be an RH-module and let g E G. Then (i) gV E g 8 V and hV V for all h f H . (ii) ( g ~ ) ' z vG. (iii) For all z, y E G, "YV "(YV). (iv) V is irreducible (completely reducible, indecomposible) if and only i f so is gV. (v) If R = F is a field, then for any field extension E / F , the E(gHg-')-modules ( g V ) E and g ( V E ) are isomorphic. . -

Proof. (i) The map f : gV --f g @ V , v H g @ v is an Risomorphism, by Corollary 1.2(iii). Given z = ghg-', h E H , we have

1. Induced modules, representations and characters

proving that gV

%g @

49

V . If h E H , then

as required. (ii) Choose a left transversal T for H in G with g = t E 2'. Then the stabilizer of t @ V is tHt-'. Hence, by (i) and Proposition 1.4,

as desired. (iii) Applying (i), we have XY

v = (zy) @ v = z(y @ V ) E "(y @ V )z "("),

as required.

(iv) This follows from the fact that W is an RH-submodule of V if and only if W is an R(gHg-')-submodule of gV. (v) Observe that ( ~ V ) and E ~ ( V Ecoincide ) as vector spaces. Let z E gHg-', X E E and w E V . Then the action of z on X 8 v E g(V') is given by 2

* (A @ w)

= g-*zg(X @ v) = X @ g-lzgw

Bearing in mind that the corresponding action in (gV), is given by z(X @ v) = X @ z

*

2)

= X @ g-'zgv,

the result follows. W 1.6. Lemma. Let H be a subgroup of G, let E f F be a field extension and let V be an FH-module. Then

( V G )Z ~(VE)'

as

EG-modules

Proof. Let { v;/iE I }be an F-basis of V and let T be a left transversal for H in G. Then

General properties of induced modules

50

are E-bases of ( V G )and ~ ( V E ) ~respectively. , Hence the map 6' : ( V G )+ ~ ( V E )defined ~ by B ( 1 €3 (t @ v;)) = t @ ( 1 €3 v;) is an Eisomorphism. Fix g E G, t E T and write g t = t'h for some t' E T , h E H . Then, for z = 1 €3 (t €3 vi), we have

as required. W

Let F be a field and let V , W be FG-modules. Then the vector space V @F W is an FG-module where the action of elements g E G is defined by g(v €3 w)= gv €3 gw and then extended to V € 3 W ~ and FG by F-linearity. Note that if pv, pw are representations of G afforded by V , W respectively and d i m ~ V< 00, dirn~W< co,then V @IF W affords the tensor product Pv 63 Pw.

1.7. Lemma. Let EIF be a f i e l d extension and let V , W be FGmodules. Then

(V @ F W ) E% VE@ E WE as EG-modules Proof. Let {v;li E I } and {wjlj E J } be F-bases of V and W , respectively. Then (1 @ (vi @ wj)li E I , j E J }

and

{(I €3 vi) €3 (1 €3 wj)li E I , j E J }

are E-bases of (V @F W ) Eand VE@ E W E ,respectively. Hence the map

extends to an E-isomorphism of (V @ F W ) Eonto VE@ E WE. It is obvious that for all v E V , w E W ,

1. Induced modules, representations and characters

51

Since for all g E G,

and 9((1@ V i ) 8 (1 63 Wj)) = (g(1 8 v;) 8 g(l €3 Wj)) = (1 @ 9%) €3 (1 8 gwj)

we have +(9(1€3 i.(

€3 Wj))= g+(l C3 (0;€3 Wj)),

as required.

For the rest of this section, G denotes a finite group and all vector spaces over a field F are assumed to be finite-dimensional. 1.8. Proposition. Let H be a subgroup of G, let V be an F H module with F-basis {vl,. . . ,urn} and let p be the matrix representation of H aforded b y V in this basis. If gl,g2,.. . ,gn is a left transversal for H in G , then (91 @ v l , * - - , g l

is an F-basis of V G with respect to which the matrix representation pG aforded b y V G is given b y ~ ~ (=9(Pij(9)) )

(1 L i , j L n>

where p;j(g) = p(grlggj) if gT'ggj E H and p;j(g) = 0 if gr'ggj

9 H.

Proof. That the elements (9;@ vj} constitute an F-basis of V Gis a consequence of Corollary 1.2(i). We may write p(h) = (yij(h)), where hv; = Cj y,;(h)vj. To find the matrix pG(g), fix j E { 1,. . . ,n}and write ggj =gkh(h E H).Then g(gj C3 vt) = ggj 8 vt = gkh C3 vt = gk C3 hvt m

s=l

m s=l

General properties of induced modules

52

Now extend the coefficients -yst to the whole of G by setting yst(z) = 0 for x E G- H. Because for a fixed j there is only one k with gk'ggj E H, we have m n S(9j

63 4 =

y Y s t ( 9 ; 1 g 9 j ) ( 9 i (23 us), s=l i = l

as required.

Let V be an FH-module and let x be the character of H afforded by V . Then the character of G afforded by the induced module V Gis called the character induced from x; we shall use the notation xG for such a character. 1.9. Proposition. Let x be a character of H and let 91,. . . , g n be a left transversal for H in G. Set ~ ( x=) 0 for all x E G - H. (i) X G ( g ) = CZ1 x ( g i l g g i ) and IHIXG(9)= x(z-'gz) (9 E G )

c

XG

(ii)

If charF does not divide \HI, then

XG(d

=

V1-l

c

x(z-19z)

(9 E G )

zEG

(iii) If charF = 0 , then for any given g E G ,

where h l , . . . h , are the representatives for the conjugacy classes of H contained in the conjugacy class Cg of g and, b y convention, X G ( g ) = 0 if H n C, = 0. )

Proof. (i) Let x be afforded by an FH-module V. Then, in the notation of Proposition 1.8, we have trpii (9)= X ( g t l g g i ) and hence i=l

i=l

53

1. Induced modules, representations and characters

as required.

Since the value of xG does not depend on the basis of V G with respect to which xG is computed, it must also be independent of the choice of the left transversal for H in G. Bearing in mind that G can be written as a disjoint union of IHI left transversals for H in G, it follows from (1) that IHIXG(d =

c

x(s-lg4

(2)

XEG

as asserted. (ii) This is a direct consequence of (2). (iii) If H n Cg= 8, then the assertion follows from (i). Assume that H n Cg # 0. For any h in H n Cg,let X h denote the conjugacy class of H containing h. Then

H n cg = x h l u X h , u . - .u x h , for some h l , . . . ,hm in H . Since x is constant on each

xh,,

we have

Now if h f H n Cg, then as s runs over G, s-'gz = h for exactly C c ( g ) values of 2. Applying (2) and (3), we deduce that

as desired. We now apply Proposition 1.9(iii) to exhibit an orthogonality relation proved by Osima (1952). 1.10. Proposition. Let H be a subgroup of G, let F be an algebraically closed field of characteristic 0 and let X I , . . . ,xr be all irreducible characters of H over F . Then, for all g E G, h E H ,

General properties of induced modules

54

Proof. If H n Cg = 8, then g is not conjugate to h. Hence the result follows from Proposition 1.9(iii). We may thus assume that Cgn H # 8. Denote by h l , hz, . . . ,h , the representatives for the conjugacy classses of H contained in Cgsuch that h E { h l , . . . ,h,} whenever g is conjugate to h. Since the number of conjugacy classes of H is equal to r , we may choose the representatives h,+l,. . . , h, for the conjugacy classes of H not contained in Cgwith h E {h,+l, . . . ,h,} whenever h is not conjugate to g . By Proposition 1.6.6(i), we have

2

Xk(hi)Xk(hi')

= IcH(hi)lhij

k=l

(4)

Now h = h, for some s E { 1,. .. ,r } and h is conjugate to g if and only if 1 5 s 5 m. Applying Proposition 1.9(iii), we derive

and, by (4),the latter is equal to IC,(g)l if 1 5 s 5 rn and 0 if s So the proposition is true. 2.

> rn.

Formal properties of induced modules

Throughout this section, R denotes a commutative ring and G an arbitrary group. 2.1. Lemma. (Transitivity of the induction). If L, H are subgroups of G with H L, then for any RH-module V ,

s

(VL)G

Proof. We have

2VG

2. Formal properties of induced modules

55

as required.

2.2. Lemma.

Let V be an RH-module. Then the map

is an injective homomorphism of RH-modules. Furthermore, the image

1 @ v of V in V G is a direct summand

Of(VG)H.

Proof. Since R H is a direct summand of the RH-module RG, RH @ R H V is a direct summand of the RH-module VG.Since the map

is an isomorphism of RH-modules, the result follows. H

We shall refer to the injective RH-homomorphism V + V G ,v H 1@ 2) as the canonical injection. With the aid of this injection, we now provide a universal characterization of V G . Let H be a subgroup of G, let V be an RHmodule and let f : V + V G be the canonical injection. Then, for a n y RG-module W and for any Q E HomRH(V, W H ) ,there exists one and only one $ E HomRG(VG,W ) such that tp = 1C, o f . 2.3. Proposition.

Proof. Since x@v = z ( l @ v ) , x E R G , v E V ,any RG-homomorphisr V G + W is uniquely determined by its restriction to 1 @ V . Thus there exists at most one such +. Consider the map RG x V + W , (2,v) H zcp(v). Then this map is balanced and therefore determines a 12 -homomorphism V G3 W , x @ v H xcp(v), x E R G , v E V . Because

General properties of induced modules

56

for all z,z1 E RG and v E V, 1c, is an RG-homomorphism. Taking into account that $(l 8 v ) = cp(v) for all v E V the result follows. W

Let S be a subring of a ring L and let M be a (left) S-module. Given 1 E L and 1c, E Homs(L,M), define lcp : L + M by for all x E L

(/y)(x) = cp(z1)

It is straightforward to verify that the additive group Homs(L,M) becomes a (left) L-module. Let H be a subgroup of G and let V be an RH-module. Since R H is a subring of RG, we may apply the foregoing construction so that N o m m ( R G , V ) becomes a (left) RG-module. We shall denote this module by GV and refer to GV as a coinduced module. 2.4. Proposition. Let H be a subgroup of an arbitrary group G, let V be an RH-module, and let W be an RG-module. Then (i) HOmRG(VG, % HOmRH( WH) as R-modules (ii)HomRG(W,G V) S HomRH(WH, V) as R-modules

w)

v,

Proof. (i) Given X E Homm(V, W H )we may, by Proposition 2.3, define A* E HomRG(VG, w) by X*(z €3 v) = .(X(v))

The map X

H

w E V,x E R G

A* is obviously R-linear. Moreover, if A * = 0, then X * ( l @ w) = X(v) = o

for all v E V

and thus X = 0. Finally, given 1c, E HomRG(VG,W), define cp : V + W by cp(v) = $(1 €3 v) for all w E V. Then cp E H o ~ R H ( V WH) , and 1c, = cp* as required. ( V), define y(X) E H O ~ R H WH, ( V ) by (ii) For any X E H O ~ R GW,G

2. Formal properties of induced modules

57

Then clearly X H p(X) is an R-linear map. If a E then the map $ ( a ) defined by

[Ilt(a>(w)l(4 = 44

H u ~ R H ( WV H ) ,,

(wE W,sE RG)

belongs to H u ~ R G ( W V ),.~Moreover, ($cp)(X) = X and Hence the map X H $(A) is a required isomorphism. H

((P$J)(CY)

= a.

Let H be a subgroup of an arbitrary group G, let V be an RH-module and let W be an RG-module. (i) If V and W are irreducible, then V is isomorphic to a submodule of WH if and only i f W is isomorphic to a factor module of V G . (ii) If V is irreducible, then for any irreducible factor module X of V", V is isomorphic to a submodule of XH. 2.5. Corollary.

Proof. (i) Note that V is isomorphic to a submodule of W H if and only if H O ~ R H ( W V ,H )# 0. On the other hand, W is isomorphic to a factor module of V Gif and only if H o ~ R ~ (WV) # ~ 0. , Now apply Proposition 2.4 (i). (ii) This is a direct consequence of (i). H We next establish the following result which will allow us to identify, under certain circumstances, induced and coinduced modules.

Let H be a subgroup of G offinite index and let V be an RH-module. Then 2.6. Proposition.

Proof. Let (91, g 2 , . . . ,gn} be a left transversal for H in G. For every $ E" v, put n

d(4)

= csi @ 4(gi') i=l

E VG /'

Then an elementary calculation shows that 8 :G V --+ V G is an Rhomomorphism. To show that 8 is an RG-homomorphism, it therefore suffices to verify that S(g4) = g O ( 4 )

for all g E G, 4 eG V

General properties of induced modules

58

For every g E G, there is an h; E H an a permutation i H r ( i ) of the

index set {1,2, . . . ,n} such that 97'9 = hjg&. It follows that n

n

i=l n

i=l n

=

proving that 8 is an RG-homomorphism. If Cy=lgi 8 $(gi') = 0, then 4(gi1) = 0 for all i E { l , .. . ) n } . Because {gT1,. . . ,g;'} is a right transversal for H in G, given g E G, we have g = hg;' for some h E H and i E (1,. . . ,n}. Hence 4(g) = h4(gT1) = 0 and so q5(Csgg) = C z g 4 ( g ) = 0, proving that 8 is injective. The elements gT',gz',. . . , g i l constitute a basis of a free left RHmodule RG. It follows that for every set {vi E VI1 5 i 5 n } , there is a 4 eG V such that 4(gc') = vi, 1 5 i 5 n. Hence 8 is surjective and the result follows. 2.7. Corollary. Let H be a subgroup of G of finite index, let V be RH-module and let W be an RG-module. Then

Proof. Apply Propositions 2.6 and 2.4(ii). Suppose that U is a submodule of an RH-module V , and let i : V be the inclusion map. Since RG is a free RH-module, RG is flat, so the canonical homomorphism

U

+

is injective. For this reason, from now on we shall identify image in V G .

U G with its

2. Formal properties of induced modules

59

2.8. Lemma.

Let &and V2 be submodules of an RH-module V. (i) &G G VF if and only if V1 G V,. (ii) Vp = V : i f and only if V1 = V,.

(iii) (Vl + V2)G= KG + .v : ( i ~ ) (& n v ~= )hG~n ~ 2 ~ . (w) If V = V, @ V,, then VG = V F @ CG.

Proof. We first observe that (ii) is a consequence of (i), that (v) follows from (iii) and (iv) and that (iv) is a consequence of the fact that RG is a free RH-module. To prove (i), note that V1 C V2 obviously implies YGG Vp. Let T be a left transversal for H in G containing 1. Then each element of VG can be uniquely written in the form CtET taut with finitely many vt E V not equal to 0. Therefore if 01 E V1 is such that l @ v l E V p then v1 E V2. Thus qG VF implies & C VZ,proving (i). To prove (iii), observe that the inclusion VF V? C (V1 V2)G follows from (i). The opposite inclusion being a consequence of the

+

+

equality

t 69 (v1

+

v2)

= t €3 211

+ t €3

02

(v1 E

v1,v2

E V2,t E T )

the result follows. The following theorem contains some important formal properties of induced modules.

Let H be a subgroup of an arbitrary group G and let U A V 3 W be a sequence of homomorphisms of RH-modules. Then the following properties hold: (i) The sequence 2.9. Theorem.

is exact if and only if the corresponding sequence of RG-modules

is exact. (ii) Suppose that the sequence (1) is exact. Then (1) splits if and

General properties of induced modules

60

only if (2) splits. (iii) I f U is a submodule of V , then V G / U G (V/U)". Proof. (i) That exactness of (1) implies that of (2) is a consequence of the fact that RG is a flat RH-module (see Proposition 1.2.8(iii)). Conversely, suppose that (2) is exact. By Proposition 1.2.9(i), I ( e r ( l 8 p ) = (I(erp)G

and

Im(1 €9 A) =

Because I(er(1 €9 p ) = Im(1 €9 A), it follows from Lemma 2.8(ii) that K e r p = ImA. A similar argument shows that Ker(1 €9 A) = 0 implies KerX = 0 and that Im(1 €9 p ) = W Gimplies I m p = W . Hence the sequence (1) must also be exact. (ii) Suppose that 0 + U 5 V 1: W -+ 0 is a split exact sequence and let y : W t V be a splitting homomorphism. Then 1 €9 y : WG+ V G is obviously a splitting homomorphism for the induced sequence. Conversely, assume that the exact sequence

of RG-modules splits. Let T be a left transversal for H in G containing 1. Consider the mappings

v

1,

WSWG

VG

W H l ( 8 W

CtfT

t (8 V t

U1

Then c and T are well defined RH-linear maps. Let $I : W G+ V Gbe a splitting homomorphism for (2) and let w E W . Then, on the one hand, for some vt E +(I €9 w)= t B vt

C

t€T

and, on the other hand,

v

2. Formal properties of induced modules

Hence

20

61

= p ( q ) and therefore

This shows that ~ $ :0W V is a splitting homomorphism for (1) and the required assertion follows. (iii) Apply (i) to the natural exact sequence 0 -+ U -+ V -+ V / U + 0. H

2.10. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. I f V G is completely reducible, then so is V.

Proof. Let W be a submodule of V and let

0

-+

w

-+

v

-+

v/w+ 0

be the natural exact sequence. The corresponding sequence 0 + WG+

vG (v/w)G0 --f

-+

splits by the hypothesis. Hence, by Theorem 2.9, the sequence 0+

w v + v/w+ 0 -+

splits, proving that V is completely reducible. Let S be a ring and let V be a left S-module. Given a (left or right) ideal I of S,let IV denote the set of all finite sums of the form

Then IV is an additive subgroup of is in fact a submodule of V .

V and if I is a left ideal, then IV

2.11. Proposition. Let H be a subgroup of an arbitrary group G and let X C Y be left ideals of R H .

General properties of induced modules

62

(i) ( Y / X ) GE A . Y / A- X fii) Let I be a right ideal of RG such that I 2 RG . X . If V is an RH-module, then there is a canonical injective map of additive groups

$J : I ( V G )+ ( X V ) G

Proof. All tensor products below are to be taken over R H . (i) Consider the diagram

0 -+ R G @X

+ R G @Y -+

R G @ ( Y / X ) -+ 0

1s

If 0 --* RG * X

--+

RG. Y

-+

RG * Y / R G* X

+

0

where all maps are canonical. Then the diagram is commutative and since RG is a flat RH-module, both rows are exact, and f , g are isomorphisms (see Proposition 1.2.8). This establishes the required assertion. (ii) Denote by X the natural homomorphism V + V / X V . Then the sequence

o -+ R G B X V --+

R G @v ' 9 R G B (V/XV)

is exact. Let x E I , y E RG, and v E V . Then xy E I can be written in the form

o E RG . X, so xy -+

Thus

and hence I(RG @ V ) C I(er(l@ A), as required. For the rest of this section, G denotes a finite group and F an arbitrary field. All FG-modules are assumed to be finite-dimensional

63

2. Formal properties of induced modules

F-spaces.

Let U and V be any FG-modules. Then the intertwinning number for U and V is defined to be

Let U be a completely reducible FG-module, and let V be an irreducible FG-module. We say that a nonnegative interger rn is the multiplicity of V as an irreducible constituent of U if a decomposition of U into irreducible submodules contains exactly rn submodules isomorphic to

V. 2.12. Lemma. Let U and V be completely reducible FG-modules. (i) i(U,V)= i(V,U). (ii) I f V is irreducible and F is a splitting field for FG, then i(U, V ) is equal to the multiplicity o f V as an irreducible constituent of U . (iii) I f F is a splittingfield for F G , then U is irreducible if and only if2(U,U) = 1. Proof. (i) Suppose that U and V are irreducible. If U

V , then

V , then obviously HornFG(U, V ) 2 Horn=~(V, U ) . This proves the case where U and V are irreducible. In the general case, write

If U

where the { U j } and { V k } are irreducible. Because

we have

i ( U 1 e Uz, U3) = i(U1,U3) for any FG-modules U,, Uz, and

U3.

+ i(U2,

Similarly,

u3)

64

General properties of induced modules

Thus

proving (i). (ii) If V is irreducible, then by the above

On the other hand, since F is a splitting field for FG,

proving (ii). (iii) Write U = mlU1 @ - . @ mtUt with U l , . .. ,Ut pairwise nonisomorphic and irreducible. Then, by (2), i(U, Us)= rn, for all s E {1,...,t } . It follows that

i(U,U ) = m:

+ - - + rn:

whence the result.

As a preliminary to the next theorem, we introduce the following definitions. Let A be a finite-dimensional algebra over a field F , and let V be an A-module. By the socle Soc(V) of V , we understand the sum of all irreducible submodules of V. We shall refer to the factor module V / J ( V ) as the head of V . 2.13. Lemma. (i) V / J ( V ) is completely reducible, and for any submodule W such that V/W is completely reducible, J ( V ) W . (ii) Soc(V) is completely reducible, and for any completely reducible submodule W of V, W C Soc(V). (iii) J ( V ) is the intersection of the kernels o j t h e homomorphisms V 4 W , where W ranges ouer all irreducible A-modules. (iv) For any given A-modules V and W , W is isomorphic t o an irreducible constituent of V /J ( V ) if and only if W is isomorphic to an irreducible factor module o j V .

2. Formal properties of induced modules

65

Proof. (i) This is a direct consequence of Lemma 1.4.3 and Proposition 1.4.1. (ii) Apply Proposition 1.3.15. (iii) This is a direct consequence of the definition of J ( V ) . (iv) Apply (i). 2.14. Theorem. (Nakayama reciprocity). Let F be an arbitrary field, let H be a subgroup of G and let V and W be irreducible F H and FG-modules, respectively. If m is the multiplicity of W as an irreducible constituent in the head of V G and n is the multiplicity o f V as an irreducible constituent in the socle of W,, then m - i(W,W )= n i(V,V ) In particular, m = n if F is a splitting field for F G and FH

Proof. We first observe that HomFG(U, W )= 0 for all irreducible FG-modules U with U 9 W . Applying Lemma 2.13(i), (iii), we therefore deduce that

HomFc(VG,W ) Z H O ~ F G ( V ~ / J (WV)~ ) , Z horn^^( W,W )@ - . -@ HomFG(W, W ) (m times) Thus i(VG,W ) = m . i(W,W ) . By Lemma 2.13(ii), the image of any FH-homomorphism V -+ WHis contained in the socle S of WH.Accordingly,

and therefore i(V,W H ) = n - i(V,V ) . Bearing in mind that, by Proposition 2.4(i), i(VG,W ) = i(V,W H )the , result follows. 2.15. Corollary. (Frobenius reciprocity). Let H be a subgroup of G , let F be a n algebraically closed field such that charF does not divide /GI, and let V and W be irreducible FH and FG-modules, respectively. Then the multiplicity of V as an irreducible constituent of W H is equal to the multiplicity of W as an irreducible constituent of VG.

66

General properties of induced modules

Proof. Apply Maschke’s theorem and Theorem 2.14. Let F be a splitting field for F G of characteristic 0 and let Cf(G, F ) be the F-vector space of class functions f : G + F . By defining

we obtain a symmetric nonsingular bilinear form Cf(G, F)x C f(G, F ) -+ F, called the inner product. If X I , . . . ,xr are the irreducible characters of G, then X I , . . . ,x r form an orthogonal basis for Cf(G, F ) . This can be easily seen by applying Proposition 1.6.6. Note also that, for any character x of G over F,

and a; =< x,xi >. Applying Corollary 2.15, we immediately derive 2.16. Corollary. (Frobenius reciprocity for characters). Let F be an algebraically closed field of characteristic 0, let H be a subgroup of G and let x and 0 be irreducible characters ofH and G, respectively. Then

< X G , O >=< x,e, >

3.

Annihilators of induced modules

Throughout this section, G denotes an arbitrary group and R a commutative ring. Unless explicitly stated otherwise, RG-modules are assumed to be left RG-modules. Let H be a subgroup of G and let V be an RH-module. We should like to read properties of RG from VGand conversely, properties of V G from RG. One very valuable tool for obtaining some of this exchange of information is propvided by “annihilators”, defined bellow. Let M be an RG-module. Then the annihilator of M , written ann(M),is defined by

a n n ( M ) = {z E RGlzM = 0)

3. Annihilators of induced modules

67

The annihilator of a right RG-module is defined similarly. Note that in both cases the annihilator is an ideal of RG. Our main result in this section describes ann(VG) exclusively in terms of ann( V ) ;most of the material presented is contained in a work of Kniirr(1979 ). Let H be a subgroup of G. Then the natural projection

R:RG+RH is defined by Y

It is clear that T is an R-linear map; in fact property, namely

R

satisfies a stronger

3.1. Lemma. Let H be a subgroup ofG and let R : RG be the natural projection. (i) R is a homomorphism of left and right RH-modules. (ii) For any left ideal I of RG, I E RG R ( I ) . (iii) For any right ideal I of RG, I R ( I ) - RG.

+

RH

-

Proof. (i) Let T be a right transversal for H in G such that 1 E T . Owing to Lemma 1.1, the map t H 0, 1 H 1, 1 # t E T , extends to a homomorphism R' : RG + RH of left RH-modules. Since 7r = R', we conclude that R is a homomorphism of left RH-modules. A similar argument, by taking T to be a left transversal, shows that R is a homomorphism of right RH-modules. (ii) Let T be a left transversal for H in G. By Lemma 1.1, each a E I can be uniquely written in the form

a =xtat

(at E RH)

tCT

with finitely many nonzero at. For each t f T , we have

t-'a E I

and at = r(t-'a) E w(I)

Hence a E RG . R( I ) , proving (ii). (iii) Choose T to be a right transversal for H in G and apply the argument in (ii).

General properties of induced modules

68

3.2. Lemma. Let N be a normal subgroup of G and let T be a transversal for N in G. Then (i) For any ideal A of R N ,

RG - A = $tET t A - t

(direct sum of RN-modules)

where t A = tAt-' (ii) For any family {A,JiE I } of ideals of R N ,

RG - (fI;czAi) = niczRG * Ai (iii) For any ideals A, B of R N , A RG*B.

c

B if and only if RG - A

c

Proof. (i) Since N d G , t A is an ideal of R N . In particular, t A RN and hence, by Lemma 1.1,

C t A * t= $ t c T t A * t t€T

is a left ideal of RG contained in RG-A. Since for t E N , tA.t = t A = A , the required assertion follows. (ii) By (i), we have RG - A; = e t E tA;~* t and therefore

niEzRG - A; = =

t ) = $tET (nicztAi) t (nicZAi) ' t

niEz(etETtAi @tET

= RG * (niEzAi)

as asserted (iii) By (i), we have

-

Hence, if RG A c RG - B , then A c B. Conversely, if A C B , then certainly RG A C RG - B. If RG - A = RG - B , then by the above, A = B , a contradiction. Thus RG A C RG B , as required. H

-

-

For any additive subgroup A of RG, let I d ( A ) denote the sum of all two sided ideals of RG contained in A. Thus Id(A) is the unique largest ideal of RG contained in A.

3. Annihilators of induced modules

69

3.3. Lemma. Let H be a subgroup of G, let I be an ideal of R H and let a : RG + RH be the natural projection. (i) Id(RG I ) = ann(RG/RG. I ) = ngEGRG(gI) where gI = gig-'. (ii) Id(RG.I) is the unique largest two sided ideal J of RG satisfying

(iii) Id( I RG) = Id( RG - I ) . (iv) If H is a normal subgroup of G, then *

I d ( RG * I ) = RG( n g E G ’ I ) Proof. (i) Because ann(RG/RG.I ) is an ideal of RG, the equality

Id(RG - I ) = ann(RG/RG - I ) is a consequence of the definition of Id(RG.1). Since I d ( R G - I ) RG-I and I d ( R G . I ) is an ideal of RG, we have

Id(RG.I ) Accordingly,

cg( R G .I ) = RG(’I)

Id(RG - I )

c ngEGRG(gI)

But ngEGRG(gI)is obviously an ideal of RG contained in RG I . Thus

as required.

(ii) Let J be an ideal of RG such that a ( J ) C I . Then, by Lemma 3.l(ii),

J~RG*T(J)GRG.I and so J C I d ( R G . I ) . On the other hand, let J be any ideal of RG contained in RG I . Then T ( J ) 2 a(RG - I ) G I , since a is a homomorphism of right RH-modules (Lemma 3.l(i)) and I is a left ideal of RH.

(iii) Applying Lemma S.l(iii) and the argument of (ii), we conclude that I d ( I RG) is also the unique largest two-sided ideal J of RG

-

General properties of induced modules

70

satisfying ? r ( J ) 2 I . Thus I d ( I *RG) = Id(RG * I ) as required. (iv) This is a direct consequence of (i) and Lemma 3.2(ii). H We are now ready to describe the annihilators of induced modules.

3.4. Theorem. Let H be a subgroup of an arbitrary group G , let V be a left RH-module and let I = ann(V). (i) ann(V G )= Id( RG I ) = ng,GRG(gI). (ii) If H d G, then ann(VG)= RG(flseG”). (iii) If V G is completely reducible, then J ( R G ) G RG(gI) for all g E G. The converse is true provided R G / J ( R G ) is artinian.

-

Proof. It is clear that (iii) is a consequence of (i). Hence, by Lemma 3.3(i),(iv), it suffices to verify that ann(VG)= Id(RG * I ) . To this end, let J C RG.1 be an ideal of RG. Since IV = 0, Proposition 2.11(ii) tells us that J ( V c ) = 0. Hence J & ann(VG)and therefore

Id(RG * I )

ann(VG)

To prove the reverse containment, let T be a left transversal for H in G. Then, by Corollary 1.2, V G = @I V . If 5 E ann(VG),say 5 = &Ttzt, xt E RH, with finitely many xt # 0, then for all o E V ,

Hence each Q E I and therefore 5 E RG.1. It follows that ann(VG)G RG I and thus ann(VG) Id(RG I ) , as desired. H

-

We close by proving a number of applications of the above result. As a preliminary observation, we first record 3.5. Lemma. Let H be a subgroup of an arbitrary group G. Then

J(RG)

RG*J ( R H )

if and only if J ( R G ) 5 J ( R H ) * RG

71

3. Annihilators of induced modules

Proof. Apply Lemma 3.3(iii) for I = J ( R H ) . H 3.6. Theorem. Let H be a subgroup of an arbitrary group G and let R be a commutative ring such that R H / J ( R H ) is artinian. Then

the following conditions are equivalent: (i) For a n y irreducible RH-module V , ( V G )is~completely reducible. (ii) J ( R H ) . RG C RG J ( RH ). (iii) J ( R H ) RG = RG J ( RH ) . Moreover, if these conditions are satisfied, then for any given positive integer n, ( R G * J ( R H ) ) " = RG * J ( R H ) " Proof. (i)+(ii): P u t V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is completely reducible, so by assumption ( V G ) ~ is completely reducible; hence J ( R H ) G ann(VG). Now J ( R H ) = ann(V) and thus, by Theorem 3.4(i), ann(V G )= Id( RG - J ( R H ) ) Hence J ( R H ) R G

s I d ( R G - J ( R H ) ) and therefore J ( R H ) - RG

RG-J(RH). (ii)

+ (iii): It follows from (ii) that RG - J ( R H )- R G

C RG

*

J(RH)

whence

RG J ( R H ) *

RG - J ( R H ) * R G G Id(RG - J ( R H ) ) = I d ( J ( R H ) RG), *

by Lemma 3.3(iii). Hence RG. J ( R H ) J ( R H ) - R Gand the required assertion follows. (iii) =$ (i): It is clear from (iii) that R G . J ( R H ) is an ideal of RG. Let V be an irreducible RH-module and let I = ann(V). Then J ( R H ) 5 I and so RG.J(RH)c R G . 1

General properties of induced modules

72

Applying Theorem 3.4(i), we conclude that

J ( R H ) C RG - J ( R H ) C I d ( R G - I ) = a n n ( V G ) Since R H / J ( R H ) is semisimple artinian, the latter implies that (V')H is completely reducible. The final assertion follows by induction on n, using (iii). 3.7. Theorem. Let H be a subgroup o f a group G and let R be a

commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) For any irreducible RH-module V , the RG-module V G is completely reducible. (ii) J ( RG) RG J ( R H ) . (ii) J ( R G ) C J ( R H ) - RG. Proof.

That (ii) is equivalent to (iii) is a consequence of Lemma

3.5. (i) =+(ii): Put V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is a completely reducible RH-module, hence by hypothesis V G is completely reducible. Therefore J ( R G ) V G = 0. But, by Proposition 2.ll(i),

VG RG/RG. J ( R H ) and thus J ( R G ) R G - J ( R H ) . (ii) + (i): Let W be any irreducible RH-module and let V be as in (i). Since W is irreducible, W S R H / M for some maximal left ideal A4 of R H . Hence W C V/U where U = M / J ( R H ) . Since W G VG/UG (Theorem 2.9(iii)), we are left to verify that V Gis completely reducible. But J ( R G ) C_ RG J ( R M ) implies that J ( R G ) V G = 0. Since by assumption R G / J ( R G ) is artinian, we conclude that V Gis completely reducible, as required. 3.8. Lemma. Let H be a subgroup of G. If for any irreducible RG-module V , the RH-module VHis completely reducible, then J ( R H )

J ( R G ) . The converse is true under the additional assumption that R H / J ( R H ) is artinian.

3. Annihilators of induced modules

73

Proof. Let I be any maximal left ideal of RG and let V = R G / I . By assumption VHis completely reducible. Hence J ( R H ) V = 0 and so J ( R H ) I . Thus J ( R H ) 5 J ( R G ) , as required. Conversely, suppose J ( R G ) and that R H / J ( R H ) is artinian. Let V be that J ( R H ) any irreducible RG-module. Since J ( R G ) V = 0, we must also have J ( R H ) V = 0. But R H / J ( R H ) is artinian, hence VH is completely

s

reducible.

Let H be a subgroup o f a group G and let R be a commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) J ( R G ) = RG' - J ( R H ) . (ii) J ( RG) = J ( R H ) - RG. (iii) For any irreducible RH-module V , V G is completely reducible, and for any irreducible RG-module M , MH is completely reducible. Furthermore, under the stronger assumption that G is finite and R is artinian, each o f t h e above conditions is equivalent to (iv) For any irreducible RG-module M , (MH)' is completely reducible. 3.9. Theorem.

Proof. The equivalence of (i), (ii) and (iii) follows from Theorem 3.7 and Lemma 3.8. Furthermore, it is clear that (iii) implies (iv). Assume that G is finite and R is artinian. Then both RG and R H are artinian and hence so are R H / J ( R H ) and R G / J ( R G ) . Let V be an irreducible RH-module. Then V Gis a finitely generated module over an artinian ring RG, hence V Gis artinian. It follows, from Corollary 1.3.2, that V Ghas an irreducible submodule, say M . Then, by Corollary 2.7,

and therefore there is an exact sequence

By Theorem 2.9, the latter gives rise to the exact sequence

General properties of induced modules

74

Hence (iv) implies that V G is completely reducible. Finally, let M be any irreducible RG-module. Then (iv) implies ( M H )is~ completely reducible, hence MH is completely reducible by Corollary 2.10. 4.

Clifford's theorem

Let N be a normal subgroup of an arbitrary group G and let R be a commutative ring. Given an RN-module V and g E G, let gV be the RN-module whose underlying R-module is V and on which N acts accordingly to the rule

n * v = (g-lng)~

(v E V )

where n * v denotes the module operation in gV and nw the operation in V. Such an RN-module gV is called a conjugate of V. 4.1. Lemma. Let V be an RN-module, let U be an RG-module and let g E G. (i) g 8 V is an RN-submodule of(VC)N and g @ V 2 gV. (ii) V is irreducible (indecomposable, completely reducible) if and only if so is gV. (iii) If W is an RN-submodule of UN, then gW is an RN-submodule of UN a n d g W E

gw.

(iv) (")G E V G . (v) 'V E V and "(YV) "YV for all x, y E G, i.e. G acts on the isomorphism classes of RN-modules by conjunction. Proof. It is clear that g @ V is an RN-submodule of ( V G ) ~ . Hence, by Lemma 1.5, we need only verify (iii). It is obvious that gW is an R-submodule of UN. If w E W and n E N, then n(gw) = g(g-'ng)w E gW.Thus gW is an RN-submodule of U N . Finally, the map gW + gW,w H gw is obviously an RN-isomorphism.

Let

V be an RN-module and let H = (9 E GIV 2 W}

4. Clifford's theorem

75

Then, by Lemma 4.l(v), H is a subgroup of G. We shall refer to H as the inertia group of V ;in case H = G we shall say that V is G-invariant. We are now ready to establish the following classical result. 4.2. Theorem.

(Cliflord (1937)). Let N be a normal subgroup of an arbitrary group G, let R be any commutative ring and let U be an irreducible RG-module. If GIN is infinite, assume that UN is artinian. Then (i) UN is completely redzlcible and, in particular, UN contains as irreducible submodule, say V . (ii) If H is the inertia group of V , then (G : H ) < 0;) and there exists a positive integer e such that UN g e

(@tErtv)

where T is a left transversal for H in G and { t V ( t E T } are all nonisomorphic conjugates of V . (iii) If W is the sum of all submodules of UN isomorphic t o V , then W is an irreducible RH-module such that

W N ~ and ~ V U g WG (Here, of course, eV denotes the direct sum of e copies of V ) .

Proof. If G I N is infinite, then by Corollary 1.3.2, U N contains an irreducible submodule V . Hence UN is completely reducible, since U = CsE~gV and each gV is an irreducible RN-module (Lemma 4.1). We may therefore assume that GIN is finite, say GIN = { g l N , . . . , g k N } . Because U is irreducible,

U = RGu = RN(g1u) +

* * *

+ RN(9kt.d)

for any 0 # u E U . Thus U N is a finitely generated RN-module. Applying Proposition 1.3.4, we conclude that UN has a maximal submodule, say M . We now set

MO

= ngEGgM

Owing to Lemma 4.l(iii), each gM is an RN-submodule of U N . Moreover, if g M M' c U for some RN-submodule M' of U , then

A4 5 g - W '

c 9-'u

76

General properties of induced modules

Thus M = g-'M', so M' = gM and therefore gM is a maximal RNsubmodule of U N . Hence each uN/gM is an irreducible RN-module. If g E G, then g = gin for some i E (1,. . . ,k}, n E N and gM = giM. Therefore k Mo = n;=,g;M

But Mo is an RG-submodule of U with Mo C M # U , hence Mo = 0. It follows that there is an injective homomorphism

and consequently UN is completely reducible, as required. (ii) and (iii) . Because CgEGgV is a nonzero RG-submodule of U , we have

u=cgv g€G

where, by Lemma 4.1(ii), (iii), each gV is an irreducible RN-module isomorphic to g V . If GIN is infinite, then by hypothesis there are only finitely many direct summands gV appearing in U N . But every conjugate gV of V is a direct summand of U , hence there are only finitely many nonisomorphic conjugates of V. This shows that (G : H ) = n < 00 and that if { 1 = 91,. . . ,g,} is a left transversal for H in G, then glV,. . . ,gnV are all nonisomorphic conjugates of V . Let W;be the sum of all submodules of U N isomorphic to giV (hence W = W l ) ,1 5 i 5 n. Then, by Proposition 1.3.19(ii),

Hence, to prove (ii), we are left to verify that each W; contains the same number of irreducible direct summands. Now for each g E G, gU = U and the modules gWi, gWj, i # j , have no composition factor in common. Thus W, = giW which proves (ii). It is clear that W is an RH-module and, by the above,

U = W @ g2W

@

* *.

@ g,W

(direct sum of R-modules)

Hence { W,g2W, . . . ,gnW} is a system of imprimitivity for U on which G acts transitively. Because H is the stabilizer of W , it follows from

4. Clifford’s theorem

77

Proposition 1.4(ii) that U Z W G . Finally, assume that X # 0 is an RH-submodule of W . Then

U‘ = x f3 gzx f3 *

*

f3 gnx# 0

is an RG-submodule of U . Therefore U = U’and so X = W , proving that W is irreducible. This completes the proof of the theorem.

4.3. Corollary. Let N be a normal subgroup of G of finite index and let R be any commutative ring. Then J ( R N ) J ( R G ) .

Proof. Apply Theorem 4.2 and Lemma 3.8. Let N be a normal subgroup of a finite group G, let F be a field and let U be an FG-module, dimFU < 00, such that U = CgEGgVfor some irreducible submodule V of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . (i) UN Z e ( & T t V ) , where e is a positive integer, T is a left transversal for H in G and {“It E T } are all nonisomorphic conjugates of V . (ii) W is an FH-module such that WN 2 eV and U S W G . (iii) U is irreducible if and only if W is irreducible. 4.4. Proposition.

Proof. Except for the irreducibility of W , properties (ii) and (iii) of Theorem 4.2 were deduced from the fact that U = x g E G g V . Hence the properties (i) and (ii) of the present proposition also hold. By the foregoing, we are left to verify (iii). If U is irreducible) then W is irreducible by Theorem 4.2(iii). Conversely, suppose that the FH-module W is irreducible. Let X be an irreducible submodule of U and let Vo denote an irreducible submodule of X N . Because VO is a n irreducible submodule of U N and UN is the sum of irreducible FN-modules gV, g E G, it follows that V, 2 gV

Z

for some g E G

gV

Now X = CgEGgVO,since X is irreducible. On the other hand,

g-lvo2

9-1

v

0 -N 9-1

(”)

v

General properties of induced modules

78

by virtue of Lemma 1.5. Hence, replacing & by g-lVO, we may assume that V E Vo, in which case H is also the inertia group of Vo, Let Wo be the sum of all submodules of X , isomorphic to Vo. By Theorem 4.2(iii), Wois an FH-module such that X E W:. Now Wo is a nonzero submodule of W , so W = Wo, by the irreducibility of W . Thus, by (ii), we have

x 5% w,"= WG5%u Hence U is irreducible and the result follows.

Let N be a normal subgroup of a finite group G, let E I F be a field extension and let U and V be, respectively, an irreducible FG-module, dimFU < 00, and an irreducible FN-submodule of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . If VE is an irreducible EN-module, then (i) UN E e ( e t 'E V )~where e is a positive integer, T a left transversal fro H in G and {'V(t E T } are all nonisomorphic conjugates of V . (ii) (UE)N e ( e t 'E (VE ~) ) and {'(VE)JtE T } are all nonisomorphic conjugates of VE. (iii) W is an irreducible FH-module and W E is an EH-module such that ( W E ) N 2 eVE , UE (WE)G 4.5. Corollary.

(iv) U E is irreducible if and only if WE is irreducible. In particular, if V and W are absolutely irreducible, then so is U . Proof. Because U = CgEGgVand

it follows that

By Deuring-Noether theorem (see Huppert and Blackburn (1982)), for any 2, y E G , "V E u V if and only if ("V), E (YV),. Because

5. Dual and contragredient modules

79

g ( V ~2 ) ( ~ V ) E(Lemma 1.5(v)), it follows that H is the inertia group of VE. Bearing in mind that W = & H h V , we also have WE = C h E H h Vand ~ the latter is, of course, the sum of all submodules of (UE)N isomorphic to VE. Applying Theorem 4.2 and Proposition 4.4, we conclude therefore that (i)-(iv) hold except possibly for the value of e in (ii). The correctness of value of e being a consequence of UN Z e ( B L E'rV ) ,the equality (UE)N= ( U N ) Eand the isomorphismg(VE) (gV)E, the result follows. 5.

D u a l a n d contragredient modules

Throughout this section, R is a commutative ring, G a finite group and A is an R-algebra. By convention, an A-module means a left A-module. Let V be an A-module. Then the dual $' of V is a right A-module whose underlying additive group is H o r n ~ ( v , R )with the action of A given by (fa)(.)

= f(m) for all

f E H o r n ~ ( V , R ) , Ea A , v E V

Similarly, if V is a right A-module, then the dual $' of V is an A-module whose underlying additive group is H o r n ~ ( VR , ) with the action of A given by (af)(v) = f ( v a )

for all

f E H o r n ~ ( VR, ) ,a E A, v E V

If V, W are A-modules and f E H o r n ~ ( VW , ) ,then is defined by ( j + ) ( v ) = +(f(v>> for all

vE

v,$E r/i/

f* E H o r n ~ ( l / ic) /, (1)

If V , W are right A-modules and f E HomA(V, W ) ,then f^ E H o ~ A (9 ~ ) , is defined analogously. Let U be an A-module. If V is a submodule of U , define

vL= {$ E fil+(V)= 0) Then V L is obviously a right submodule of the right A-module is also clear that if V, C V, then V' 2

K ' .

fi.

It

General properties of induced modules

80

5.1. Lemma. Let Then

U ,V , W be A-modules such that U = V 63 W .

o=W'@V'

and W L g Q ,

vl z2

w

Proof. It is clear that W' n V' = 0. Given f E 0,we may write f = f1 f 2 where f1, f 2 E 0 satisfy f l ( W )= f2(V) = 0. Thus 0 = W' @ V*. Finally, if f E W', then f H f l V is obviously a n isomorphism of W' onto p. Similarly, if f E V I , then f H flW is a n isomorphism of V* onto

+

w.

5.2. Lemma. Let V and W be A-modules. (i) Iff : V -+ W is a surjective A-homomorphism, then f : W -+ 6' is an injective A-homomorphism. (ii) The map V -+ v H v*, where v*($) = $(v) for all 1c, E P, is an A-homomorphism.

(vr,

Proof. (i) This is a direct consequence of the definition o f f given by (1). and (ii) The given map is obviously additive. If ZI E V , It, E a E A, then

as required. In what follows, given a n A-module V , we write VRfor V regarded as an R-module. Assume that VRis free with a finite R-basis vl, . . . ,v,,. Then P is free with a basis 61,.. . ,6, given by 6;(vj) = &j. We shall refer t o C 1 , . . . ,Gn as the dual basis of 01,. . . ,v,. 5.3. Lemma. Let V be an A-module such that VR is free with a finite R-basis vl,. . . ,v,,, and let G1,.. . ,G, be the dual basis of v1,. . . , v,. If a E A and av; = Cj"= rjivj , with rj; E R, then 6 j a = r$j.

cjn=l

81

5. Dual and contragredient modules

Proof. For all i, k E { 1 , 2 , . . . ,n}, we have

Thus 6;a =

cy=lr;j6j, as required.

5.4. Lemma. Let V and W be A-modules such that VR and WR are finitely generated free R-modules. (i) The map V --$ v H v*, where v*($) = $(v) for all $ E $' is an A-isomorphism.

($'I,

A

(ii) I f f E HornA(V, W ) ,then j ( v * ) = f(v)* for all v E V . Furthermore, i f f is an isomorphism of V onto W , then f is an isomorphism o j w onto

Q.

Proof. (i) By Lemma 5.2(ii), the given map is an A-homomorphism. Let q , . . . ,v, be an R-basis of V and let Gl,. . . ,f i n be a dual basis of V. Then vT(Gj) = Gj(vi) = 6jj which shows that TI;, . . . ,v;t is an R-basis of

c. Thus the given map is an isomorphism.

(ii) For any given v E V and ?j E W , we have

proving the first assertion. Assume that f is an isomorphism of V onto W . If vl, . .. ,v, is an R-basis of V , then maps the dual basis f ( v l ) ,. . . ,f(vn) onto 61,. . . ,Gn. Thus jis an isomorphism of r/i/ onto h

1

h

v. rn

5.5. Lemma. Let V be a right A-module such that V is R-free. Then ann(V) = ann(C/)

+

Proof. Assume that z E ann(V)and let be an arbitrary element of Then, for all v E V , (z$)(v) = vz) = $(O) = 0, so z E ann(P). Conversely, suppose that z E ann(V). Then, for all $ E and all v E v,

c.

$1

$(4 = (.$)(.I

Q

=0

General properties of induced modules

82

Since V is R-free, we conclude that vz = 0 for all v E V . Hence z E ann(V),as required. Turning our attention to induced modules over group algebras, we now prove 5.6. Proposition. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

(vGj 2 (QG Proof. Let T be a left transversal for H in G. By Corollary 1.2(j), a typical element of V G can be uniquely written in the form

Ct8vt

E V)

(Vt

tET

with finitely many ut # 0. Because T-l is a right transversal for H in G, each element of (p)G can be uniquely written in the form

c

$t

8 t-l

($t

E

t€T

with finitely many

gt # 0. Consider the map 8 : (P)" -+(vGj

defined by

Then 8 is obviously an R-homomorphism. If 8 (CtET $t 8 t-') = 0, then $t(vt) = 0 for all t E T , vt E V . Hence each $t = 0 and so 8 is injective. If X E (V"], then

v,

hence t9(C At 8 t - l ) = X where Xt(vt) = X(t 8 vt). But each At is in and so 8 is surjective. We are therefore left to verify that for z = $ @ t - l ,

5. Dual and contragredient modules

83

Thus O ( z g ) = O ( z ) g and the result follows. Let V be an RG-module. An element v E V is said to be Ginvariant (or simply an invariant element) if g v = v for all g E G. In what follows Inv(V) denotes the set of invariant elements of V . It is clear that Inv(V) is the largest R-submodule of V on which G acts trivially. 5.7. Lemma. Let V and W be RG-modules. Then the RGmodule H o ~ R ( V W, ) becomes an RG-module under the following action of G:

(sf)(.) = s ( f ( 9 - W

(9 E G, ZI E

v,f E H O ~ R ( Vw) ,

Fu rt h e rmo re, Inv( HomR( V,W ) )= H O ~ R GV,(W ) Proof. The first statement is obvious. To prove the second, note that f E Inv(Horn~(V,W) if and only if f ( v ) = ( g f ) ( v ) = g ( f ( g - ' v ) ) for all v E V , g E G. The last condition is equivalent to f E H o ~ R G ( V W, ) . The special case of the RG-module HornR(V,W ) ,in which W = R and g r = r for all r E R, g E G deserves a separate mention. In this case, HomR(V,R ) = as an R-module and the action of G on is given by

c

( g f ) ( v >= f ( g - l v )

(v E V , g E G,f E

c)

General properties of induced modules

04

We shall denote this module by V*and refer to V' as the contragredient of

v.

5.8. Lemma. Let V be an RG-module which is R-fiee on afinite basis, say vl,. . . , v n and let r : G + G L n ( R ) be the matrix representation of G agorded by V with respect to v1,. . .,vn. Then, with respect to the R-basis $ 1 , . . . ,V;, of V* the matrix representation of G agorded b y V*is given b y g Ht r(g-*) (9 E G) where T ( g - ' ) is the transpose of q g - ' ) .

Proof. Since the left action of g on l;i in V* is the same as the right action of g-' on .L;i in P, the result follows by virtue of Lemma

5.3. Our next aim is to provide an analog of Proposition 5.6 for contragredient modules. This will be achieved as an easy consequence of the following general result.

.

5.9. Proposition Let H be a subgroup of an arbitrary group G, let V be an RG-module and let W be an RH-module. Then (i) om^( W,v)]' HornR( wG,V ) (ii) [ H O ~ ~ ( w)]" V ~ , 2 H O ~ R ( V w") ,

Proof. Let T be a left transversal for H in G. (i) A typical element f E [ H o ~ R ( W VH)]" , can be uniquely written in the form f=Ct@ft

(ft

E H o m ( W 7 VH))

t€T

with finitely many ft # 0. Consider the map

w,V H ) I G + HomR( W G V, )

8 : [HOrnR( defined by

5. Dual and contragredient modules

85

(only finitely many wt # 0). Then 8 is clearly an R-homomorphism. If 6(f) = 0, then O ( f ) ( t 8 w)= t f t ( w ) = 0 for all t E T ,w E W . Hence v) each f t = 0 and therefore 0 is injective. If X E H o r n ~ ( W ~ ,then (&T t 8 w t ) = E t E T tXt(wt),where xt(wt) = x(t 8 wt).Since each At is in horn^( VH),we see that 8(f) = for f = &T t 8 At. Hence 6 is surjective. Fix g E G and, for each t E T , write g t = t'ht with ht E H and t I+ t' a permutation of 2'. Hence, if f = CtETt @ f t ,then g f = CtETt ' 8 htft. Moreover,

w,

and

proving (i). (ii) A typical element f E [ H o r n ~ ( WIG v ~ , can be uniquely written in the form

with finitely many

ft

# 0.

Consider the map

8 : [ H O r n R (V H , W)IG--+ H O r n R (

v,W C )

General properties of induced modules

86

defined by O(f>(v) =

ct 8

ft(t-'v)

tCT

Then 8 is obviously an R-homomorphism. If S(f) = 0, then ft(t-'v) = 0 for all t E T , v E V . Hence f t = 0 for all t E T , so f = 0 and thus 8 is injective. If X E H o r n ~ ( VW"), , then X(v) = & - t8 Xt(t-'v) for some At E H o r n ~ ( vW~ ), with finitely many X t # 0. Hence e(z) = X for z = CtET t 8 A t ) proving that 0 is surjective. Fix g E G and, for each t E T , write g t = t'ht with t H t' a permutation of 2'. Then, for all 'u E V , (9W))(.)

= ge(f)(g-''u) =

c

9t 8 ft(t-lg-lv)

t U

and

proving (ii). 5.10. Corollary. Let H be a subgroup of an arbitrary group G and let W be an RH-module. Then

(W*)"c (W")* Proof. This is a special case of Proposition 5.9(i) in which V = R and G acts trivially on V . 5.11. Corollary. The group ring RG is self-contragredient, that is, (RG)*E RG as left RG-modules. Proof. Let H = 1 and let W be the RH-module R on which H acts trivially. Then W" = RG 8~ R E RG and W* E W.Hence, by Corollary 5.10, (RG)*E (W')" Z W GE RG

87

6. Induction, restriction, and outer tensor products

as required. H

Let V be a finitely generated projective RGmodule. Then V* is also projective. 5.12. Corollary.

Proof. By hypothesis, RG @ . . . @ RG = V @ W for some RGmodule W . Hence, by Corollary 5.11, V* is a direct summand of R G @ . . . @ RG. Thus V* is projective. H 6.

Induction, restriction, and outer tensor products

The aim of this section is twofold: first, to relate the induction and restriction processes, and second, to prove that the operation of forming outer tensor products commutes with the operation of forming induced modules. Throughout, R denotes a commutative ring and G an arbitrary group. Let V and W be RG-modules. Then the R-module V @ R W is an RG-module, where the action of the elements g,g E G is defined by

(v E v , w E W )

g(v@w) =gv@gw

and then extended to V @ R W and RG by R-linearity. We shall refer to V @ R W as the (inner) tensor product of V and W . It is clear that V @ R W S' W @ R V. Suppose now that G1 and G2 are groups and that & and V, are RGl and RG2-modules, respectively. Then the outer tensor product V, # V2 of V, and V2 is an R(G1 x Gz)-module whose underlying R-module is V, @R V2 with the module operation given by ( z , y ) ( v l8 v2) = X v 1 @ yv2

(z

E Gl7y E

G27v1

E

K,v2

E

b)

and extended to R(Gl x G 2 )and V, @ R V2 by R-linearity. Another interpretation of the outer tensor product V, # V2 comes from the isomorphism RGI @ R RG2 R(G1 x G2). Indeed, V, @ R V, is an RG1 @ RG2-module by means of (U

8 b)(v1 @ ~

2 = ) U V ~8 bv2

(a E

RG1, b E RG2, ~1 E V , 7

~2

E V,)

Hence, if we identify RG1 @ RG2 with R(GI x G2), the module above is nothing else but the outer tensor product of V, and V2.

General properties of induced modules

88

6.1. Theorem. Let H be a subgroup ofG, let U be an RH-module and let V be an RG-module. Then

(U€3 vH)G Proof.

= U G €3 v

as RG-modules.

Let T be a left transversal for H in G. Then the map

f :UG x

v + (U€3RVH)G

given by

is balanced. Hence f determines a unique R-homomorphism

v

€3 G (U €3 VH)G €3 u ) €3 2, H t €3 ( u €3 t-lv)

{ UG (t

Similarly, the map ( U ~ V H -,)UG@V, ~ t@(u@t-lv)H (t@u)@vis an R-homomorphism which is inverse to f*. Thus f* is an R-isomorphism. Fix g E G and write gt = tlh with tl E T , h E H . Then

f*(g[(t€3 U ) €3 v]) = f * ( ( t l€3 hu) €3 gv) = t i €3 (hu 8 t;'gh) and

Hence f* is an RG-homomorphism, as required. 6.2. Theorem. Let Hi be a subgroup of a group G;, i = 1,2, and let V, be an RH;-module, i = 1,2. Then

Proof. We may identify R(H1 x H2) with R H I C ~ R R H and ~ R(G1 x G2) with RG1 € 3 RG2. ~ With these identifications, ( a €3 b ) ( q €3 vz) = av1 @ bv2 for all v1 E V1, v2 E V2, a E RH1, b E RH2 and a similar

7. Mackey theorems and their applications

formuls holds for the action of RGl €3 RG2 on result, we have to establish an isomorphism

89

KG1#bGz. To prove the

V = (RGi € 3 RG2) ~ € ~ ‘ R H ~ @ R(V,#v2) H~ (RGi @ R H ~h)# (RG2 @ R H ~VZ) = W of (RG1 €3 RG2)-modules. Let Ti, i = 1,2, be a left transversal for H;

in G;. Then, by Corollary 1.2,

and

kv = @ ( t l , t z ) ~ ~ l x€3~ Vi) z ( t €3 i ( t 2 8 V2)

(direct sum of R-modules)

f “tl €3 t2) €3 (01 €3 0 2 ) ] = (tl €3 01) 0 (t2 €3 v2) is obviously an R-isomorphism, we see that f induces an R-isomorphism f* : V --f W. We are thus left to verify that for any given 91 E GI, 92

E Gz,

To this end, write g;t; = sihi with

s;

E T,, hi E H , i = 1,2. Then

as required. W

7.

Mackey theorems and their applications

Throughout, unless explicitly stated otherwise) G denotes an arbitrary group and R a commutative ring. Our aim is to prove some classical

General properties of induced modules

90

results of Mackey and to provide some applications of them. In particular, we present a criterion for two induced modules to be disjoint and a criterion for the irreducibility of induced modules. The section ends with the proof that if V and W are irreducible RN-modules, N 4 G, then VG S W Gif and only if V Z gW for some g E G. In what follows, we put t &Ht-'nS = (tV)tHt-lnS

7.1. Theorem (Mackey Decomposition (Mackey (1958)). Let H and S be subgroups of G , let T be a full set of double coset representatives for ( S ,H ) in G and let V be an RH-module. Then

Proof. Let {g;Ii E I} be a left transversal for H in G. Then, by Corollary 1.2 and Lemma 1.5(i),

V G= @iefgi8 V

(direct sum of R-modules)

and gi €3 V

"V

(as R(g;Hg;l-modules)

(1)

Put X = {g; €3 Vli E I } . Then G and, in particular S , acts on X . Moreover, 9; €3 V and gj €3 V lie in the same S-orbit if and only if gi and gj belong to the same double (S,H)-coset. For each t E T , let Wt denote the sum of the gj €3 V for which gj E S t H . Then each Wt is an RS-module and

(vG)s= @tETWt Setting I4 to be the restriction of tV to R ( t H t - l to verify that Wt E (K)". Let J C I be such that

n S ) , we are thus left

Then S acts transitively on the set {gj@VIj E J } and under this action the stabilizer o f t €3 V is {S

E Slst 8

v = t Q V } = {S

E Slt-lst E H } = t H t - l n s

91

7. Mackey theorems and their applications

Hence, by (1) and Proposition 1.4(ii), Wt Z' (&)' as required. W

7.2. Theorem. (Mackey Tensor Product Theorem (Mackey (1958)). Let H and S be subgroups of G , let V be an RH-module and let W be an RS-module. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. Invoking Theorems 6.1 and 7.1, we have

On the other hand, by Theorem 6.1,

The desired conclusion is therefore a consequence of the transitivity of induction. In what follows, given an RG-module V , we write I n v ( V ) for the R-module of G-invariant elements of V .

7.3. Lemma. Let H be a subgroup of G of finite index, let T be a left transversal for H in G with 1 E T and let V be an RH-module. Then, the m a p I n v ( V ) + Inv(VG) &Tt

@ 'U

is an R-isomorphism.

Proof. It is clear that if w is H-invariant, then

&Tt

@

v is G-

invariant. Since the given map is obviously an injective R-homomorphism, we are left to verify that any G-invariant element of V Gis of the form

General properties of induced modules

92

CtETt 8 v for some v E I n v ( V ) . So assume that x E V Gis G-invariant and write

x=Ct@vt

(vt

EV)

t€T

Then the equalities t x = z and hx = x ( h E H ) imply, respectively, vt = v1 and hvl = vl, t E T , as required. W In what follows, for any RG-module V , V' denotes the contragredient of V .

7.4. Lemma. Let V be a finitely generated R-free RG-module and let W be an RG-module. Then

V*8 W

HomR(V, W )

as RG-modules

Proof. Let v l , . . . ,v, be an R-basis of V and let v;, . . . ,vz be the dual basis of V*. Define 1c, : V*8 W + H o r n ~ ( VW , ) by

[+(f €3 41(4= [f(v>lw

(v E

v,w E W,f E V*)

Then 1c, is obviously an R-homomorphism. If 1c, (CZ,vr €3 w;) = 0, then wj

C V8 ~

= {IC,

20;

1

}(~j) =

o

proving that 1c, is injective. If X E Horn~(V, W ) ,then X = II, (CZ, v t €3 X(v;) and thus 1c, is an R-isomorphism. Finally, fix g E G , v E V,w E W and f E V*. Then

7.5. Lemma. Let F be a field, let G be a finite group and let V and W be finitely generated FG-modules. Then

dirnFlnv(V* 8 W ) = i ( V , W )

7. Mackey theorems and their applications

Proof.

93

Applying Lemmas 5.7 and 7.4, we have

as required. W

As an easy application of the above, we now prove

7.6. Theorem. (Mackey Intertwining Number Theorem (Mackey (1958)). Let F be an arbitrary field, let H and S be subgroups of a finite group G and let V and W be finitely generated F H and FS-modules, respectively. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. We have

where the last equality follows from Lemma 7.5 and the obvious fact that '(V') = ('V)'. In what follows, all modules are assumed to be finitely generated over their ground rings. Let F be a field, let G be a finite group and let V and W be FG-modules. We say that V and W are disjoint if they have no composition factors in common.

General properties of induced modules

94

7.7. Lemma. Let F be a field, let G be a jnite group such that charF does not divide IGl and let V and W be FG-modules. Then V and W are disjoint if and only if i(V,W ) = 0. Proof. By Maschke's theorem, we may write V = CBVj and W = (4) and {Wk} are irreducible FG-modules. Hence

$Wk, where the

i(v,W ) = C i ( 4 ,Wk) j,k

Since y $?! Wk if and only if i ( y ,Wk) = 0 , we see that i(V)W ) = 0 if and only if vj $?! Wk for all j , k. But the latter is equivalent to the disjointness of V and W , hence the result. We are now ready to provide two applications of Theorem 7.6. The first gives a criterion for two induced modules to be disjoint and the second provides a criterion for the irreducibility of induced modules.

7.8. Theorem. Let G be a finite group, let F be a field with charF not dividing IGl and let H and S be subgroups of G. If V and W are F H and FS-modules, respectively, and T is a full set of double coset representatives for ( S ,H ) in G, then V G and W G are disjoint if and only if for all t E T , the F(tHt-' n S)-modules t q H t - l n S and WtHt-lnS are disjoint. Proof. Apply Lemma 7.7 and Theorem 7.6.

7.9. Corollary. Further to the hypotheses and notation of Theorem 7.8, assume that both V Gand W Gare irreducible. Then V G$?! W G i f and only if f o r all t E T , the F(tHt-' fl S)-modules t&Ht-lnS and WtHt-lnS are disjoint. Proof. Since V G and W Gare irreducible, they are disjoint if and only if V G W G .Now apply Theorem 7.8. 7.10. Theorem. Let H be a subgroup of afinite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an FH-module. Denote b y T a full set of double coset representatives for

7. Mackey theorems and their applications

95

( H ,H ) in G. Then V Gis irreducible if and only i f V is irreducible and for all t E T - H , the F(tHt-' n H)-modules t&Ht-'nH and &Ht-'nH are disjoint. Proof.

Applying Theorem 7.6 for W = V and S = H , we have

Also, by Lemma 2.12(iii), V Gis irreducible if and only if i(VG,V G )= 1. Since for t E H , t K H t - l n H S K H t - I n H (Lemma 1.5), it follows that V G is irreducible if and only if

and

i(V,V ) = 1 Now apply Lemmas 2.12(iii) and 7.7. H

A particular case where H 4 G deserves special attention and is recorded in the following: 7.11. Corollary. Let N be a normal subgroup of a finite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an irreducible FN-module. Then V G is irreducible if and only i f V g V for a l l g E G - N . Proof.

Direct consequence of Theorem 7.10.

Applying Theorem 7.10 to one-dimensional representation of H , we obtain

7.12. Corollary. Let H be a subgroup o f a finite group G , let F be an algebraically closed field with charF not dividing IGJand let p be a one-dimensional representation of pi. Then the induced representation pG is irreducible if and only iJ for each g E G - H , there exists x E gHg-' n H such that p ( z ) # ~ ( g - l z g ) .

General properties of induced modules

96

Proof. Let V be an FH-module which affords p . Then, for any g E G, the representation g p of gHg-' given by "(z) = p(g-'zg) is afforded by g V . Since, for any given g E G, the F(gH9-l n H ) modules '&Hg-InH and &Hg-lnH are irreducible, they are disjoint if and only if they are nonisomorphic. In terms of representations, the latter is equivalent to the existence of z E gHg-' nH such that p ( z ) # g p ( z ) = p(z-'gz). The desired conclusion is therefore a consequence of Theorem 7.10. W We close by presenting two results pertaining to an arbitrary group G and an arbitrary commutative ring R. The following observation will clear our path.

7.13. Lemma. Let S be an arbitmy ring, let V be an S-module such that V = @,EIK whem each V , is irreducible, and let W be an irredu'cible submodule of V . (i) W Z I.j. for some j E I . (ii) If all the are nonisomorphic, then W = I.j. for some j E I . (i) Let J C I be such that the 5 ,j E J are all representatives of the isomorphism classes of V,. For each j E J , denote by Wj the sum of all V; with V; E %. By Proposition 1.3.19(i), the Wj,j E J are all homogeneous cornponenets of V . Hence W Wj for some j E J . Since W is irreducible, W C Xl @ - @ X , for some finitely many S-modules X I , . . . ,X,, isomorphic to 5. Thus W E % by the Jordan-Holder theorem. (ii) If all the V; are nonisomorphic, then Wj = 5 and W C Since the result is established. the latter is possible only in the case W = 6 ,

Proof.

- -

K.

W

7.14. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V be an irreducible R N module. If gV 9 V for all g E G - N , then V G is irreducible.

Proof.

Let W be a nonzero submodule of

VG. Then

W N

is a

8. Counting induced modules and characters

97

submodule of ( V G )and ~ (VC)N

= $ t u t €3

v

(direct sum of RN-modules)

where T is a transversal for N in G. Hence ( V G ) N (and therefore W N ) is completely reducible. Let X be an irreducible submodule of W'. Then X is an irreducible submodule of ( V G )and, ~ by hypothesis and Lemma 1.5(i), the t €3 V,t E T , are nonisomorphic. Thus, by Lemma 7.13(iii), X = t o @ V for some to E T . For any g E G , g X Wwhich implies that t €3 V C W for all t E T . Thus Vc = W as we wished to show. W

7.15. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V and W be irreducible RN-modules. Then V GE W G i f and only i f V 2 gW for some g E G. Proof. If V 2 gW, then by Lemma 1.5(ii), V G (gW)GE WG. Conversely, assume that V G W G .If T is a transversal for N in G, then and (W'), = $t& 8 W (V')N = $t& €3 V Since (VG)' 2 (W C )and ~ 1€3 V is an isomorphic copy of V contained in (VG)', V is isomorphic to a submodule, say X of ( W c ) N . Since, by Lemma 7.13(i) and 1.5(i), X is isomorphic to t €3 W E tW for some t E T , the result follows. w 8.

Counting induced modules and characters

Throughout, N denotes a normal subgroup of a finite group G and F an arbitrary field of characteristic p 2 0. Our first aim is to find the number of nonisomorphic FG-modules induced from the irreducible FN-modules. We begin by recording some preliminary information. All FG-modules bellow are assumed to be finitely generated. Let V be an FG-module. We say that V is a permutation module if there exists a basis {vl,. . . ,vn} of V on which G acts as a permutation group.

8.1. Lemma. Let V be a permutation module and let B = {vl,. . . ,v,} be a basis of V on which G acts as a permutation group.

General properties of induced modules

98

(i) If B1,. . . ,B, are G-orbits of B and H; is the stabilizer of any given element in Bi,l 5 i 5 r then

V

E

$i=l(l~i)G and

r = dirnFInv(V)

(ii) v E V'. (iii) IfB' = {ui,.. . ,vk} is another basis o f V on which G acts as a permutation group, then B' and B have the same number of G-orbits. Moreover, ifcharF = 0 , then for each g E G, the number of elements of B fixed b y g is equal to the number of elements of B' fixed by g . Proof. (i) This is a direct consequence of Proposition 1.4 and Lemma 7.3. (ii) Apply Corollary 5.10 and (i). (iii) The first statement follows from (i). To prove the second, assume that churF = 0 and fix g E G. Denote by M and M' the matrices of the linear transformation g with respect to B and B', respectively. Then M and M' are similar and hence t r M = trM'. Because g permutes the elements of B and B', trM and trM' are, respectively, the numbers of elements of B and B' fixed by g. This completes the proof of the lemma. Given an FN-module V and g E G, denote by x and characters of N afforded by V and gV respectively. Thus

"(4= x(9-lx9)

for all

5

g x

the F -

EN

Let n be the least common multiple of the orders of the p'-elements of G and let e be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers p , taken modulo n, for which e H e p defines an automorphism of F ( E )over F . Recall that, by definition, two p'-elements a, b E G are F-conjugate if z-lbx = up for some x E G and some p E I,. Owing to Proposition 1.6.7, the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G. 8.2. Lemma. Let x be a p'-element of N and let T, and I(, be the F-conjugacy classes of G and N, respectively, with x E T and x E K .

8. Counting induced modules and characters

99

Then

Tz = UgEGKg-Izg Proof. Let m be the least common multiple of the orders of the $-elements in N , so that n = mk for some natural number k and S = E~ is a primitive m-th root of unity over F . Assume that s E T,. Then s = g - ' s p g for some g E G and some p E I,. If p = r ( m o d m ) , 0 5 r 5 m - 1, then s = g-ls'g. The automorphism E H cfi of F ( E ) over F induces the automorphism 6 H 6p = 6' of F(6) over F . Hence r E I, and s E Kg-lZg. Since every automorphism of F ( 6 ) over F extends to that of F ( E )over F , it follows that if two elemets of N are F-conjugate in N , then they are F-conjugate in G. This proves the opposite inclusion, as desired. 8.3. Lemma. Let V and W be irreducible FN-modules and let x v and xw be the characters afforded b y V and W , respectively. Then V G W G if and only if x v = gxw for some g E G.

Proof. This is a direct consequence of Theorem 7.15 and Proposition 1.6.3. W Let x be the character of an F-representation p of G. If the $-elements a , b E G are F-conjugate, then x ( u ) = x(b). 8.4. Lemma.

By hypothesis, s-'bs = a' for some x E G and some Proof. r E I,. Let u be the automorphism of F ( E )over F such that a(&)= E'. We may choose a nonsingular matrix M such that p ( u ) = M-'diag(c*',

where k = degp and

~ ( b=) x ( a r ) = as required. W

.. . ,E @ ~ ) M

. . ,cyk are positive integers. Thus + . . . + &*k' - a(&*]+ * . + E a r ) = o(x(a))= x ( a ) ~1,.

Let x be the character of an F-representation of G and let K be an F-conjugacy class. Owing to Lemma 8.4, we may define x ( K ) unambiguously by setting

x(10 = x ( g )

for

g EK

General properties of induced modules

100

Recall that, by Propositions 1.6.7 and 1.6.3, the number of irreducible F-characters of G is equal to the number of F-conjugacy classes of G. We are now ready to prove the following crucial result. 8.5. Theorem. Let xl,. . . ,xm and K1,. . . ,Km be the irreducible F-characters of G and the F-conjugacy classes of p'-elements of G, respectively. Let A be a group acting on {XI, . . . ,xm}and { Kl, . .. ,K m } such that

x;(Kj) = "x;("Kj)

for all

a E A , i , j E { l ) .. . ,rn}

Then the number of A-orbits of { X I , . . . ,Xm} is equal to the number of A-orbits of {KI,. . . ,K,}. Furthermore, if charF = 0 , then for each a E A, the number of elements of {XI,... ,xm}fixed b y a is equal to the number of elements of {Kl, . . . ,K,} fixed by a.

Proof. We denote by V and W the permutation FA-modules corresponding to the action of A on { X I , . . . ,x m ) and { K l , . . . Km} respectively. Let pv and pw be the matrix representations of A afforded by V and W with respect to the bases {xl,.. . ,x,} and {Kl,. . . ,Km}, respectively. Put M = (x;(Kj)), 1 5 i , j 5 rn. By Proposition 1.6.3, the characters X I , . .. ,Xm are F-linearly independent and hence M is nonsingular. Fix a E A and write pv(a) = (a;j),pw(a) = ( b ; j ) . Then, for any # xi. given j E (1,. . .,rn}, a;, = 1 if c x j = xi and a;, = 0 if Similarly, b;j = 1 if "Kj = K; and b;, = 0 if "Kj # K;. Put )

(cij) = pv(a)M

and

(dij) = M p w ( a )

k=l

where the last equality follows from the assumption that x;(Kj) = "xi( " K j ) . On the other hand,

8. Counting induced modules and characters

101

which shows that pv(a)M = M p w ( a ) for all a E A. Since M is nonsingular, we conclude that V 2 W . Thus, by Lemma 8.1(i), the number of A-orbits of { X I , . .. ,x,} is equal to the number of A-orbits of {KI,. . . ,K,}. Finally, assume that charF = 0 and let xv and x w be the characters of V and W . Since V W , we have x v ( a ) = Xw(a) for all a E A. But x v ( a ) is the number of elements of {xl,.. . ,x,} fixed by a , while Xw(a) is the number of elements of { K l , . . . ,K,} fixed by a , hence the result. We shall say that two F-characters a,,f?of N are G-conjugate if /? = ga for some g E G.

Let N be a normal subgroup of G and let F be an arbitrary field of characteristic p 2 0. (i) The number of G conjugacy classes of irreducible F-characters o f N is equal to the number of F-conjugacy classes of p'-elements of G contained in N . (ii) If charF = 0 or charF does not divide IN1 and F is algebraically closed, then for any g E G the number of irreducible F-characters x of N with g x = x is equal to the number of F-conjugacy classes I( of p'-elements of N such that gKg-' = I<. 8.6. Corollary.

Proof. (i) Let { X I , . . . ,xn} and { K l , . . . ,K,} be the sets of irreducible F-characters of N and F-conjugacy classes of p'-elements of N , respectively. Then G acts on { x l , .. . x,} and { K l , . . . ,I<,} by sending, for each g E G , x; to g x ; and I(; to gK; = 9Kig-l. Since, for all g E G , g x i ( W j ) = "i(gKjg-1) = Xj(Kj) it follows from Theorem 8.5 that the number of G-orbits of { X I , . . . ,xm} is equal to the number of G-orbits of { K l , .. . ,Km}. But the former number is equal to the number of G-conjugacy classes of irreducible F-characters of N , while the latter number, by Lemma 8.2, is equal to the number of F-conjugacy classes of p'-elements of G contained in N . Thus (i) is established. (ii) The case charF = 0 follows from (i) and Theorem 8.5. Assume that charF does not divide IN[ and that F is algebraically closed. Then

102

General properties of induced modules

as is well known (see Curtis and Reiner (1981, p.420)) that the number of xi with g x ; = x; is independent of the characteristic of F . Hence we may assume that charF = 0 and the result follows. W

8.7. Corollary. (Karpilovsky (1978)). Let N be a normal subgroup of a finite group G and let F be an arbitrary field of characteristic p 0. Then the number of nonisomorphic FG-modules induced from the irreducible FN-modules is equal to the number of F-conjugacy classes of p'-elements of G contained in N .

>

Proof. By Lemma 8.3, the number of G-conjugacy classes of irreducible F-characters of N is equal to the number of nonisomorphic FG-modules induced from the irreducible 3"-modules. The desired conclusion is therefore a consequence of Corollary 8.6(i). H The groups G satisfying the hypothesis of the theorem below are commonly known as Frobenius groups. 8.8. Theorem. Let F be an algebraically closed field of characteristic p 2 0 let N be a normal subgroup of G with charF not dividing IN[ and assume that CG(z) N for every 1 # x E N . Then (i) If V is a nontrivial irreducible FN-module, then V G is irre-

ducible. (ii) The number of irreducible FG-modules induced from the irreducible FN-moduEes is equal to the number of nonidentity conjugacy classes of G contained in N . Proof. (i) Let x be the character of N afforded by V . Owing to V. Theorem 7.14, it suffices to show that for any g E G - N , gV Hence, by Proposition 1.6.3, we need only show that for any g E G- N , 9% # x. Applying Corollary 8.6(ii), we are thus left to verify that for any nonidentity conjugacy class C of N , gCg-' # C. Let x E C and assume, by way of contradiction, that gxg-l E C. Then 9x9-l = nxn-' for some n E N . Hence g-'n E CG(Z) N and therefore g E N, a contradict ion. (ii) Let r be the number of conjugacy classes of G contained in N . By Corollary 8.7, there exist exactly r , say FG,. . . ,YG,of nonisomor-

9. Relative trace maps

103

phic FG-modules induced from the irreducible FN-modules. By (i), there are precisely r - 1 irreducible modules among the qG,1 5 i 5 r , hence the result.

We close by proving th following result. 8.9. Theorem. (Osima (1952)). Let H be a subgroup of G, let F be an algebraically closed field of characteristic 0 and let xl,.. . ,xT be all distinct irreducible F-characters of H . Then the number of linearly independent F-characters of G among the induced characters xf,.. . ,x,G is equal to the number of conjugacy classes of G which contain an element of H .

Proof. Let 91, g2, . . . ,gs be the representatives for the conjugacy classes of G containing an element of H and let rn be the dimension of the vector space spanned by x?, . . . ,x:. Replacing each gi by its conjugate, if necessary, we may assume that gi E H , 1 5 i 5 s. By Proposition 1.1.9, each x: takes value 0 on each conjugacy class of G which contains no element of H . This shows that rn = rankA, where A is the r x s matrix defined by

(i and j are row and column index, respectively). Setting B = ( x ; ( g J y l ) ) , 1 5 i 5 r , 1 5 j 5 s, it folloews from Proposition 1.1.10 that tAB = diag(n,, 122,. . .,n,) where n; = ICc(gi)l, 1 5 i 5 s, and t A is the transpose of A . Because tAB is nonsingular, the rank of A is equal to s. Thus rn = s and the result follows. 9.

Relative trace maps

Throughout, G denotes an arbitrary group, H a subgroup of G of finite index and R a commutative ring. Given an RG-module V , we write I n v ( V ) for the R-module of G-invariant elements of V . In particular,

General properties of induced modules

104

1nv(V') denotes the R-module of H-invariant elements of the module V. 9.1. Lemma. Let T be a left transversal for H in G and let V be an RG-module. Then the map

Trg : Inv(V')

--+

Inv(V)

given by

is an R-homomorphism which is independent of the choice of T . Proof. Assume that T' is another left transversal for H in G. Then, for each t E T , we may find ht E H such that T' = {th,]tE T } . Hence, for any v E Inw(V')

Ctv = C t h t v = C t'v tCf

t€T

t'ET'

proving that Trg is independent of the choice of T. If g E G , then is another left transversal for H in G, hence

gT

gTrg(v) = C ( g t ) v= Trg(w) tQ

and thus T r s ( v )E Inv(V). Since Trg is obviously an R-homomorphism, the result follows. H In what follows, we refer to Tr$ as the (interior) relative trace map. For the convenience of future reference , we first record some elementary properties of relative trace maps. 9.2. Lemma. Let S

c H be subgrozlps of G offinite index. Then

T r z = Trg 0 T r f . (ii) Tr:(Inv(Vs))c T r z ( I n w V ~ ) . (a)

Proof. (i) Let X and Y be left transversals for H in G and S in H , respectively. Then XY is a left transversal for S in G. Hence, for

9.

Relative trace maps

105

any v E Inv(Vs),

as required.

(ii) By (i), we have

as asserted. H 9.3. Lemma. Let H be a subgroup of G of finite index and let V be an RG-module. Then, for any given g E G, v E Inv(VH) if and only zf gv E Inv( &Hg-l ). Moreover

and, in particular,

Proof. The first statement follows from the fact that, for any h E H , v E V , hv = v if and only if (ghg-')(gv) = gv. Let T be a left transversal for H in G. Because gTg-' is a left transversal for 9Hg-' in G, we have

as required. H 9.4. Lemma. Let H be a subgroup of G offinite index and let V be an RH-module. Then the map T r g : I n v ( ( V G ) H )+ I n v ( V G ) restricts to an R-isomorphism of I n v ( l @ V )onto I n v ( V G ) . In particular, Trg is surjective.

106

General properties of induced modules

Proof. Let T be a left transversal for H in G. Then

The desired conclusion is therefore a consequence of Lemma 7.3, since I n w ( V ) 2 I n v ( l 8 V ) via v H 1 8 v.

Let H be a subgroup ofG offinite index and let an RG-module V be isomorphic to a direct summand o f ( V ~ ) Then ~ , the map T r ; : I ~ ~ ( v H+ ) I~V(V) 9.5. Lemma.

is surjective. Proof. Our assumption ensures that Inv(V ) / T r g ( I n v (V H )is ) isomorphic to a direct summand of Inv((VH)G)/Tr$(Inv( (V H ) ~ ) H Since, ). by Lemma 9.4, the latter is 0, the result follows. 9.6. Lemma. Let H , K be subgroups of a finite group G and let V be an RG-module. (i) For all v E I n u ( v K ) , T r g ( v )= CtETT r g t - l n H ( t v ) , where T as a set of all double coset representatives for ( H ,I() in G. (ii) Trg(Inu(VK)) CtET TTgt-l.H(Inv(V,Kt-lnH)).

Proof. (i) Fix t E T . It is clear that if 91,. . . ,g8 is a left transversal for tKt-l n H in H , then g l t , . . . ,g,t is a complete set of representatives of the left cosets of I( in H t K . Since t v E InvtKt-l(V) InqKt-lnH(V), it follows that

which clearly implies the assertion. (ii) This is a direct consequence of (i).

Of special interest will be the case where the relative trace map is applied to the RG-module V = H o r n ~ ( UW , ) where U and W are

10. Induction and relative projectivity

107

RG-modules. Then VH = H o ~ R ( U H , W Hand, ) by Lemma 5.7,

I ~ ~ ( V= H )H o ~ R H ( ~WHH, ) I n v ( v ) = HomRG(U,W) Thus, by Lemma 9.1,

Trg : HomRH(UH, W H )+ HOmRG(U,

w)

is an R-homomorphism. Note also that, by the definition of Trg and the action o f t E T on f E HomRH(UH, WH) V , we have

T r g ( f )= C t f t - ' t@

9.7. Lemma. Let G be an arbitrary group, let H be a subgroup of finite index and let U , V , and W be RG-modules. (i) For a n y f E HomRH(U, v ) and g E HomRG(V, W ) , W39.f)=9T4(f)

(ii) For any f E HomRG(U, V ) and g E HomRH(V,W), T r Z ( 9 f )= T G ( 9 ) f

(iii) Tr$(EndRH(U)) is an ideal of EndRG(V). Proof. (i) Let T be a left transversal for H in G. Then tgt-' = g for all t E T , hence T r g ( g f )=

C t(9f)t-l = g C t f t - l t€T

=gTrg(f)

t€T

(ii) This is proved by a similar argument. (iii) Apply (i) and (ii) for the case U = V = W . W 10.

Induction and relative projectivity

Throughout, G denotes an arbitrary group and R a commutative ring. Let H be a subgroup of G and let V be an RG-module. We say that V is H-projective if every exact sequence of RG-modules

O+U+W+V+O

108

General properties of induced modules

for which the associated sequence of RH-modules

splits is also a split exact sequence of RG-modules. The RG-module V is said to be H-injective if every exact sequence of RG-modules

for which 0

VH

WH -+

UH

--f

0

is a split sequence of RH-modules, is also a split sequence of RGmodules. Hence, if R is a field, then an RG-module is projective (respectively, injective) if and only if it is l-projective (respectively, l-injective). Our aim is to provide various characterizations of H projective modules and record a number of applications. AS a point of departure, we first establish the following two preliminary results.

10.1. Lemma. Let H be a subgroup ofG offinite index, let T be a lest transversal for H in G containing 1 and let V be an RG-module. Define f : (VH)G

v

by

f

Ct@vt =Ctvt (tET

(Vt

E V)

tET

Then f is a surjective RG-homomorphism such that Kerf is a direct summand of ( ( V H ) ~ ) H .

Proof. It is clear that f is a surjective R-homomorphism. To show that f is an RG-homomorphism, fix g E G. Then, for any t E T , gt = t’h, for some t’ E T and h, E H. Thus

109

10. Induction and relative projectivity

and therefore

)

= gf C t @ v t ET

(t

proving that f is a surjective RG-homomorphism. Let i : VH -+ (VH)' be the canonical injection. It is clear that ~(VH n )K e r f = 0. Because

C t ~ v t - 1 8Ctvt ~ ~ e r j , tET

(tET

we conclude that ((vH)')

= Icerf

i(vH)

as asserted.

10.2. Lemma. Let H be a subgroup o f G offinite index, let T be a left transversal for H in G and let V be an RG-module. The map

f : V -+ (VH)'

defined b y f(v) =

ct

@ t-lv

tET

is an injective RG-homomorphism which is i n ~ e ~ e ~ of~ ethen tchoice ofT and f ( V ) , is a direct summand of ((V","),.

Proof. Assume that T' is another left transversal for H in G. Then, for any given t E T , there exists ht E H and t' E T' such that t = t'ht. Thus

C t @ t-'v = C t'ht 8 htl(t')-'v tET

=

C t' 8 (t')-'v, t'ET'

t6T

proving that f is independent of the choice of T . If g E G , then

f (gv)

=

C t 63 t-'(gv) = 9 C ( g - ' t ) 8 (t-lg)v tET

=

sfW,

tQ

110

General properties of induced modules

proving that f is an RG-homomorphism (since f is clearly an Rhomomorphism). Bearing in mind that

(VH)'

= @tcTt (8 V H ,

f is an injection. Note further that, by Corollary 1.'2@),W' = @teT,tgHt@ V ' is an RH-submodule of ( ( V H ) " ) ~ We . claim that ((VH)')>,

= w'@f(V)H

Indeed, if f ( v ) = CtfTt @t-'v E f ( V )n W', then v = 0. On the other hand, if t' E T n H, then

ct

@ vt -

ct

@ t-lt'vt1 E

W',

tET

tfT

thus completing the proof. H We are now ready to prove the following result essentially due to Higman( 1954). 10.3. Theorem. Let G be an arbitrary group, let R be a commutative ring and let H be a subgroup ofG offinite index. Then, for any

RG-module V , the following conditions are equivalent: (i) V is H-projective. (ii) V is isomorphic to a direct summand of(VH)G. (iii) V is isomorphic to a direct summand of W G ,where W is an R H -module. (iv) E n d R ( V ) is H-projective. (v) T r g : EndRH(VH) 4 EndRG(V) is surjective. (vi) There exists $ E EndRH(VH)) such that T r s ( $ ) = 1. (vii) V is H-injective. Proof. (i)+(ii): By Lemma 10.1, there is an exact sequence of RG-modules

0 u (VH)' v 0 such that the associated sequence of RH-modules splits. By hypothesis, V is H-projective, so K e r f is a direct summand of ( V H ) ~ Thus . ( v H ) ~=

K e r f @ V'

111

10. Induction and relative projectivity

where V' 2 V . (ii) + (iii): Obvious. (iii) + (vi): Let us show first that any RG-module V of the form V = W G ,where W is an RH-module, satisfies (vi). To this end, denote by T a left transversal for H in G containing 1 and define 1c, : V + V bv

It is easily verified that 1c, is an RH-homomorphism. Moreover, for all w E W , we have

(t' €3 w)=

c t$(t-'t'

€3 w)= t' €3 w

(t' E T )

t€T

Thus II, satisfies (vi). Turning to the general case, we may harmlessly assume that

W G=

v Cf3 V'

(direct sum of RG-modules)

Let T : W G --+ V be the projection map and let II, satisfy (vi) with respect to W G . Then T 0 II, induces an RH-homomorphism V -+ V , and for all v E V , we have

since 7~ is an RG-homomorphism. The desired implication follows. (vi)+ (vii): Suppose that V is an RG-submodule of the RG-module U such that UH = VH@ W for some RH-submodule W . Let 7r : U,y + VH be the projection map and let 'p E HornRc(U,V) be defined by 'p = TrE(ll,o T ) , where II, E E ~ R H ( V H satisfies ) (vi). If v E V , then ~ ( v= ) v and hence U = V @ Ker'p.

(vii)+ (i): Assume that V is H-injective. Owing to Lemma 10.2, V is isomorphic to a direct summand of ( V H ) G . Applying implication (iii)+ (vi), there exists II, E E ~ R H ( V such H ) that TT$($)= 1. Now assume that f : U + V is a surjective RG-homomorphism for which

U = Kerf

Cf3

W

(direct sum of RH-modules)

112

General properties of induced modules

The restriction f 1 = f IW is an RH-isomorphism of W onto VH. Set cp = and put 8 = Trz(cp$f). Then 8 E EndRG(U) and

fc'

(f 0 e)(4

=

c

ftcp$ft-'u

t€T

= Ctfcp$t-lu tET

= C t $ t - ' f u = f(u) t€T

proving that U = 8 ( U ) 8(u) =

+ K e r f . If 8(u) E 8 ( U ) n K e r f , then

c

c tcp+

ft-lu =

t ET

tcp+t-'fu = 0

tET

Thus U = O(U) @ K e r f , proving (i). (ii)+ (iv): Since V is isomorphic to a direct summand of ( V H ) it~ , follows that EndR(V) is isomorphic to a direct summand of H o m R ( ( v ~ ) ~ , But, by Proposition 5.9(i),

H o m ~( (V H ) ~V, )

H o m ~VH, ( V H )=~(EndR(V ) H ) ~

hence EndR(V) is H-projective, by the equivalence of (i) and (ii). (iv)+ (v): This is a direct consequence of Lemma 9.5 and the implication (i)+ (ii). (v)+ (vi): Obvious. H We now record a number of consequences of the result above. 10.4. Corollary. Let H be a subgroup of G of finite index n such that n is a unit of R. Then any RG-module is H-projective.

Proof. Let V be an RG-module and let $ : V +(v) = n-lv Then

for all

+ E EndRH(VH) and for all v E V,

(E

t$t-l

1

2)

= 2)

2)

--+

E

V

V be defined by

10. Induction and relative projectivity

113

where T is a left transversal for H in G. This shows that TT$($)= 1 and the result follows by virtue of Theorem 10.3(vi). W 10.5. Corollary. Let H be a subgroup of G offinite index and let V be an RG-module which is R-free offinite rank. Then V is H projective if and only if

is surjective. Proof.

Owing to Lemma 7.4,

Now apply Theorem 10.3. 10.6. Corollary. Let G be an arbitrarygroup, let H be a subgroup of finite index and let V , W be RG-modules. Then, for any given X E H o m R c ( V , W ) ,the following are equivalent: (i) X E Tr$(HomRH(V, W ) ) . f (ii) Assume that U is an H-projective RG-module and that U -+ f WH + 0 splits. W + 0 is an exact sequence for which UH -+ Then there exists ?,h E HomRG(v with = f o $. (iii) Assume that U is an H-injective RG-module such that 0 -+ f f V U is an exact sequence for which 0 -+ VH -+ UH splits. Then there exists q!~E HomRG(U, with = $ o f .

u)

w)

Proof. We will demostrate that (i) is equvalent to (ii). The dual of our argument will easily establish the equivalence of (i) and (iii). (i)+ (ii): Since UH f, WH-+ 0 splits, there exists p E HomRH(W, such that f p = 1. By hypothesis, there exists Xo E HomRH(V, W )such that X = T r $ ( X o ) . Setting $ = Tr$(pXo) E H o w L R G ( V , U we ) , obtain

u)

f$ = T r g ( f&) = T r g ( X o )= A,

(by Lemma 9.6(i))

General properties of induced modules

114

as required.

(ii)+ (i): Since U is H-projective, it follows from Theorem 10.3(vi) that T r $ ( A,) = 1 for some E EndRH( U ) . Hence

Tr%f A,$>

= f T r Z (A,)$ = f$=A,

(by Lemma 9.6(i), (ii))

as required. We close by recording a number of useful properties of H-projective modules and some applications.

10.7. Theorem. Let S 5 H be subgroups of G offinite index and let V be an RG-module. (i) Let V = @y=lK, where each E is an RG-module. Then V is H-projective if and only if each K is H-projective. (ii) IfV is S-projective, then V is also H-projective. (iii) If V is H-projective and VH is S-projective, then V is Sprojective. (iv) For any RH-module V , the RG-module VG is H-projective. (v) If W is an S-projective RH-module, then any RG-module which is a direct summand of W G is S-projective.

Proof. (i) Suppose that V is H-projective. Then, by Theorem 10.3(vi), there exists $ E E ~ ~ R H ( Vsuch H ) that Tr$($) = 1. Let ~ i 1 ,5 i 5 n, be the projection of V onto T/: defined by the decomposition V = @K. Then K ~ O $E E n d R H ( T / : ) H and obviously Tr$(r;o$) = 1. Thus, by Theorem 10.3(vi), each V , is H-projective. Conversely, if each $; E E ~ ~ R H ( Eis )such H that Tr$($i) = 1, then t,b E EndRH(V'&) defined by $lK = $; obviously satisfies T r $ ( $ ) = 1. Thus V is H projective. (ii) Let 0 + U -+ W + V -+ 0 be an exact sequence of RG-modules such that the associated sequence 0 -+ U, t WH + VH + 0 splits. Then the sequence 0 + US -+ WS -+ VS -+ 0 also splits. Because V is S-projective, the sequence 0 t U ---f W + V + 0 splits and therefore V is H-projective. (iii) Owing to Theorem 10.3(ii), we have (VH)G = V' @ X and

10. Induction and relative projectivity

(Vs)H = V" @ Y , where V'

%

115

V and V" 2 VH.It follows that

( vs)G ( v H ) G

@

Y G= v' @ x @ Y G

and thus V is isomorphic to a direct summand of ( V S ) ~Now . apply Theorem 10.3(ii), (iv) This was established in the course of the proof of Theorem 10.3. (v) Owing to Theorem 10.3(ii), ( W S )=~W ' $ X , where W' E W ,so that (Ws)G % W G@ XG.By (iv), (WS)G is S-projective and therefore, by (i) any RG-module which is a direct summand of WGmust be Sprojective. H 10.8. Theorem. Let H be a subgroup of G of finite index n. (i) If V is a projective RG-module, then V H is a projective R H module. The converse is true if n is a unit of R. (ii) If V is an RH-module, then the induced module V G is projective if and only i f V is projective. (iii) If R as a field of characteristic p > 0 and P is a Sylow p subgroup of a finite group G, then an RG-module V is projective if and only i f Vp is projective.

Proof. (i) Let V be a projective RG-module. Then there exists a free RG-module X with X = V @ W for some RG-module W . Since X H = VHfBWHand X H is a free RH-module, we conclude that VHis projective. Conversely, assume that V is an RG-module such that VH is a projective RH-module and that n is a unit of R. Let

be an exact sequence of RG-modules. Then the associated sequence

splits, since VH is projective. By Corollary 10.4, V is H-projective, hence the sequence

O-+U+WtV+O also splits. Thus V is a projective RG-module. (ii) Let T be a left transversal for H in G. Then, by Corollary

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General properties of induced modules

1.2( ii) ,

(VG),=(l@V)$

c (tET,W

)

t@V

(direct sum of RH-modules)

where, of course, 1 8 V is identifiable with V . Suppose now that VGis projective. Then, by (i), ( V G ) is ~ also projective. Since V S 1 @ V is a direct summand of (VG),, we deduce that V is projective. Conversely, assume that V is projective. If V is a free RH-module, then the RG-isomorphism

implies that VG is a free RG-module. Thus V G is projective. In the general case, we have W 2 V @ U for some free RH-module W and some RH-module U . Then W GS VG$UGand hence V Gis projective. (iii) This is a direct consequence of (i).

A ring S is said to be local if S / J ( S ) is a division ring. In what follows, we write U(S)for the group of units of S. We next record certain properties of local rings, some of which will be required for subsequent investigations. 10.9. Lemma. For any ring S , the following conditions are equivalent: (i) S is local. (ii) S has a unique maximal left ideal.

(iii) J ( S ) = s - U ( S ) .

(iv) The set of nonunits of S is a left ideal.

Proof. (i)+(ii): Let I be a maximal left ideal of S. Then I / J ( S ) is a proper ideal of S / J ( S ) and thus I = J ( S ) . (ii)+(iii): Let J be a unique maximal left ideal of S. Since J ( S ) is the intersection of all maximal left ideals of S , we have J ( S ) = J , and so J is an ideal. The inclusion J ( S ) S - U(S)being obvious, suppose that x E S - U(S).If Sx # S , then Sx lies in a maximal left ideal of S , whence Sx C J ( S ) , so x E J ( S ) . On the other hand,

10. Induction and relative projectivity

117

if S x = S then yx = 1 for some y E S . Clearly y 4 J ( S ) , otherwise 1 = y E J ( S ) . Hence S y = S , so zy = 1 for some z E S and hence z = x. Thus z E U ( S ) ,a contradiction. (iii) (iv): 0bvious. (iv)+(i): Let 5 @ J(S)and let I = S - U ( S ) . If M is a maximal left ideal of S , then M C_ I # S , whence M = I = J ( S ) . Hence x is a unit and thus x J ( S ) is a unit of S / J ( S ) . Consequently, S / J ( S )is a division ring as required.

+

+

10.10. Corollary. Let S be a local ring. Then (i) The only idempotents o f S are 0 and 1. (ii) For any ideal I o f S , S / I is a local ring. Proof. (i) Let e # 1 be an idempotent of S. Then s is a nonunit and hence, by Lemma 10.9, e E J ( S ) . Thus, by Proposition 1.4.10, e = 0. (ii) Let J / I be a maximal left ideal of S / I . Then J is a maximal left ideal of S , hence J = J ( S ) by Lemma 10.9. Thus S / I has a unique maximal left ideal and, by Lemma 10.9, S / I is local. 1

We next provide necessary and sufficient conditions under which group algebras of finite groups are local rings. 10.11. Proposition. Let G # 1 be afinite group and let R be a commutative ring. Then RG is local if and only if R / J ( R ) is a field of prime characteristic p and G is a p-group. Proof. Assume that RG is local and put F = R / J ( R ) . Since both R and FG are homomorphic images of RG, they are local by Corollary lO.lO(ii). Now the augmentation ideal I ( G ) of FG is a maximal left ideal, hence I ( G ) = J ( F G ) is nilpotent. By Maschke’s theorem, we conclude that charF = p > 0 and pl /GI. If G has an element g of prime order q # p , then g - 1 E I < g > is not nilpotent by Maschke’s theorem. This is impossible since I < g >C_ I ( G ) , hence G is a p group. Conversely assume that G is a p-group and R/ J ( R ) is a field of chracteristic p . It suffices to show that R G / J ( R G ) R / J ( R ) . If

118

General properties of induced modules

J ( R ) = 0, then R is a field and I ( G ) has an R-basis { g - 111 # g E G } consisting of nilpotent elements. Hence I ( G ) is nilpotent by Theorem 1.5.12. But then J(RG) = I ( G ) and RG/J(RG)Z R, proving the case J ( R )= 0. Turning to the general case, we first note that J(R)G J ( R G ) by Corollary 4.3. Hence, by Corollary 1.4.4(ii), J ( R G / J ( R ) G )= J ( R G ) / J ( R ) G Since RG/J(R)G2 ( R / J ( R ) ) Git, follows from the special case above that RG/J(RG) R / J ( R ) ,as required. The following preliminary observation will allow us to apply the above result.

10.12. Lemma. Let S be a local ring. Then any finitely generated projective S-module is free. Proof. Let V be a finitely generated projective S-module, and let

o1,. . . ,o, be a minimal generating set for V . Assume that we are given

a relation

We claim that all such s; E J ( S ) . Indeed, if si 6 J ( S ) , then it is a unit in S , and we can solve for the corresponding oi as an S-linear sum of the vis. This, however, is impossible since otherwise the generator o; can be deleted. Thus all s; are in J ( S ) , as claimed. Let W be a free S-module freely generated by wl,. . . ,w, and map W onto V by sending w;to 0;. By the foregoing, the kernel of this map, say K ,is contained in J ( S ) W . By hypothesis, V is projective so that W = IC @ V’ with V’ V . Because K 2 W/V’ it must be finitely generated. Moreover, J ( S ) W = J ( S ) K $ J ( S ) V ’ and hence the inclusion I( C_ J ( S ) W forces I( = J ( S ) K . Thus I( = 0 by Nakayama’s lemma (Proposition 1.4.9) and W 2 V , as required. The following result is essentially due to Dickson (1907).

10. Induction and relative projectivity

119

10.13. Theorem. Let G be a finite group and let P be a Sylow p-subgroup of G. Assume that R is a local ring such that R/ J ( R ) is of characteristic p and that V is a projective RG-module which is R-free of finite rank n. Then [PI divides n.

Proof. By Theorem 10.8(i), Vp is a projective RP-module. Since Vp is finitely generated and, by Proposition 10.11, R P is local, it follows from Lemma 10.12, that Vp is free. Thus Vp is isomorphic to a direct sum of, say r , copies of R P . But each R P is R-free of rank lPl, hence n = IPIr, as required. We close by recording the following result. 10.14. Theorem. Let R be an arbitrary commutative ring and let N be a normal subgroup of G of finite index n. Then the following conditions are equivalent: (i) J ( R G ) = RG J ( R N ) . (ii) n is a unit of R. (iii) For any irreducible RN-module V , V G is completely reducible. *

Proof. The equivalence of (i) and (ii) follows from Theorems 3.9 and 4.2. (ii)=+(iii): Let T be a transversal for N in G. Then V G= etETt@V where, by Lemma 1.5(i),(iv), each t 8 V is an irreducible RN-module. Let Mr be an RG-submodule of V Gand let 0

+

w + V G+ V G / W+ 0

be the natural exact sequence. Then the sequence

0+

w, + ( V " ) ,

+(VG/W)N +0

splits, since ( V " ) , is a complete reducible RN-module. The desired assertion now follows by virtue of Corollary 10.4. (iii)=+(ii): Let I be any maximal ideal of R. It suffices to show that n = n - 1 I . To this end put V = R / I . Then the irreducible R-module V is also an irreducible RN-module under the trivial action of N . Furthermore, we obviously have

V G (R/I)(G/N)

120

General properties of induced modules

Since, by hypothesis, V Gis completely reducible, it follows that ( R / I ) ( G / N ) is semisimple. Thus, by Proposition 1.6.2, n $ I and the result is established. 11.

Unique decompositions

In this section, we shall collect a number of general results which will be used repeatedly in our subsequent investigations. Throughout, unless explicitly stated otherwise, R denotes an arbitrary ring.

11.1. Lemma. Let V be an indecomposable R-module which is both artinian and noetherian. Then every f E E n d R ( V ) is either a unit or nilpotent.

Proof. Let L, and X n be, respectively, the image and the kernel of f",n 2 1. We assert that

to a sufficient large n; if substantiated, it will follow that f is nilpotent, provided f is a nonunit, hence the result. Since V is both artinian and noetherian, there exists n 2 1 such that Ln = L2n and X n = X2n. Then, for any x E V , we may find y E V for which u(x)= u2(y),where u = f". Thus

Finally, given x E L, n X,, we have u(x)= 0 and x = u(y) for some y E V . Then y E X2,, = X,, whence x = u(y) = 0, proving (1). Let R be an arbitrary ring and let V # 0 be an R-module. We say that V is strongly indecomposable if EndR(V) is a local ring. 11.2. Lemma.

If V is strongly indecomposable, then V is inde-

composable. The converse is true if V is both artinian and noetherian.

Proof. Assume that V is strongly indecomposable and that V = V' $ W is a direct decomposition of V . If .n : V + V' is the projection

121

11. Unique decompositions

map, then x is an idempotent of the local ring E n d R ( V ) , so x = 0 or T = 1, by Corollary lO.lO(i). Hence either V' = 0 or V' = V , proving that V is indecomposable. Conversely, assume that V is indecomposable and both artinian and noetherian. By Lemma 10.9, it is enough to show that every nonunit f E E = EndR(V) is in J ( E ) . Let g be an arbitrary element of E . Then gf is a nonunit and hence is nilpotent, by Lemma 11.1. Thus 1 - gf is a unit of E , so by Proposition 1.4.10, f E J ( E ) as we wished to show. W The following result is Azumaya's generalization of a classical theorem of Krull and Schmidt. 11.3. Theorem.

Let R be a ring, let V be an R-module and let

where each of the R-modules K, Wj is strongly indecomposable. Then m = n and, after possibly reordering the Wj, 2 W; for all i E { l , .. . , m } .

Proof. We argue by induction on min{m,n}. If m = 1 or n = 1, then the result is obvious since V is indecomposable. Let ei : V 3 K and fj : V + Wj be the projection maps, 1 5 i 5 m, 1 5 j 5 n. Note that el = elfjel and that, by hypothesis, E n d ~ ( 6is) local. Thus e l f j e l is a unit of EndR(V1)for some j . By renumbering the Wj, we may therefore assume that e l f l e l is an automorphism of Vl = elV. Thus flV1 C W1 and the kernel of f1 on V, is 0. Our next aim is to show that

@z2K.

To this end, suppose that II: E flVl n Then el2 = 0 and 5 = f i e l y for some y E V . Consequently, e l f l e l y = e1II: = 0 and therefore e1y = 0, since elfie1 is an automorphism on Vl. Hence II: = f l e l y = 0 , proving that flV, n = 0. Now fix x E V . Then e l z E elV = elflelV and hence e l x =

@z2K

122

General properties of induced modules

elflelw for some w E 5

- flelw E

W . It follows that el(z

-

f i e l w) = 0 and

Hence m

5

= flelw

+ (a: - flelw) E f1V + C V , i=2

proving (2). Since flV, C W1, it follows from (2) that

Wl = flvl @ {Wl n @Z1W But W1 is indecomposable and flV,

# 0, so Wl = fl&

S

Vl and

v = Wl @ ( @ 3 3 Consequently,

@Z2K= V/W1 = @jn=2Wj

and therefore, by induction, m = n and after possible reordering the W j , V , S W j f o r a l l i E { 1 , ...,m}. Let V be an R-module. We say that V has the unique decomposition property if the following two properties hold: (i) V is a finite direct sum of indecomposable modules; (ii) If V = = @YzlWj, where each K, Wj is indecomposable, W; for all then m = n and, after possibly reordering the Wj, i E (1,. . . , m } .

@clK

11.4. Corollary. Let R be a noetherian ring such that each indecomposable R-module is strongly indecomposable. Then any finitely generated nonzero R-module has the unique decomposition property.

Proof. Because R is noetherian, it follows from Corollary 1.3.7 that any finitely generated R-module V # 0 is noetherian. Hence, by Proposition 1.3.11, V is a finite direct sum of indecomposable modules. The desired conclusion is therefore a consequence of Theorem 11.3. 11.5. Corollary. (Krull - Schmidt theorem). Let V # 0 be an Rmodule which is both artinian and noetherian. Then V has the unique decomposition property.

12. Projective covers

123

Proof. By Proposition 1.3.11, V is a finite direct sum of indecomposable modules. Since these indecomposable modules are again artinian and noetherian, it follows from Lemma 11.2 that they are strongly indecomposable. Now apply Theorem 11.3.

11.6. Corollary. Let R be an artinian ring. Then any finitely generated R-module V # 0 has the unique decomposition property. Proof. Because R is artinian, the module V is both artinian and noetherian, by virtue of Corollary 1.4.15. The desired conclusion is therefore a consequence of Corollary 11.5. 12.

Projective covers

Our aim in this section is twofold: first to provide some general information pertaining to projective covers of modules and second to record a result concerning projective covers of induced modules. In what follows, R denotes an arbitrary ring and all modules are assumed to be finitely generated left modules. We know that each Rmodule is a homomorphic image of a projective module. For some modules V a stronger assertion is possible. A homomorphism

f:V-+W of R-modules is said to be essential if for every proper submodule V' of V , f(V') # f ( V ) . We say that an R-module P is a projective cover of V in case P is projective and there is an essential epimorphism

P-tV 12.1. Lemma. (i) A homomorphism f : V + W of R-modules is essential if and only i f K e r f is superfluous. (ii) If K , . . . , V, are superfluous submodules of V , then so is Vi *. t (iii) If P; is a projective cover of V;, then P I @ - .$P, is a projective cover of I/, @ , . . @ V, . (iv) A projective cover of an irreducible module is indecomposable.

v,.

+

124

General properties of induced modules

Proof. (i) Assume that f : V + W is an essential homomorphism. If W is a submodule of V such that V = W K e r f , then f(V) = f(W) and so W cannot be a proper submodule of V . This shows that V = W and hence Kerf is a superfluous submodule of V . Conversely, assume that K e r f is superfluous. If V' is a submodule of V with f(V) = f (V'),then V = V' K e r f . Since K e r f is superfluous, we must have V = V' and therefore f is essential. (ii) Suppose that W is a submodule of V such that

+

+

+ +- + + +- + +

Then V1 (V2 - . V, W ) = V and, since & is superfluous, we - * V, W = V . Now apply induction on n. have V2 (iii) If fi : P; -+ is an essential epimorphism, then

is an epimorphism whose kernel @y=llcerfiis superfluous, by virtue of (ii). Hence, by (i), @ fi is essential, which implies that @Pi is a projective cover of @K. (iv) Suppose that $ : P + V is an essential epimorphism, where V is an irreducible R-module. If P = P' @ P", for some proper submodules P' and P", then $(PI) = $(P") = 0 since 0 is the only proper submodule of V . But then V = $(PI) $(PI') = 0, a contradiction. Hence P is indecomposable, as asserted. W

-+

One of the consequences of the following lemma is that if a module does have a projective cover, then it is unique up to isomorphism.

12.2. Lemma. Assume that an R-module V has a projective cover P and let M be an R-module such that V is a homomorphic image of M . (i) If M is projective, then P is isomorphic to a direct summand of

M. (ii) If M has a projective cover L , then P is isomorphic to a direct summand of L. (iii) If M is another projective cover of V , then it4 % P .

12. Projective covers

125

Proof. (i) Fix an epimorphism cp : M -+ V and an essential epimorphism $ : P + V. Since M is projective, there exists a homomorphism 8 : M + P such that cp = $ 0 8. Because $ is an essential epimorphism and $08 is an epimorphism, 8 is also an epimorphism. But P is projective, hence there is a splitting homomorphism f : P + M and so M = Irn f @ Ker8. Since f is injective, (i) is established. (ii) Assume that L is a projective cover of M and let f : L + M be an epimorphism. Then cpf : L + V is an epimorphism, hence by (i) P is isomorphic to a direct summand of L. (iii) Suppose that M is a projective cover of V and choose 'p in (i) to be essential. It will be demonstrated that 0 is also essential, from which will follow that M = Imf 2 P. Assume that M' is a proper submodule of M and put P' = 6(M'). Then

and so P'

# P.

Hence 8 is essential, as required. 1

From now on, we write P ( V ) for a projective cover of V . Owing to Lemma 12.2(iii), if P ( V ) exists, then it is unique up to isomorphism. 12.3. Theorem. Let R be an artinian ring and let V be an Rmodule. (i) V has a projective cover. (ii) A projective R-module W is a projective cover of V if and only

i f W / J ( R ) Ws V / J ( R ) V . In particular, (a) IfV is completely reducible, then V 2 P ( V ) / J ( R ) P ( V ) . (b) I f U and V are completely reducible R-modules, then U E V if and only i f P ( U )C P ( V ) . (iii) If V is projective and i f V / J ( R ) V= @L1K is a decomposition as a direct sum of irreducible R-modules, then V Z @Y."=,(K)is a decomposition as a direct sum of indecomposable R-modules. (iv) I f V and W are projective R-modules, then V S W i f and only

if V / J ( R ) V2 W / J ( R ) W Proof. (i) Let L be a projective module such that V = L / S for some submodule S of L. For any submodule W of S , let fw : L / W + L / S be

126

General properties of induced modules

the canonical homomorphism. Next choose W to be minimal in S such that fw is essential; such a submodule exists, since fs is essential and S is artinian. We claim that W is a direct summand of L ; if sustained, it will follow that L/W is a projective module and hence that L / W is a projective cover of V . Let W’ be a submodule of L minimai among those whose projection cp : W’ -+ L/W is surjective, and let n : L + L/W be the projection map. Since L is projective, there exists a homomorphism n’: L -+ W’ with ?r = cp o n’,and the minimality of W’ forces ?r’(L)= W’. Denote by L‘ the kernel of n’.The projection f~ factors onto

L/L’+ L/W

-+

L/S

and the two factors are essential. Because L’ is contained in W , the minimality of W implies that W = L’, i.e. that Q is an isomorphism. It follows that L = W @ W’, as required. (ii) Let W be a projective module, let f : W + V be an essential epimorphism and let 71 : V + V / J ( R ) Vbe the projection. Since n is essential, so is nf and thus Kernf E J(R)W (Lemma 12.l(i) and Proposition 1.4.5). On the other hand

and therefore J(R)W

K e r n f . Thus Kernf = J(R)W and so

W / J ( R ) W2 V / J ( R ) V Conversely, assume that W is a projective R-module such that the above isomorphism holds. Let f : V + V / J ( R ) V be the projection map and let g : W + V / J ( R ) Vbe the homomorphism induced by the isomorphism W / J ( R ) WE V / J ( R ) V . Since f is an essential epimorphism and since W is projective, there is an epimorphism g’ : W -+ V with fg‘ = g . Hence

Kerg’ C Kerg = J(R)W is a superfluous submodule of W . Thus W is a projective cover of V , as required.

127

12. Projective covers

(iii) By Propositions 1.4.5 and 1.4.8, J(R)V is a superfluous submodule of V . Since V is projective, it follows that P ( V / J ( R ) V )= V . Invoking Lemma 12.l(iii) and (iv), we deduce that

where each P( E) is indecomposable. (iv) If V E W , then obviously V / J ( R ) VZ W/J(R)W. Conversely, assume that V and W are projective R-modules such that V / J ( R ) VE W / J ( R ) W .By the Krull-Schmidt theorem, we may write

and

W / J ( R ) W= w163 . . * @ W, where each we have

rZ:, Wi is irreducible and

2 Wi, 1 5

i 5 n. Then, by (iii),

as required. H

Assume that R is an artinian ring and let n 2 1 be such that

where the U; are principle indecomposable R-modules. Owing to the Krull-Schmidt theorem, the Ui are uniquely determined up to isomorphism and the order in which they appear. Furthermore, the above decomposition determines a complete set {el,. . . ,e,} of primitive idempotents in R such that Vi = Re;. Except when J ( R ) = 0, the principal indecomposable R-modules form only a small subclass of all indecomposable R-modules. However, it is the class which plays a very important role due to the following result.

Let R be an artinian ring. (i) The following conditions are equivalent: (a) V is a projective cover of an irreducible R-module.

12.4. Theorem.

128

General properties of induced modules

(b) V is a projective indecomposable R-module. (c) V is a principal indecomposable R-module. (ii) If Rel, , ,. ,Re, are all nonisomorphic principal indecomposable R-modules, then R e l / J ( R ) e l , .. .,Re,/J(R)e, are all nonisomorphic irreducible R-modules. Proof. (i) The implications (a)+(b) and (c)+(a) are consequences of Lemma 12.l(iv) and Theorem 12.3(iii), respectively. Now assume that (b) holds. Since V is projective, there exists n 2 1 such that R" = V @I V', where R" denotes a direct sum of n copies of R. Now the indecomposable components of R" are principal indecomposable Rmodules and V is indecomposable. Hence (c) follows by virtue of the Krull-Schmidt theorem. (ii) Invoking (i), we may find irreducible R-modules Vi, . . . ,V m such that Rei = P ( x ) , in which case K E Rei/J(R)ei by Theorem 12.3(ii). Re, and so i = j . If V is any irreducible RIf then Rei module, then by (i), P ( V ) E P ( K ) for some i E (1,. . . ,n } . Thus, by Theorem 12.3(ii), V Z V , as required.

4,

We close by proving the following result.

12.5. Proposition. Let H be a subgroup of a finite group G, let F be a field and let V and W be FH and FG-modules, respectively. Then (i) P(VG)is isomorphic to a direct summand of P ( V ) G . (ii) P ( W H )is isomorphic to a direct summand of P ( W ) H . (iii) If H is normal, then for all g E G , P(") = " ( V ) . In particular, V is G-invariant if and only if P ( V ) is G-invariant.

Proof. (i) By the definition of P ( V ) ,there is an FH-epimorphism P ( V ) --+ V . Hence, by Theorem 2.9(i), there is an FG-epimorphism P(V)" V G . By Theorem 10.8@), P(V)Gis projective. Thus, by Lemma 12.2(i), P ( V G )is isomorphic to a direct summand of P(V)". (ii) Since there is a homomorphism of P ( W ) onto W and hence of P ( W ) H onto W H ,P(WH)is isomorphic to a direct summand of P ( P ( W ) H ) .Because P ( W ) is projective, so is P(W)Hby Theorem 10,8(i). Hence, by Lemma 12.2(i), P ( W H )is isomorphic to a direct

12. Projective covers

129

summand of P ( W ) H . (iii) Let cp : P ( V ) + V be an essential epimorphism. Then cp can also be regarded as an epimorphism " ( V ) + gV. Because W is a submodule of P ( V ) if and only if it is a submodule of " ( V ) , the latter epimorphism is essential. It therefore suffices to prove that if V is projective, then so is gV. So assume that V is projective and write FH @I - @I FH = V @ W for some FH-module W . Then

--

and, since obviously

g(

F H ) S F H , the result follows.

This Page Intentionally Left Blank

131

Chapter 3 Induction from normal subgroups In this chapter we provide a detailed information on the process of induction from normal subgroups. Our first major result (Theorem 2.1) reduces the study of V G to the case where V is G-invariant. Under this assumotion we then show that E ~ ~ R G is ( Va crossed ~) product of GIN over E n d ~ ~ ( vThis ) . will allow us to discover circumstances under which V G is totally indecomposable or absolutely indecomposable. As a consequence of our main results, we derive a classical theorem known as the Green Indecomposability Theorem. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to provide a criterion for V G to be homogeneous. Our next major result asserts that if G is a finite group and S a finite-dimensional semisimple algebra over a field F , then any F-algebra which is a crossed product of G over S is symmetric. As an application, it is shown that if V is a completely reducible FN-module ( N is a normal subgroup of G), then the F-algebra E n d r ; . ~ ( Vis~ ) symmetric. This fact will allow us to derive a number of important facts concerning induction from irreducible modules and their projective covers. The chapter ends with the discussion of the Loewy length of induced modules.

132

1.

Induction from normal subgroups

Complete noetherian local rings

Throughout this section, R denotes a commutative ring. We begin by recording the following piece of information. A metric space ( M , p ) is a set M together with a map

p:MxM-+IR such that for all z, y, z E M , (9 P ( Z , Y) = P(Y, 4 2 0. (ii) p ( z , y ) = 0 if and only if x = y. 6;;) P ( 5 , Z ) I P b , Y ) P ( Y , Z ) . A Cuuchy sequence in M is a sequence {z;} of elements of M such that for any e > 0, there exists a positive integer k such that

+

P(2",4

<E

m,n > k

for all

A sequence {xi} is said to converge to provided

lim;,,p(z,z;)

3:

E M , written limxi =

z,

=0

Assume that I is an ideal of a ring S such that

Then, for any distinct z,y E S, there exists a unique m 5 -y

1 0 such that

E I" - Im+l

where by convension Io = S. We define PI : S x S 3 IR by

It is then easily verified that (S,pl) is a metric space. We refer to S as being complete in the I-udic topology if each Cauchy sequence in (S,P I ) converges. Let { s i } be a sequence in S. The following properties are direct consequences of the definitions: (i) {s;} is Cauchy if and only if for any n such that i,j > m implies s; - sj E I".

> 0, there exists m > 0,

1. Complete noetherian local rings

133

(ii) limsi = s if and only if for any n > 0, there exists m > 0 such that i > m implies s; - s E I". (iii) If limsi = s and lims: = s', then lirn(si si) = s s' and

+-

+

lim(sis:) = ss'. 1.1. Lemma. Assume that R is a noetherian ring and that A is an R-algebra which is finitely generated as R-module. If I is an ideal of R such that I J ( A ) , then fl;=,I"A = 0. In particular, if R is a noetherian local ring with maximal ideal P = J ( R), then n =:, P" = 0.

Proof. Let L be an arbitrary ideal of R and let V be a finitely generated R-module. It is well known (see Bourbaki (1961)) that L W = W for W = n;=,L"V. Applying this fact for V = A , L = I and W = nr=,I"A, we conclude that IW = W and hence that J(A)W = W . Because W is a finitely generated A-module, it follows from Nakayama's lemma that W = 0. So the lemma is true. 1.2. Lemma. Let A be an R-algebra which is finitely generated as R-module. Then J(R)A C J ( A ) .

Proof. Let V be any irreducible A-module. Then V = Av for each nonzero v in V , hence V is a finitely generated R-module. Now J ( R)V is an A-submodule of V , hence is 0 or V . But J ( R ) V # V , otherwise by Nakayama's lemma we find that V = 0. Hence J ( R ) V = 0 and so J ( R ) annihilates every such V . Thus the same is true for J ( R ) A and so J ( R ) A J ( A ) .

c

1.3. Lemma. Let R be a local ring with maximal ideal P and residue class field F = R / P . Suppose that A is an R-algebra which is finitely generated as R-module and put A = A / P A . Then (i) J ( A ) = J ( A ) / P A and A / J ( A ) S A / J ( A ) as F-algebras. (ii) For any finite group G , R G / J ( R G ) F G / J ( F G ) .

c

Proof. (i) By Lemma 1.2, P A J ( A ) and so, by Corollary 1.4.4(ii), J ( A ) = J ( A ) / P A . The latter obviously implies the second assertion.

134

Induction from normal subgroups

(ii) Put A = RG and note that

A = R G / P ( R G ) ( R / P ) G= FG Now apply (i). W

Assume that R is a local noetherian ring with maximal ideal P . Owing to Lemma 1.1, n : y n = 0. We say that R is complete if R is complete in the P-adic topology. 1.4. Lemma. Let R be a complete noetherian local ring with maximal ideal P , and let A be an R-algebra which is finitely generated as R-module. Then, for any ideal J C J ( A ) of A , A is complete in the J-adic topology.

Proof. By Lemma 1.2, P A C J ( A ) ,while by Lemma 1.3(i), J" C_ J(A)" P A for some n 2 1. Thus we may assume that J = PA. Note that, by Lemma 1.1, the intersection of all (PA)" = P"A is zero. Write A = Ro1 . . . Ro,, the o; E A, and let { w ; } be a Cauchy sequence in A. Given an integer n > 0, denote by m(n) the largest integer such that for all i , j 2 n wi - wj E P"(")A

+ +

(if no such m ( n ) exists, set m(n) = n ) . Bearing in mind that {wi} is Cauchy, we have limn-,mm(n)= 00. Let

and write

We then have

135

1. Complete noetherian local rings

where

and t=n

so for any t , {bnt} is a Cauchy sequence in R. Since R is complete, limb,, = b, for some b, E R. Setting w = C;=l btvt, it follows that

thus completing the proof.

We have now accumulated all the information necessary to prove the following result. In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Z Se2 as left S-modules if and only if e l S 2 e2S as right S-modules (see Jacobson (1956)). 1.5. Theorem. Let R be a complete noetherian local ring, let A be an R-algebra which is finitely generated as R-module and let J C J ( A ) be an ideal of A . For each x E A, let 5 be the image o f x in A = A / J . (i) Each idempotent E E A can be lifted to an idempotent e E A, that is e = E . Moreover, e is primitive if and only if is primitive. (ii) If e l , e2 are idempotents of A, then Ael E Aez as left A-modules if and only if Ael 2 A Z 2 as left A-modules (equivalently, e l A e 2 A as right A-modules if and only i f e l A E e2A as right A-modules). (iii) If I = E~ e2 t.- . en is a decomposition of I into orthogonal idempotents in A, then there exists orthogonal idempotents e l , . . . ,en E A such that 1 = el . . e, and e; = ci

-+

+

+ +

Proof. (i) Given an idempotent e E A, choose x1 E A with 51 = E . Then nl = x: - x1 E J . Once x ; and n; are chosen, put x;+1 = x ;

+ n; - 2x;n;,

n;+l =

2

- xi+l

(1)

Induction from normal subgroups

136

Then n; E J2iand thus {xi} is a Cauchy sequence. By Lemma 1.4, A is complete in the J-adic topology, so limx; = e for some e E A. Then e2 - e = limn; = o and, moreover, e = 51 = c , so e is a required idempotent of A. Let e be an idempotent in A and let E = €1 6 2 , where €1 and ~2 are orthogonal idempotents in A. It will next be shown that there exist orthogonal idempotents el, e2 in A with

+

To this end, choose a E A such that si = €1 and put z1 = eae. Then 51 = E s i E = €1 so we may consider the Cauchy sequence {z;} given by (1). Since z1 E eAe, each x; E eAe and thus ex; = x; = x;e for all i. Accordingly, setting e; = Zimzi, we deduce that el is an idempotent of A such that E l = €1 and (by taking limits) eel = el = ele. Letting 2 = e2 # 0, 22 = €2: and e1e2 = e2e1 = 0, e2 = e - el, we now have e2 proving (2). If E is not primitive, then so is e, by virtue of (2). The converse being obvious, it follows that e is primitive if and only if so is 2. (ii) Assume that e1,e2 are idempotents in A. Any isomorphism Ael 2 Ae2 of left A-modules carries J e l onto Je2 and induces an isomorphism A E 1 A22 of left A-modules. Conversely, let f : /IE1 --+ Ac!~ be an isomorphism of left A-modules and let 3 = f-'. Because A = Aei @ A ( l - e,), each Ae; is a projective A-module. Thus we can find A-homomorphisms f : Ael -+ Ae2 and g : Ae2 -+ Ael lifting f and g,'respectively. We claim that 6 = gf is an automorphism of Ael,; if sustained, it will follow by symmetry, that fg is an automorphism of Ae2 and hence that f is an isomorphism. therefore (8 - 1)Ael C PAel. Setting It is obvious that 4 lifts p = 1 - 8, we have PkAel C Jkel. Thus 1 p t p2 t . . . is a well defined endomorphism of Ael, and is a two-sided inverse of 1 - p. Because 1 - p = 8, this demonstrates that 8 is an A-automorphism of Ael, as desired. (iii) The case n = 1 being obvious, we argue by induction on n. So assume that n > 1 and that the result holds for n-1. Put 6 = en-l +E,,

sf,

+

137

1. Complete noetherian local rings

so that 1 = ~1

+. . . + en-2 + 6 is an orthogonal decomposition. By the

induction hypothesis, we may find an orthogonal decomposition:

- 1 it follows that Applying (2) to the decomposition ?t = 6 = ~ ~ +c,, e= en for some orthogonal idempotents en-l, en in A such that En-1 = En-1 and = c,. Thus 1 = el+- - .+en is the desired orthogonal decomposition.

+

As an easy consequence of the above result, we derive 1.6. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then A is local i f and only if 1 is the only nonzero idempotent of A .

Proof. If A is local, then 0 and 1 are the only idempotents of A , by Corollary 2.10.10(i). Conversely, suppose that 1 is the only nonzero idempotent in A . Let P be the maximal ideal of R and put A = A / P A . By Lemma 1.3(i), A / J ( A ) A / J ( A ) and so it suffices to show that A is local. By Theorem 1.5, I is the only nonzero idempotent of A. Since A is a finite-dimensional algebra over a field, the result follows.

The next lemma will enable us to take full advantage of the results so far obtained.

1.7. Lemma. Let R be a noetherian ring and let V be a finitely generated R-module. Then the R-algebra E n d R ( V ) is a finitely generated R-module.

+ +

Proof. We may write V = Rvl Rv, for some vi E V . Let W be a free R-module with basis wl,. . . ,w, let 'p : W 3 V be the R-homomorphism such that 'p(w;) = v,, and put IC = Ker'p. Since E n d R ( W ) Z M,(R), E ~ ~ R (isWa )finitely generated R-module. Put Eo = {f E E n d R ( W ) J f ( K ) I(}. I f f E Eo, then f induces ) an S E E n d ~ ( vby -

f(.

+ IC) = f(.) + K

a

(z E

W)

138

Induction from normal subgroups

For any given g E E n d ~ ( v write ),

and define g E E n d ~ ( wby )

j=1

If Cj bjwj E I(, then Cj bjvj = 0 and

so

proving that g E Eo. It follows that g H g is a surjective R-homomorphism Eo + EndR(V). Now EndR(W) is a finitely generated R-module and R is noetherian. Hence Eo and EndR(V) are finitely generated Rmodules, as required. W 1.8. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then every nonzero finitely generated A-module has the unique decomposition property.

Proof. By Corollary 1.3.8, A is noetherian. Hence, by Corollary 2.11.4, it suffices to verify that any given finitely generated indecomposable A-module V is strongly indecomposable. Because V is inde. to composable, 0 and 1 are the only idempotents of E n d ~ ( v )Owing Lemma 1.7, E n d ~ ( v(and ) hence E n c l ~ ( V )is) a finitely generated Rmodule. The desired conclusion now follows by appealing to Corollary 1.6.

139

2. Reduction to G-invariant modules

2.

Reduction to G-invariant modules

Throughout this section, N denotes a normal subgroup of a finite group G. All modules are assumed to be finitely generated over their ground rings. It will be also assumed that R is a commutative ring of one of the following types: (i) R is artinian. (ii) R is a complete noetherian local ring. These assumptions are made to guarantee that if A is an R-algebra which is finitely generated as R-module, then any nonzero A-module has the unique decomposition property (see Corollaries 2.11.6 and 1.8). Let V be an indecomposable RN-module and let H be the inertia group of V. Our aim is to show that the decomposition of V G is determined uniquely by that of V H .This will allow us to concentrate on the case where V is G-invariant, i.e. gV E V for all g E G. 2.1. Theorem. (Ward (1968) and Willems (1976)). Let V be an indecomposable RN-module, let H be the inertia group o f V and write

where the K are the indecomposable RH-modules. Then (i) V G = KG @ - - .03 Vf, where the yG are indecomposable RGmodules such that qGE yGimplies 2 4. (ii) If, for some i E (1,. . . ,s } , is irreducible, then KG and V are irreducible.

.@KG

Proof. (i) The equality V G= Kc@. . follows from the facts that induction is transitive (Lemma 2.2.1) and preserves direct sums (Lemma 2.2.8). To show that each yGis indecomposable, suppose that

XG=X@Y is is a direct decomposition. We know, from Corollary 2.1.2, that identifiable with a direct summand of ( y G ) ~Hence, . by the KrullSchmidt theorem, we may assume that I$ is a direct summand of X H ,

say

X*=I/:@X'

(2)

Induction from normal subgroups

140

Observe also that ( V H )is~the direct sum of isomorphic RN-modules of the form h 8 V with h E H . Invoking the Krull-Schmidt theorem, we deduce from (1) that

for some positive integer m;. Accordingly, for any g E G, we obtain RN-isomorphisms (9c3 V t ) N

9 €9

(K)N 2

m;(g €9 V )

which in turn implies

(yG)N

(4)

= ( @ t c T t €9 v i ) N 2 mi(t €9 V )

where T is a left transversal for H in G. Restricting (2) to N, it follows from (3) that X N % mzv @ X & But X is an RG-module, so

proving that for each t E T , t €9 V appears as a direct summand of X N with multiplicity at least mi. Bearing in mind that for distinct t l , t 2 E T , tl 8 V and t 2 8 V are nonisomorphic RN-modules, we de~ isomorphic to a direct summand of XN. duce from (4) that ( q ' ) is But X N is a direct summand of (V;.')N, hence by the Krull-Schmidt theorem, X = KG. This shows that KG is indecomposable. Assume by way of contradiction that KG S '3/1 but V;: V,. Because is identifiable with a direct summand of (K')H, we may write

(KG)* = V , 63 q'

and

( y G =) vj~ @ q'

where K' and are RH-modules. Invoking the Krull-Schmid theorem, we see that K is isomorphic to an RH-direct summand of 5'. It follows that ( K ) N is isomorphic to an RN-direct summand of Applying (3)) we therefore deduce that V is isomorphic to an RN-direct summand of ( 5 ' )and ~ that

(y)~.

(FG)N = ( Y ) N @ ( 6 ' ) N 2 mjV 63 ( 6 ' ) N

3. Group-graded algebras and crossed products

141

However, by (4), the multiplicity of V as an indecomposable direct summand of ( y G is ) ~exactly mj. This contradiction proves the required assertion. (ii) Suppose that there exists an i E (1,. . . ,s} such that r/: is irreducible. Then by Clifford's theorem (Theorem 2.4.2), ( K ) N is completely reducible. Thus, by (3), V is irreducible. Let W be an irre(such W exists, by virtue of Proposition ducible factor module of 1.3.4). Then, by Corollary 1.2.5, V , is isomorphic to an RH-submodule of W,. Consequently, is isomorphic to an RN-submodule of W,. Therefore, by (3), W N contains a submodule X Z m;V. Thus, for any g E G, WN contains the RN-submodule gX Z m;(g@I V ) . Let T be a left transversal for H in G. Then the RN-module t @I V , t E T , are irreducible and nonisomorphic and they appear as direct summands of WN with multiplicity at least mi. On the other hand, by (4), the multiplicity of each t @I V as a direct summand of ( y G ) N is exactly m;. Now W = v G / X for some submodule X of V,' and WN 2 ( K G ) ~ / XThus ~.

KG

(v)~

and, by the foregoing, X = 0. The conclusion is that W = irreducible, as desired. H

YG is

2.2. Corollary. Let V be an irreducible RN-module, let H be the inertia group of V and let ( H : N ) be a unit of R. Then V G is comple t ely reducible .

Proof. Invoking Theorem 2.1, we may harmlessly assume that H = G. Now apply Theorem 2.10.14. H 3. Group-graded algebras and crossed products

This section contains a number of general results which will enable us to scrutinize the structure of the endomorphism ring of induced modules. Our exposition is strongly influenced by a work of Dade (1970a). Throughout, A denotes an algebra over a commutative ring R. If X , Y are R-submodules of A , then X Y denotes the R-submodule

Induction from normal subgroups

142

of A consisting of all finite sums (2;E

x,y;E Y )

Let G be a multiplicative group. Then A is called a G-graded algebra if there is a family { A h E GI of R-submodules of A indexed by the elements of G such that the following conditions hold: (direct sum of R-modules). (a) A = @gEGAg for all x , y E G. (b) A A , A,, We shall refer to (a) as a G-grading of A and to A, as the g-component of A. When (b) is replaced by the stronger condition, namely (4 A A y = A,, for all x,y E G we say that A is a strongly G-graded algebra. From now on, A denotes a G-graded algebra over R and U ( A )denotes the unit group of A.

c

3.1. Lemma. (i) A1 is a subalgebra of A with 1 E Al. (ii) For each g E G, A, is an ( A l ,Al)-bimodule under left and right multiplication b y the elements of Al. (iii) A is strongly G-graded if and only if 1 E A,A,-l for all g E G.

Proof. (i) By definition, Al is an R-submodule of A and, by (b), Al is multiplicatively closed. It therefore suffices to show that 1 E Al. Invoking (a), we may write

l=Ca, SEG

where a, E A, for all g E G, and all but finitely many of a, are zero. Now fix some h 6 G and aL E A h . Because G is a group, (a) implies that A = @g€GAgh (1) Owing to (b), the product agai lies in Agh for all g E G. Therefore, aI h = l . a h = I

c

,€G

a g 4

3. Group-graded algebras and crossed products

143

is precisely the expansion of a; in the decomposition (1). But ah already lies in the summand A1.h = A h in (1). Hence all the agai for g # 1 must be zero and ala: must be u i . This proves that a1 acts as a left identity on A h for all h E G. In view of (a), we deduce that al is a left identity for the algebra A. Thus al E Al is the identity element of A, as desired. (ii) This is a direct sequence of (b). (iii) Suppose that A is strongly G-graded. Because 1 E Al, it follows from (a)that 1 E AgAg-1 for all g E G. Conversely, assume that 1 E AgAg-l for all g E G. Invoking (b), we then have = 1 - A,, = AxA,-~Ax,

A,,

A x A x - ~ z= y A,A, for all x , y E G. Hence A,A, = A,, for all z,y E G, as asserted. W

We refer to a unit u of A as being graded if u E A, for some g E G. In this case, we say that g is the degree of u and write g = deg(u) = degA(u) The set of all graded units of A will be denoted by GrU(A). 3.2. Lemma. (2) I f u E G T U ( A )is of degree g , then u-l is of degree 9 - l . (ii) GrU(A) is a subgroup o f U ( A ) and the map deg : GrU(A)-+ G

is a group homomorphism with kernel U(A1). (iii) The map G r U ( A ) + Aut(A1) u H i,

{

where

i&)

= uzu-l

for all

x E A1

Induction from normal subgroups

144

is a homomorphism. (iv) Right multiplication b y any u, E A,

n U ( A ) is an isomorphism

of left Al-modules.

Proof. (i) We first write u-l = CxEGu,with a, E A, and with A,A, A,, for all x E G, it finitely many nonzero a,. Since ua, follows that uu-l = C I E G u a ,is a unique expansion for ua-l in the decomposition

c

A = fBxEGAgz = @XEGAX Now, by Lemma 3.1(i), uu-' = 1 lies in A1 = A,A,-1. Hence the az for # 9-l must be zero, and 1 must equal uag-l. Therefore u-' = a,-' lies in A,-], as asserted. (ii) That GrU(A)is a subgroup of U ( A )follows from ( i ) and the fact that A,A, 5 A,, for all x,y E G. Since A,A, A x y ,the given map is a homomorphism with kernel Al n U ( A ) . Owing to (i), A1 n U ( A ) U(Al) and since the opposite inclusion is trivial, the required assertion is established. (iii) By hypothesis, u E A, for some g E G. Thus, by (i) we have

x

uA1u-l 5 A,AlA,-i

A1

Bearing in mind that GrU(A)is a group, (iii) follows. (iv) Since u, is a unit of A , right multiplication by ug is an Alisomorphism of Al onto Alu,. Because u, e A,, we have Alu, C A,. Owing to (i), we also have

A, = A,u,'u,

C AgAg-lUgC Aiug

The conclusion is that A, = Alu, and the result follows. W

We next introduce a very important class of group-graded algebras, namely crossed products. Owing to Lemma 3.2(ii), the sequence of group homomorphisms

1

+ U(A1)+.

GrU(A)%' G + 1

(2)

3. Group-graded algebras and crossed products

145

is always exact, except possibly at G. We say that a G-graded algebra A is a crossed product of G over A1, written A = A1 * G, provided the sequence (2) is exact. Thus A is a crossed product of G over A1 if and only if for any g E G, there exists 3 E A, n U(A). In the special case, where (2) is an exact splitting sequence, we shall refer to A as a skew group ring of G over Al. Thus A is a skew group ring of G over A1 if and only if for each g E G, there exists E A,nU(A) such that zy = zy for all z,y E G

A G-graded algebra A is called a twisted group ring of G over A1 if for all g E G, there exists g E A, n U(A) such that g centralizes Al. In a particular case, where Al Z(A) and A, n U ( A ) # 8 for all g E G, we shall refer to A as a twisted group algebra of G over Al. Finally, if Al * G is a twisted group algebra and a skew group ring, A1 * G is notihng else but the group algebra of G over Al. Our next aim is to provide a condition under which a strongly Ggraded algebra A is a crossed product of G over Al. Let A be a strongly G-graded algebra whose subalgebra Al is a local ring. Then A is a crossed product of G over A1 . 3.3. Proposition.

Proof. Since A,A, = A,, for all

J(A1) C A1 = A,A,-1 Thus we may find a, E A, and

2,y

E G, we have

for all g E G

E A,-I for which

Because A1 is a local ring, we have U(Al) = Al - J(Al). Hence

is a right inverse to a, in A. The idempotent a, lies in A,-] A, Because 0 # 1 = l2 = (u,u,.-,)2 = ug(ag-lug)ag-l tI

C Al.

Induction from normal subgroups

146

we have a;-lag # 0. Therefore a$)-lag= 1 and thus a;-, is a two sided inverse to a, in A. We conclude that

A,

n U ( A )# 8

for all g E G

which shows that A is a crossed product of G over A l . W We now proceed to describe crossed products more explicitly. Let A be a crossed product of G over Al. A map Q

: G -+

Aut(A1)

is called an automorphism system for A if, for all g E G, there exists a unit g of A in A, with I = 1 such that

for all x E Al

a ( g ) ( s ) = gxg-'

(3)

Now suppose that B is an R-algebra. Given maps Q

: GtAut(B)

(Y

:

GxG+U(B)

we say that (G,B , Q, a ) is a crossed system for G in B if, for all 2,y , z E G, and b E B, the following properties hold:

where ,b = a ( g ) ( b )for all g E G, b E B.

Let a G-graded R-algebra A be a crossed product of G over A l . For a n y g E G, fix a unit g of A in Ag with I = 1, let n : G + Aut(A,), be the corresponding automorphism system given b y (3) and let (Y : G x G U(A1) 3.4. Proposition.

--j

be defined b y +,Y)

= ZY"Y

(7)

3. Group-graded algebras and crossed products

147

Then the following properties hold: (i) A is a strongly G-graded algebra with A, = A1ij = jA1. (iz) ( G ,A1,o, a ) is a crossed system for G over A1 (to which we refer as corresponding to A) (iii) A is a free (left and right) Al-module freely generated b y the elements g , g E G (iv) For all x,y E G and r1, 7-2 E Al,

Conversely, for any R-algebra B and a n y crossed system ( G ,B , 0,a ) for G over B , the free B-module C freely generated b y the eIements g , g E G with multiplication given b y (8) (with r1,r2 E B ) is a Ggraded R-algebra (with C, = Bg for all g E G) which is a crossed product of G over C1 = B and having (G,B, 0,a ) as a corresponding crossed system.

Proof. Assume that u E A, n U ( A ) for some g E G. Owing to Lemma 3.2(i), u-' E A,-] n U ( A ) and hence 1 = uu-l E A,A,-I. Invoking Lemma 3.1(iii), we conclude that A is a strongly G-graded algebra. By Lemma 3.2(iv), A, = A1g and the argument of that lemma applied to left multiplication shows that A, = gA1, proving (i). Let x,y E G and b E Al. Then

"( Yb) = z(yby-')Z-'

= a ( x ,y)ZybZy-lcu(x, y)-' = ~ ( zy ,) " y b a ( x ,y ) - 1 ,

proving (4). For any x , y , z E G, we also have

(.y). -

.(Y.)

__

= a ( x ,y ) z y z = a ( x ,y ) a ( z y ,2)= q " ( y , z ) y z = "cu(y,z)ZyZ= Z a ( y , Z ) a ( x , y Z ) q z ,

proving ( 5 ) . Since = 1, (6) is also true, proving (ii). Property (iii) is a direct consequence of (i). To prove (iv), fix x,y E G and r l , r2 E Al. Then ( r 1 z ) ( r 2 y= ) rl(%r2ii-')(.y) = r1 " r 2 a ( x ,y)-,

Induction from normal subgroups

148

proving (8). To prove the converse, it suffices to show that the multiplication given by (8) is associative. To this end, fix x , y , z E G and rl,r2,r3 E B. Then

( r i z )[(r2Y)(r@)]= ri "[r2'r3a(Y,z ) ] a ( zY, z ) m = rl "r2"( Yr3)" a ( y , z ) a ( z ,y z ) ~ (bY(4)) = rl I r z a ( ~ , yzYr3a(x,y)-1 ) =a(y,z ) a ( z ,y z ) z y ~ (by ( 5 ) ) = r1 "r2a(z,y) "Yr3a(z,y)-la(z,y)a(zy, z ) z y ~ =

[ ( T i2) (

f a Y)] (r3z).

as required.

Returning to our study of G-graded algebras, we now introduce the following important notion. Let A be a G-graded R-algebra. An ideal I of A is called a graded ideal if

I = @ g E G ( I n Ag) Assume that B is another G-graded R-algebra. A homomorphism

f:A+B of R-algebras is said to be graded if f ( A g )C B, for all g E G.

3.5. Lemma. (i) If I is a graded ideal of A, then A / I is a G graded R-algebra by setting ( A / I ) , = ( A , I ) / I for all g E G. (ii) I f f : A -+ B is a graded homomorphism of G-graded Ralgebras, then Kerf is a graded ideal. Moreover, if f is surjective, then there is a graded isomorphism A / K e r f --+ B . (iii) An ideal I of A is graded i f and only i f I is the kernel of a graded homomorphism A -+ B of G-graded R-algebras.

+

3. Group-graded algebras and crossed products

149

as asserted. (ii) First of all, we know that Kerf is an ideal of A. Let z E Kerf and write x = CgEG x,, xg E A,. Then

0 = f (4=

cf

(z,>

,EG

and

f ( 4E 4

Since B = $,&Ideduce ,, that each z, E Kerf and therefore Kerf is a graded ideal. Now assume that f is surjective and let

f* : A / K e r f + B be the induced homomorphism of R-algebras. Then, for all g E G,

proving that f* is a graded isomorphism. (iii) If I is a graded ideal of A , then the natural map A -+ A / I is a graded homomorphism with kernel I . The converse is a consequence of (ii). H An ideal X of A1 is called G-invariant if

A,XA,-i = X

for all g E G

In the special case where A is a crossed product of G over A1 and ij E U ( A ) n A,, we have A,XA,-I = 3Xg-l. Thus in this case X is G-invariant if and only if gxg-1 =

x

for all g E G

We are now in a position to describe all graded ideals of A . 3.6. Theorem. (Dude (1970~)). Let A be a strongly G-graded R - a lg e b ra . (i) If I is a graded ideal of A and X = I n Al, then X is a Ginvariant ideal of A1 such that

I = AX = X A

and

I n A, = A,X = X A , for all g E G

Induction from normal subgroups

150

(ii) For any G-invariant ideal X of Al, I = X A = AX is a graded ideal o f A such that

I n A, = AgX = XA,

for all g E G

(iii) J(A1) is a G-invariant ideal of A1 and hence A - J ( A l ) = J(A1)- A is a graded ideal of A. Proof. (i) By hypothesis, I = e g E ~ where I g I, = I n A, for all g E G. In particular, we must have X = 1 1 . Given z,y E G, we have

A,Iy = A,(I n AY)C A,I n A,Ay In&,

c

T

Thus A,-lIacy I y and therefore

Consequently, AxIy = I,, and, in particular, A,I1 = I, for all g E G. A similar argument shows that IlA, = 1, for all g E G. Given g E G, we then also have

AgIIAg-l= IgAg-1= 11, proving that Finally,

11

is G-invariant.

as required. (ii) By assumption, AgXAg-l = X for all g E G, so multiplication on the right by A, gives A,X = XA, for all g E G. Since A = $,EGA,, we have

I = AX = $,,GA,X

= $,EGXA, = X A ,

proving that I is a graded ideal of A such that I n A, = A,X = X A , for all g E G.

151

3. Group-graded algebras and crossed products

(iii) Owing to (ii), it suffices to show that J(A1) is a G-invariant ideal of A1. Let V be an irreducible Al-module and let L be the annihilator of V . For any given g E G, it will be shown that A,LA,-I is the annihilator of some irreducible Al-module. Because J ( A l ) is the intersection of all such annihilators, the desired assertion will follow. Choose a maximal left ideal I of A1 such that V E A1/I. Since A,A, = Azy, for all 2,y E G and 1 E Al, the map X + A,X sends bijectively the set of all left ideals X of A1 onto the set of left Alsubmodules of A,, with Y 3 A,-IY as the inverse map. These maps clearly preserve inclusion, hence A,I is an Al-submodule of A,. An elementary calculation shows that the annihilator of the irreducible Al-module A,/A,I is precisely A,LA,-l, as required. H We have now come to the main result of this section. 3.7. Theorem. Let A be a strongly G-graded R-algebra. (i) A/A J(A1) is a strongly G-graded R-algebra by setting +

(A/A - J ( A I ) ) ~= (A,

+ A - J(Al))/A - J(A1)

f o r all g E G

Mo re0 v e r,

(A/A J(A1))l

A,/&%)

and, provided G is finite,

(ii) If A is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al, then A/A J(A1) is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al / J(A,). (iiz) If A is a crossed product of G over A1 and A1 = R. 1~ J(A1), then A/A - J(A1) is a twisted group algebra of G over R . ~ A / (- R 1~n

-

+

J(A& Proof. (i) The fact that A / A . J ( A l ) is a G-graded R-algebra with the given g-component is a consequence of Lemma 3.5(i) and Theorem 3.6(ii), (iii). Because A is strongly G-graded, so is A/A J(Al). By

Induction from normal subgroups

152

definition,

and therefore

On the other hand,

and therefore

proving that ( A / A - J ( A l ) )2, Al/J(A,). If G is finite, then A.J(A1)C J ( A ) ,by virtue of Corollary 4.2.3 in Karpilovsky (1987a). (ii) Assume that A is a crossed product of G over A,. Then, for any g E G, there exists a unit g of A in A,. Hence ij A . J(A1) is a unit of A / A - J(A1)in ( A / A J ( A l ) ) ,for all g E G. This proves that A / A J ( A l )is a crossed product of G over A l / J ( A l ) ,by applying (i). Suppose that A is a skew group ring of G over A,. Then, for any g E G, there exists a unit g of A in A, such that

+

5g = zy

But then for all

2,y

for all x , y E G

E G,

proving that A / A J(A1) is a skew group ring of G over A,/J(A,). Now assume that A is a twisted group ring of G over Al. Then, for any g E G, we may choose g E U(A)n A, such that 3 centralizes Al. But then, for each g E G, ij+A.J(A,) centralizes (Al+ A . J ( A ) / A - J ( A , ) ) , proving that A / A - J(A1)is a twisted group ring of G over A,/J(Al). Finally, assume that A is a twisted group algebra of G over A,. Then A1 G Z(A) and hence

153

4. The endomorphism ring of induced modules

Thus A / A - J ( A l ) is a twisted group algebra of G over A l / J ( A , ) , as required. (iii) Our assumption guarantees that

and therefore A / A . J ( A l )is a twisted group algebra of G over AlIJ(A1). Since the homomorphism R . 1~ + A l / J ( A l ) ,r - 1 H r 1 J ( A 1 ) is surjective with kernel R 1~ n J ( A l ) ,the result follows.

- +

4.

The endomorphism ring of induced modules

The proof of most of the results of this section can be found in Dade (1980),(1981), although many of the essential ideas are contained in previous works of Conlon (1964) and Ward (1968). Throughout, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Let V be an RN-module. Recall that V is said to be G-invariant if gV Z V for all g E G. If W is another RN-module, we say that V weakly divides W if there exists a positive integer k and an injective homomorphism f : V -+ kW = W

@

W

(k copies)

such that f ( V ) is a direct summand of kW. Weshall say that V and W are weakly isomorphic if each weakly divides the other. Of course, this is an equivalence relation among RN-modules. Finally, we shall refer to V as being weakly G-invariant if it is weakly isomorphic to each g V , g E G. 4.1. Lemma. Let V be an RN-module. (i) The map

is an injective homomorphism of R-algebras whose image consists of all $ E EndRG(V") for which $(1@ V ) C 1 @ V .

Induction from normal subgroups

154

€3 V is a n RN-isomorphism, t h e n f can be uniquely extended t o a n RG-isomorphism V G+ V G .

(ii) Iff : 1 €3 V

+g

Proof. (i) It is plain that the given map is an injective homomorphism of R-algebras. Moreover, for all f E E n d m ( V ) ,l @f sends 1@V into 18V. Suppose that $ E EndRG(VG)is such that $(1€3V) l @ V . Then there exists f E EndRN(V)such that $ ( l @v) = 1 @ f ( v ) for all v E V . Because $ is uniquely determined by its restriction to 1 €3 V , we have $ = 1 €3 f , as required. (ii) Thanks to a universal characterization of induced modules (Proposition 2.2.3) f can be uniquely extended to an RG-homomorphism $ : VG4 V G .Since for any z E G,

it follows that 1c, is surjective. To show that $ is also injective, denote by T a transversal for N in G. Then, any given w E V G ,can be uniquely written in the form

If $(w)= 0, then

c tf

(1€3 vt) = 0

tCl'

Because t f (1 €3 vt) E tg €3 V and Tg is a new transversal for N in G, we have t f (1 €3 vt) = 0. Thus 1 €3 vt = 0 and therefore w = 0, as we wished to show. Let V be an RN-module. We say that V can be extended to an RG-module if there exists an RG-module V * , whose additive group coincides with that of V while the multiplication

* : RG x V* -, V*

in

V*

satisfies z*v=zv

forall

~ E V , ~ E R(1) N

We are now ready to prove the following fundamental result.

4. The endomorphism ring of induced modules

s

155

4.2. Theorem.

Let V be an RN-module, let E = EndRG(VG), = EndRN(V) and, for any g E G , define EgN by

(i) E is a GIN-graded R-algebra with E g as ~ its gN-component, g E G and the identity component of E is identifiable with S . (ii) E is strongly GIN-graded i f and only i f V is weakly G-invariant, while E is a crossed product of G I N over S if and only if V is G -

invariant. (iii) E is a skew group ring of GIN over S if and only if V can be extended to an RG-module. In fact, there is a bijective correspondence between all extensions V* o f V to RG-modules and all splitting homomorphisms y for 1 + U ( S ) + G r U ( E ) 2 G I N + 1, in which V* corresponds to y if and only if 1 @ g * v = y ( g N ) ( g@v)

for all

E V,g E G ( 2 )

-

(iv) If V is weakly G-invariant, then E J ( S ) = J ( S ) .E is a graded ideal of E contained in J ( E ) and such that E I E - J ( S ) is a strongly G/ N-graded R-algebra with ( E / E- J ( S)),, = (E g N +E. J ( S ) ) / E .J ( S ) , g E G . Furthermore, i f V is G-invariant, then E I E . J ( S ) is a crossed product o f G / N over S / J ( S ) . (v) If V is G-invariant and S = R - 1s J ( S ) , then E I E - J ( S ) is a twisted group algebra of G I N over R l s / ( R 1s n J ( S ) ) S / J ( S ) .

+

-

Proof. (i) Observe that if X N = y N , then x @ V = y 8 V which shows that E g N is well-defined. Let T be a transversal for N in G. Then (direct sum of RN-modules) V G= $ t c T t @I V and so if rt : V G+ t @I V is a projection map, then

Now assume that f E E and let gt : 1 @ V + t 8 V be the R N homomorphism that is the restriction of rt 0 f to 1 @ V . Owing to a

Induction from normal subgroups

156

universal characterization of induced modules (Proposition 2.2.3), there exists ft E E such that

It follows that ft E Et-"

whence f = CtETft. Now assume that then have and

vt

&T

and, applying (3) and (4),we have

CtETft= 0 with

= 0. Thus

ft E

Et-IN. For any v E V, we

vt = f t ( 1 c3 v) E t @ v vt

= 0 and f t = 0 for all

t E T , proving that

I f f E E=N,cp E E,N, then

and thus (fcp)(l@ V)

c f(y-'

63 V ) c y - ' f ( l @ V ) E y-'z-l@

v

It follows that f ( o E E z y and ~ therfore E is a GIN-graded R-algebra with E I l as ~ its gN-component. Finally, by Lemma 4.1(i), the identity component of E is identifiable with S, proving (i). (ii) Thanks to Lemma 3.l(iii), E is strongly GIN-graded if and only if for any given g E G, there exists (o; E E g ~y5i, E E g - l ~1, 5 i 5 n = n(g), such that

It will next be shown that ( 5 ) is equivalent to V and gV to be weakly isomorphic, which will imply the first assertion.

157

4. The endomorphism ring of induced modules

If (5) holds, then any map f : l@V + n(g@V),w t-+ ($l(v), . . . ,$ n ( v ) ) is an injective RN-homomorphism. Moreover, the map (wl, . . . ,wn) H ‘pI(w1) * v n ( v n ) is a homomorphism from n(g QD V ) to 1 QD V which is a left inverse to f . This proves that V weakly divides gV. By symmetry, V weakly divides g-’V and thus gV weakly divides V , proving that V and gV are weakly isomorphic. Conversely, suppose that V and gV are weakly isomorphic. Then V weakly divides g 8 V ,so there is a positive integer n and an injective RN-homomorphism $ : 1 @ V + n(g @ V ) such that $(1 @ V ) is a direct summand of n(g 8 V ) . We may write $(z) = ($1(z),. .. ,$ n ( S ) ) , where : 1 @ V + g @ V is an RN-homomorphism, in which case $; can be regarded as an element of E g - l N . Let f : n(g V )-+ 1 8 V be an RN-homomorphism which is a left inverse to $. Then f determines RN-homomorphisms

+-+

+;

‘p;

:g @

v + 1 QD v

(1 5 i 5 n )

(which can be regarded as elements of E g ~such ) that ( 5 ) holds. If V is G-invariant, then E is a crossed product of GIN over S, by virtue of Lemma 4.1(ii). Conversely, suppose that for any g E G there exists fg E U ( E )n E g ~ Then . fg-l : 1 @ V + g @ V is an injective RN-homomorphism such that fg-ifg is an automorphism of 1@ V and hence of g 8 V . Thus fg-1 is an RN-isomorphism, as desired. (iii) Assume that V’ is an extension of V to an RG-module. Then (1) implies that there is a unique RN-isomorphism of g 8 V onto V sending g 8 w into g * w, for all g E G, w E V . This isomorphism obviously depends only on the coset g N . Hence, by Lemma 4.1(ii), there is a unique y(gN) E E g n~U ( E ) such that (2) holds. Now fix z,y E G. Then, for all v E V ,

+N)r(yN)(w @ 4 = 7 ( 4 z 7 ( ” y QD 4 = r(zN)z(lQD Y*4 (by (2))

*

= 7 ( z N ) ( zQD y v) = 1@ z * (y * v) = 1 8 (zy) = 7 ( z y N ) h €3 4

*

2,

(by (2))

which forces r(zN)y(yN) = y(zyN). This demonstrates that y is a splitting homomorphism, the only such homomorphism satisfying (2).

Induction from normal subgroups

158

Conversely, suppose that 7 : G/N + GrU(E) is a homomorphism . any g E G, v E V , let such that for all g E G, r ( g N ) E E s ~ For g * v E V* be defined by (2). Then for any n E N, 1 8 n * o = r ( N ) ( n8 v) = n 8 v = 1 8 nu and therefore n * v = nv. Furthermore, given s , y E G and v E V , we have

Thus V extends to the RG-module V*,the only such module satisfying (2). (iv) Suppose that V is weakly G-invaraint. Then, by (ii), E is strongly GIN-graded, so by Theorem 3.6(iii), E . J ( S ) = J ( S ) - E is a graded ideal of E . By Theorem 3.7(i), this graded ideal is contained in J ( E ) . The rest of the first statement is a consequence of Theorem 3.7(i). The second statement is a consequence of (ii) and Theorem 3.7(ii) . (v) Apply (ii) and Theorem 3.7(iii). 4.3. Corollary. Let V be a strongly indecomposable RN-module.

Then V is G-invaraint if and only if V is weakly G-invariant.

Proof. Because V is strongly indecomposable, S = EndRN(V) is a local ring. Now suppose that V is weakly G-invariant. Then, by Theorem 4.2(ii), E = EndRG(VG)is strongly GIN-graded. Because S is local, it follows from Proposition 3.3 that E is a crossed product of GIN over S. Thus, by Theorem 4.2(ii), V is G-invariant. Conversely, if V is G-invariant, then V is obviously weakly G-invariant, hence the result. Our next aim is to provide a condition under which J ( E ) n S = J ( S ) , where E = EndRc(VG)and S = EndRN(V). The following simple observation will clear our path. 4.4. Lemma. Let A be an R-algebra which is finitely generated

5. Relations between the decompositions of VG and End&VG)

159

as an R-module and let R / J ( R )be artinian. Then

J(A)" C J ( R ) A

for some n

21

Proof. We know, from Lemma 1.2, that J ( R ) AC J ( A ) . Hence, by Corollary 1.4.4(ii), we have

J ( A / J ( R ) A )= J ( A ) / J ( R ) A But A / J ( R ) Ais an R/J(R)-algebra which is finitely generated as an R/J(R)-module. Hence, by Corollary 1.3.8, A / J ( R ) Ais artinian. Invoking Proposition 1.4.14, we therefore deduce that J ( A ) / J ( R ) Ais nilpotent, hence the result. 4.5. Theorem. Let V be a finitely generated G-invariant R N module, let E = EndRG(VG) and let S = EndRN(V). If R is artinian or a noetherian local ring, then

J ( E )n S = J ( S ) Proof. Owing to Theorem 4.2(iv), J ( S ) C J ( E ) n S . Now V is a finitely generated R-module, hence EndR(V) (and therefore S ) is a finitely generated R-module (Lemma 1.7, Propositions 1.4.14 and 1.3.3, Corollary 1.3.7). Similarly, V Gis a finitely generated R-module and hence E is a finitely generated R-module. Thus, by Lemmas 4.4 and 1.2, J ( R ) S C J ( S ) and J(E)" & J ( R ) E for some n 2 1. Moreover, by Theorem 4.2(i), E = $ z c ~ / ~ E with z El = S. Hence J ( R ) E n S = J ( R ) S and

( J ( E )f l S)" C J ( R ) En S = J(R)S C J ( S ) It follows that ( J ( E )n S ) / J ( S ) is a nilpotent ideal of S / J ( S ) and therefore J ( E )n S J ( S ) ,as required. 5.

Relations between the decompositions of V Gand EndRG(VG)

Throughtout this section, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Given an RN-module

160

Induction from normal subgroups

V , it will be convenient to identify EndRN(V) with the subring of E n d ~ c ( Vconsisting ~) of all $ E EndRG(VG)for which $(1 @ V ) 1@ V (see Lemma 4.1(i)), In what follows we shall freely use this identification. For convenience of reference, we first record some general observations which provide a link between a module and its endomorphism ring. Let S be a ring, let V be a left S-module and let E = Ends(V). For any subset L of E and any subset W of V , let LW denote the set of all finite sums

It is clear that LV is a submodule of V . Note also that if I is a principal right ideal of E generated by $, then

We may regard V as a left E-module via 'pv = cp(v),cp E The E @ E V becomes a left E-module via

E,v E V .

Moreover, the canonical map

is an isomorphism of E-modules. 5.1. Lemma. Let S be a ring, V a left S-module and I a right ideal of E = Ends(V) which is a direct summand o j E . Then I = Ee for some idempotent e of E c n d (i) IV = e ( V ) (ii) The map I @ E V + IB

is an isomorphism of E-modules.

5. Relations between the decompositions of V G and End&VG)

161

Proof. The fact that I = E e for some idempotent e of E is a consequence of the right analog of Proposition 1.3.12. Hence (i) follows from (1). Because I is a direct summand of E , we may identify I @ E V with its image in E @IE V, in which case (2) maps I @ E V onto IV. This proves (ii) and hence the result.

5.2. Lemma. Let S be an arbitrary ring, let V be a left S-module and let E = E n d s ( V ) . (i) If V = c&V;. and ei : V 3 6 is the projection map, then el,. . . ,e, are mutually orthogonal idempotents of E with sum 1. Conuersely, if el,. . . ,e, are mutually orthogonal idempotents of E with sum 1, then V = c&V, where = e;(V) and ei : V + K is the projection

map.

(ii) K is indecomposable if and only i f e; is primitive. (iii) K Z 4 as left S-modules if and only if e;E Z ejE as right E-modules. be the Proof. (i) Assume that V = @i=lx and let e; : V + projection map. It is plain that e: = ei, 1 5 i 5 T . If i # j , then (e;ej)(V) = e;(y)= 0 , and thus e;ej = 0. Moreover, we obviously have C:==, e; = 1.

Conversely, assume that el,. . . ,e, are mutually orthogonal idempotentsofEwithsum1. Since1 = e l + - - - + e , , w e h a v e V = V,+-..+V,. If wi E V;., 1 5 i 5 r , are such that el(v1) - e,(v,) = 0, then by taking the image of both sides under e;, we obtain e;(v;)= 0, 1 5 i 5 r , Thus V = $r==,K and obviously e; : V + K is the projection map. (ii) If e; = e f , where e, f are orthogonal idempotents, then K = e(V )$ f ( V ) . Conversely, if K = V,' @ Kii and A; : V -+ p; : V + are projection maps, then & , p i are orthogonal idempotents of E with e; = A; pi. Thus V , is indecomposable if and only if ei is primitive. (iii) Assume that eiE 2 e j E as right E-modules and put I; = e;E, 1 5 i 5 r. Then, by Lemma 5.1, IiV = K and V; % V, as E-modules. as S-modules. Hence Q Z Conversely, suppose that 8 : r/: -, vj is an S-isomorphism and fix v E V. Given cp f e;E, we have Imcp C ei(V) = K, so Im(89) C 4.

+- - +

+

+

v',

162

Induction from normal subgroups

Thus ( ~ Q ) ( v = ) ej(V1) for some ~1 E V, so e j ( b ) ( v ) = ej(V1) = ( b ) ( v ) and therefore

BQ = ej(0cp) E e j E If we now define p : eiE -t ejE by p(cp) = 09, then p is obviously a homomorphism of right E-modules. A similar argument shows that the map X : ejE + eiE defined by A($) = 0-'$ is also an E-homomorphism. Because X is the inverse of p , we deduce that eiE E ejE, as desired. Let V be a free module over a ring S. In case any two bases of V have the same cardinality, we refer to this cardinality as the rank of V. 5.3. Lemma. Let V be a free module over a local ring S. Then any two bases of V have the same cardinality.

Proof. Let {v;li E I} be two bases of V. Denote by V; and tijj the images of vi and wj, respectively, in 3 = V/J(S)V. Then (6ili E I}and {tijjlj E J } are two bases of the S/J(S)-module V . But, by hypothesis, S / J ( S ) is a division ring, hence 111 = lJl, as required. The observation above, will allow us to speak about

"

the rank of

V" whenever V is a free module over a local ring. Let S be a ring and let V be a left S-module. Then V can be regarded as a right Ends(V)'-module by setting VQ

=~ ( v )

for all v E V,cp E Ends(V)'

5.4. Lemma. Let V be a G-invariant RN-module, let S = EndRN(V and let E = EndRG(VG)". For each g E G, choose 'pg E U ( E ) mapping 18 V isomorphically onto g 8 V, put 91 = 1 and define $g : V + V by 'p;l(g 8 V ) = 1 8 $ ~ ~ ( v )Then . V 8 s E is an (RG,E)-bimodule with the actions

5. Relations between the decompositions of V G and End&VG>

163

Proof. Because V is a right S-module and E is a left 5'-module, the tensor product V 8 s E is at least a 2-module. Given s E S, g E G,vE V , we have the following equality in VG:

which implies

We therefore conclude that

Given g E G, we now introduce f : V x E + V 8 s E given by It is clear that f is R-bilinear. Since, for any s E S,

the m a p v 8 e H g(v 8 e) is an R-endomorphism of V 8 s E . We now claim that, given z,y E G and v E V , where s = 'pz'py'p;i E S. Indeed, we have

Induction from normal subgroups

164

proving (4). It follows that X(Y(V

@ el) = z($J&) @ vve) = $W%/(v)) @ cpxvve

= $z($v(v)) @ ~ ' P q = e 4$Jz($Jv(v))>@ vzve = $z&) @ cpxve (by (4)) = (zY>(v@ 4

Because vl = 1 (hence &(TI) = v for all v E V ) ,we also have l ( v @ e ) = v @ e. Hence V 8 s E is a left RG-module. It is clear that V @sE is a right E-module under the given action of E. Moreover, for all g E G,u E V,e,e'E E ,

[g(v8 e)]e' = $Jg(v) cpgee' = g ( v

eel) = g[(v@ e)er],

thus completing the proof. H The following theorem is the main result of this section.

(Cline (1972), Theorem A). Let V be a G-invariant RN-module, let E = End&VG)' and let S = EndRN(V)' be identified with the subring of E consisting of all cp E E with cp(l@V)C l @ V . Then the map 5.5. Theorem.

PsE + v@e

H

e(l@v) vG

is an (RG,E)-bimodule isomorphism, where the (RG, E)-bimodule structure of V 8 s E is given by Lemma 5.4. Proof. The given map, say f, is obviously R-linear. Since for all seG f (g(v @ 4 ) = f ($Jg(v>@ cpge) = (cpge)(l@$ J g W = (vge)(cp;'(g @ 4) = 4s @ 4 = g e ( l @v) = g f ( v @ e ) ,

it follows that f is an RG-homomorphism. Because for all e,e' E E and v E V f((v

e)e') = f ( v eel) = ( e e ' ) ( l @w) = ( e ( I @v))e' = f ( v e)e',

5. Relations between the decompositions of V G and En&c(VG)

165

we deduce that f is also an E-homomorphism and hence is an (RG,E)bi mo dule homomorphism. Let T denote a transversal for N in G. Then

Because pt(l @ V ) = t @ V , it follows from (5) that f is surjective. Owing to Theorem 4.2(ii) and Proposition 3.4(iii), E is a free left Smodule freely generated by all yt, t E T . Thus any element of V 8 s E can be uniquely written in the form

Because

f ( p w t )= Ct Et TP t ( l @ v t ) and cpt( 1@ vt) E t @ V , it follows from ( 5 ) that if CtETvt €3 pt E K e r f , then pt(l @ vt) = 0 for all t E T . Thus all vt = 0 and therefore C t Evt ~€3 (Pt = 0. Hence f is injective and the result follows. As a second major result, we now prove the following theorem in which the RN-module V is not assumed to be finitely generated.

Let V be a strongly indecomposable G-invariant RN-module and let S = EndRN(V) be identified with the subring of E = EndRG(VG) consisting of all (P E E with p(l @ V ) 1 @ V . (i) I f a right ideal I of E is a direct summand of E, then I is a free S-module offinite rank. (ii) E contains a finite complete set of orthogonal primitive idempotents el,. . . ,en such that for Ij = e;E the following properties hold: I;V where each I;V is an indecomposable RG-module (a) V G= such that (IjV)N is a direct sum o f n ; copies o f V , nj = ranks(I;) (V is identified with 1 8 V ) . (b) I;V IjV if and only i f e j E Z ejE. (c) I f V is R-free offinite rank, then I;V is R-free with 5.6. Theorem.

run kR( 1;V ) = ranks (I;)runkR( V )

Induction from normal subgroups

166

Proof. (i) By Theorem 4.2(ii) and Proposition 3.4(iii), E is a finitely generated free S-module. Since I is a direct summand of E , I is a finitely generated projective S-module. But, by hypothesis, S is local, hence I is a free S-module, by Lemma 2.10.12. Finally, by Lemma 5.3, I has a uniquely determined rank which is finite since I is finitely generated. (ii) It is a consequence of (i) that E admits a finite decomposition E = $Z,E; where each E; is indecomposable. Hence E; = e;E for some orthogonal primitive idempotents e; of E with el - e, = 1. By Lemma 5.2, we deduce that

+ +

where each ei(VG)is indecomposable and e;(VG)Z ej(VG)if and only if e;E ejE. We now claim that

e;(VG)= I;V

for all i

(6)

Indeed, since V C V G and e;(VG)= I;VG,we obviously have I;V e;(VG).On the other hand, V Gis the sum of all g 8 V,g E G. Hence, to prove (6), we need only verify that ei(g 8 V

) C IiV

for all

g EG

Fix g E G and w E V . Let cp E E map V isomorphically onto g 8 V . Then g @ w = cp(v1) for some 01 E V . Hence ei(g 8 v) = ( e i c p ) ( Y ) E IiV,

proving (6). All the remaining assertions will now follow provided we show that (IjV)N is a direct sum of nj copies of V , n; = ranlcs(1;). Since Ij is a free right S-module of rank n;, E" o e; is a free (left) 9-module of E" 0 e; under the isomorphism rank n;. Moreover, the image of V of Theorem 5.5 is I;V. Hence

which immediately implies the required assertion by virtue of the fact that V @ s o So V .

6. Twisted group algebras over fields

6.

167

Twisted group algebras over fields

In this section, we shall collect a number of general observations concerning twisted group algebras. The results recorded will play a significant role in our subsequent investigations. Throughout, G denotes a finite group and F* the multiplicative group of a field F . Let Z2(G,F*)be the set of all functions

a:GxG-,F* which satisfy the following identities:

We shall refer to the elements of Z2(G,F') as cocycles. Given a , p E Z2(G7F'), define a@by the rule

It is then obvious that a@is also a cocycle and that Z2(G7F * ) becomes an abelian group. Let t : G -+ F* be such that t ( 1 ) = 1 and let 6t : G x G + F* be defined by W(z, 3)=W ( y ) t ( z y ) - ' We shall refer to S t as a coboundary. It is routine to verify that the set B2(G,F') of all coboundaries constitutes a subgroup of Z 2 ( G ,F*). The factor group

H2(G7F') = Z2(G7F * ) / B 2 ( GF, * ) is said to be the second cohomology group of G over F*. The elements of H2(G,F * ) are called cohomology classes ; any two cocycles contained in the same cohomology class are said to be cohomologous. Given (Y E Z2(G7F * ) ,we shall write &. for the cohomology class containing a. Given a E Z 2 ( G 7 F * )we , write F*G for the vector space over F

Induction from normal subgroups

168

with basis {gig E G} which is in one-to-one correspondence with G. Define multiplication in F"G distributively using

zg = a ( z , y ) w ,

Xa: = a:X

for all z, y E G, X E F. Then F*G is a twisted group algebra of G over F and, conversely, any twisted group algebra arises in this manner. Note that if a ( z , y ) = 1 for all z , y E G, then F a G Z FG. More generally, we have 6.1. Lemma. For any given (Y E Z2(G,F*),the following conditions are equivalent. (i) F"G E F G as F-algebras. (ii) F"G admits an F-algebra homomorphism into F. (iii) a is a coboundary.

Proof. (i)+(ii): This follows from the fact that the augmentation map F G + F is a homomorphism of F-algebras. (ii)+(iii): Let f : F*G + F be a homomorphism of F-algebras. Then, for all x , y E G,

f(a:)f($=

fm)= f ( + , Y ) z y )

= +,Y)f(zy)

and so a = bt, where t : G -+ F*is defined by t ( g ) = f(g). (iii)+(i): Assume that a = St for some t : G --.) F* with t ( 1 ) = 1. Then the map qh : F"G 4 F G , which is the extension of g H t ( g ) g by F-linearity gives the desired isomorphism. H 6.2. Lemma.

(i) Each element of H2(G,F*)has order dividing

IGI-

(ii) If G is a p-group and F a perfect field of characteristic p , then F G for all H2(G,F*) = 1. In particular, b y Lemma 6.1, F*G a E Z2(G,F*). FG (iii) If G is cyclic and F is algebraically closed, then F"G for all a E Z2(G,F*)(in particular, b y Lemma 6.1, H2(G,F') = 1).

Proof. (i) Assume that a E Z2(G,F'). Because 4 3 , Y)+Y,

4 =4 Y ,4 4 x 9Y

4

169

6. Twisted group algebras over fields

Thus a" = S t , as asserted. (ii) By (i), it suffices to show that if QP is a coboundary, then so is a. To this end, assume that t : G 3 F', t ( 1 ) = 1 is such that

Since F is perfect and charF = p , we have

for some p : G + F' with p(1) = 1. Thus a = Sp, as required. (iii) Let g be a generator of G, say of order n. Then g" = X - 1 for some A E F'. Because F is algebraically closed, X-' = p" for some p E F'. Thus (pij)" = 1. Since the elements I, pij, .. . ,(pij)"-' form an F-basis of F"G, we conclude that FOG Z FG.

6.3. Lemma. Let a E Z2(G,P)and let L be a field extension of F . Then the map

z@g

zg

is an isomorphism of L-algebras. Proof. Let {ijlg E G} be an F-basis of F"G with zij = a ( x , y ) Z y for all x , y E G. For each g E G, put ij = 1 8 9 . Then {jig E G) is an L-basis of L @F F"G and, for all x,y E G,

So the lemma is verified.

170

Induction from normal subgroups

We are now ready to prove the following useful result. 6.4. Proposition. Let G be a nonidentity group and let CY E Z2(G,F*).If charF = p > 0 and G is a p-group, then F"G is local.

The converse is true if F is algebraically closed.

Proof. Let charF = p > 0 and let G be a pgroup. If E is the algebraic closure of F, then F"G admits an injective homomorphism into E"G (Lemma 6.3). Since E is perfect of characteristic p and G is a pgroup, E"G Z E G by Lemma 6.2(ii). Hence, by Proposition 2.10.11, E"G is local. Thus F"G must also be local. Conversely, suppose that F is algebraically closed and F"G is local. Assume by way of contradiction that G has an element g of order n > 1 where n # 0 in F. Then, for H =< g >, we have F"H Z FH. Since e = (l/n)Ch,=Hh is a nontrivial idempotent of F H , we obtain the desired contradiction. Another useful observation is recorded in the following proposition. 6.5. Proposition. Let F be a field of characteristic p > 0 , let G be a p-group and let Q E Z2(G,F*). Then F"G/J(F"G) is a finite

purely inseparabze field extension of F.

Proof. Let E be the algebraic closure of F . Then, by Lemmas 6.3 and 6.2(iii)

Since G is a p-group and charF = p , we also have E G / J ( E G ) s E . Thus EaG/J(EaG) E E. Now put I = J(EaG) n F"G. Then I is a nilpotent ideal of F"G with F"G/I an F-subalgebra of E. Thus F"G/I is a finite field extension of F and therefore I = J(F*G). Furthermore, the field extension is purely inseparable since it is generated by the images of the element 3,s E G and these satisfy ijpn E F for some n. H Our next aim is to determine explicitly a basis for the centre Z(F"G) Q E Z2(G,F') and a basis ( Z l s E G}

of FOG. To this end, we now fix

6.

171

Twisted group algebras over fields

of F"G such that

iy = a(x,y)zy

for all x,y E G

An element g E G is said to be a-regular if

Expressed otherwise, g is a-regular if and only if

It is obvious that the identity element of G is a-regular. The following elementary properties of a-regular elements are worthwhile mentioning. 6.6. Lemma. (i) An element g E G is a-regular i f and only i f it is @-regular for a n y cocycle @ cohomologous t o a. (ii) If g E G is a-regular, t h e n so is a n y conjugate of 9.

Proof. (i) Let g be an a-regular element and let p be cohomologous to a, say @(x,y) = a ( x , y ) t ( . ) t ( y ) t ( x y ) - l for some t : G + F*. Then, for x E C G ( g ) ,

proving that g is @-regular. The converse is obvious. (ii) Let y E G and let x E C ~ ( y g y - ' ) . Then y-'xy E C G ( ~and ) therefore y - l z y . g = ji . y-1.y Since y - l x y = X ~ - ' C C for ~ some X = X(x,y) E F*, it follows that y-lzys = g y - l x y

Hence i commutes with ygy-'. Again, since ygy-1 = pyijy-l for some p = p ( g , y ) E F*, we see that i commutes with ygy-1. Thus ygy-l is also a-regular.

Induction from normal subgroups

172

Let Cgbe the conjugacy class of G containing g. We say that Cg is a-regular if g (and hence, by Lemma 6.6(ii), any element of Cg)is aregular. The set of all a-regular elements of G will be denoted by Go. We also say that the cocycle a is standard if it satisfies the following two properties: for all x E G

a ( x , x - ~= ) 1

for all g E G o , x E G

a ( x ,g ) a ( z g ,2 - 1 ) = 1

Expressed otherwise, a is standard if and only if the following two properties hold

___-

xgx

= xgx-1

for all

x E G,g E G"

6.7. Lemma. Let G be a finite group of order n , let F be an algebraically closed field of characteristic p 2 0, and let a E Z 2 ( G ,F*). If p divides n, let P be the largest normal p-subgroup of G. Then, after making a diagonal change of basis, if necessary, the following properties hold: (i) 9-' = ij-' for all g E G (ii) Z g 5 - l = xgx-l for all x E G,g E Go (iii) ~y = ~y for all x , y E P .

Proof. Owing to Lemma 6.2(ii), there exists t : P + F* such that a ( x ,y ) = t(z)t(y)t(xy)-'

for all

2,y

E

P

Let X : G + F* be defined by X(x) = t ( z ) - l if x E P and X(x) = 1 if x $ P . Then, replacing g by X ( g ) g , we may assume that (iii) holds. We now show that (ii) holds for g E P . Indeed, write 5g5-l = pxgx-l for some p E F* and put rn = ]PI. Then, by (iii),

gm whence

= xgx - 1 m = 1

6. Twisted group algebras over fields

173

Since m is a power of p , p = 1 and therefore (ii) holds. Suppose now that g E G" - P and let Cg = { g = g 1 , g 2 , . - . , g r } . Because g is a-regular, ij has r conjugates in I' = {XglX E F*,g E G}. Choose i j arbitrarily, take g 2 , . . . , g r as its conjugates in I' and leave other basis elements unchanged. Then we have conditions (ii) and (iii) holding. Due to (iii), to prove (i), we may assume that g $ P . Consider the elements of G - P and write G - P as a disjoint union

where S1 consists of the a-regular elements. For one of each pair 9,g-l E Sz with g # 9-', leave g unchanged and replace by g-' = a(g,g-')-'g-l. For each g E S2 with g = g-' replace i j by a ( g , g-')-'/Zg. Now consider g E S1 and write Cg = ( 9 = 91,. . . , g p ) . We still have above the choice of g at our disposal. If Cg # Cg-l, we choose ij,g-' asin the case g# 9-'. If Z g 2 - l = gi, then 2ij-lz-l = gi-' = g i ' , by choice of gi,gy'. Finally, let Cg = Cg-1. Thus 9-' = ygy-' for some y E G and 9-' - Y9Y (1)

----'

Replacing g by a(g,g-')-'12ij, and so all g; by c ~ ( g , g - ' ) - ' / ~ g ; , we still have (1) and also 9;' = g i - ' . This completes the proof.

6.8. Lemma. Let a E Z 2 ( G , F * ) ,let (91,. ..,st} be a full set of representatives for the a-regular classes of G and let Ti be a left transversal for Cc(g;)in G . Then the elements CzETi ZgiZ-', 15 i 5 t f o r m an F-basis for Z(F*G). In particular, if a is standard and C; is the conjugacy class of g i , then the elements

form an F-basis for Z(F*G). 7

Proof. Setting sgiz-1 = Zg;Z> z E Ti, 1 5 i 5 t , and = i j if g ----I is not regular, we have zgz - zgz-l for all z E G and all a-regular

Induction from normal subgroups

174

g E G. Thus we may assume that 3 g Z - l = xgz-' for all x E G and all a-regular g E G and we must show that the elements of (2) form an

F-basis for Z(F"G). To this end, put ki = EgECig, 1 5 i _< t . Let y = Eygg E F"G be the elements of F* such that and let

Then y is central if and only if for all x E G

9

9

Now put Suppy = (9 E GI?, # 0) and fix t E Suppy,a E Cc(t). If y is central, then comparing the coefficients of f, we get yt = ytX,,t. Hence = 1 and therefore t is a-regular. Since Zij5-l = zgz-l for all z E G and all a-regular g E G, it follows from (3) that y is central if and only if (i) Suppy consists of a-regular elements (4 r g s = 7gzg"-' = 7z-lgz9- for all z E G or, equivalently, if and only if y is an F-linear combination of k;,1 I: i 5 t. Since the elements Ici are obviously linearly independent, the result follows.

c

c

c

Let A be an F-algebra and define [A,A ] ,the commutator subspace of A , to be the F-linear span of all Lie products [x,y] = zy -yz, z, y E A. The following observation illustrates how the commutator subspace can be brought into argument. 6.9. Lemma. Let A be an algebra over a field of characteristic

P>O cz1 ay"(mod[A,A ] ) (i) ( c E 1 a*)'" (ii) I f s E [ A ,A ] , then s p E [A,A]

(a; E A , n E N)

Proof. (i) By induction argument, it suffices to show that (a

+ b)P" E a'" + bP"(mod[A,A ] )

175

6. Twisted group algebras over fields

for all a, b E A. Suppose first that n = 1. Expanding by the distributive law, we have (a+

b ) P = up

+ bp +

&la2

(4)

*"UP

where the sum is over all products a1 - + - a of p p terms, a; E { a , b } , not all equal to a or b. With each word a1a2 - - up associate its cyclic permutations

-

U1U2

*

- - U p , U2U3

* *

a

UpUl,

* *

,U p U l - - U p - l *

All these products are congruent modulo [A,A ] . This is so since for x = a;,a;, - . . u ; ~y , = a; a;,+,

-

1

a;,ql

- . . a;3-1

we have x - y = [y, 61 E [A,A ] , where y = a;, a,,

-

*

a;J-l and

S = aiJa;j+l

- a;,

Therefore the sum of these cyclic permutations is p x 1 - x p modulo [ A , A ] ,and hence it belongs to [ A , A ] .Applying (4),it follows that

and hence that

(ab - b ~ E) (~~ 6 ) ' - (ba)P(mod[A,A ] ) G [a,(ba)P-'b](rnod[A, A]) Accordingly

The required assertion now follows from (5) and (6) by induction on n. (ii) Apply (6). We next describe the commutator subspace of FaG. 6.10. Lemma. Let a be a standard cocycle and let y = E y Z z E FOG. Then y E [FOG,F"G] if and only if the sum of the coeficients yz over each a-regular class of G is zero.

176

Induction from normal subgroups

Proof. Denote by C1, ,C, the a-regular classes of G and let : F*G -+ F x F x - * x F (s times) be the F-linear mapping given

t,b

by

c

-

7 3 H (Ply ' ' ' P a )

where

Pi

=

c

7x

XECi

xEG

We must show that [F"G,F"G] = Kert+b. Observe that [F"G,F"G] is spanned by all Sy - yii with z,y E G. First assume that zy is not a-regular. Then so is yz since zy and yz are conjugate. But then

zy - yz = a ( x ,y) zy - a ( y , z)yz which shows that [ z , y ] E K e r $ . Suppose next that z y is a-regular. Then, bearing in mind that a is standard, we have

zy - yz = 5 y - z-l(zy)z = a ( z ,y )z y - a@,y)5-1"Yz = Q(Z, y ) ( a - z-'(zy)z) Hence [Z,y] E Kerzl, and therefore [F"G,F"G] KerG. To prove the opposite containment, we first demonstrate L a t i g E G is not an a-regular element, then g E [F*G,F*G].Indeed, let g be a nonregular element so that Zg # 33 for some x E cG(g). Because z commutes with g, we have zg2-l = X i j for some X E F* with X # 1. In view of the identity

g - 235-1 = (33-1 )z- - "35-1)

(7)

we deduce that (1 - X)g, and hence g, belongs to [F*G,F"G]. We are thus left to verify that if xi = CIECi y&i and if CIECi~i = 0, then x; E [ P G ,FaGI. The latter being a consequence of (7) and a vector space identity

if

xy=lX i = 0, the result follows. H As a generalization of Maschke's thpwern) we next prove

6.11. Proposition. Suppose that the characteristic of the field F does not divide the order of G. Then FOG is semisimple.

177

6. Twisted group algebras over fields

Proof. Let W be a submodule of an PG-module V . Since V is a vector space over F , its subspace W has a complement in V, say V = W@W'.Let 8 : V + W be the projection map, and let $ : V + V be defined by

$(v) = IGI-'

C z8s-lv

(v E V )

xEG

Because for all v E V,y E G,

$(yv)

C ZOZ-lijv = 1GI-l CjE6jjZ-1yv 1GI-l C i j Z ~ r ( yz, ) - ~ ~ c Y~ )( Z~ -, ~ i j - ~ i j v 1GI-l C i j Z 8 Z - l ~

= 1GI-l

xEG

=

ZEG

zEG

=

zEG

= Y+(v>,

it follows that $ is an FQG-homomorphism. Now assume that v E W . Then, for any x E G,Z-'v E W , so O(Z-lv) = Z-lv and thus ZOii-'v = v. Consequently, $(v) = v for all v E W . This implies that W is a direct summand of V and the result follows. H The next lemma is a final step in the preparation for the proof of the main result. 6.12. Lemma. Let A be a finite-dimensional algebra over an algebraically closed field of characteristic p > 0 and let S = [A,A].

Then

T = (5 E AlzPmE S

for some integer m 2 1)

is an F-subspace of A and the number of nonisomorphic irreducible A-modules equals dimF(A/T). Proof. Owing to Lemma 6.9(ii), a E [A,A]implies ap E [A,A]. Thus S C 2'. By Lemma 6.9(i), if a , b E T , then for sufficiently large m,

(a

+ b)pm = apm+ bp'"

3

0

(mod S)

178

Induction from normal subgroups

+

which shows that a b E T . Clearly, Xu E T for all X E F and a E 2'. Hence T is an F-subspace of A. Since the Jacobson radical J ( A ) of A is nilpotent, it follows from the definition of T that J ( A ) & T . Now put A = A / J ( A ) and T = T / J ( A ) .If r is the number of nonisomorphic irreducible A-modules, then there exists positive integers n l , . . . ,n, such that P

A 2 IIM,;(F) i=l

+

Now put = (S J ( A ) ) / J ( A ) Then . clearly S is the F-subspace of spanned by all ab - bu ( a ,b E A ) and

T = {z E A l s p "

ES

for some integer

A

rn 2 I}.

Let Si, 2';be the analogous F-subspaces defined for the F-algebra A; = M n i ( F ) .Then P

d i r n ~ ( A / T=) d i m ~ ( A / T=)

C dirn~(A;/T;) i=l

and therefore it suffices to verify that

Let est be the ni x ni-matrix with (s, t)th entry equal to 1 and all other entries 0. Then, for any s # t , est = esjejt - ejtesj E Si

and ess - ett = estets - etsest E S;

It follows that S; contains the n' - 1 linearly independent elements est(l 5 s , t 5 ni,s # t ) and ess - e l l ( s = 2 , . . . ,ni). Because S; C T.j, this shows that dirn~(A;/Ti)= d i m ~ ( A i-) d i m ~ ( T ;5) n: - (ns - 1) = 1 On the other hand, the elements of 5'; are all n; x n;-matrices with trace 0. In particular, ell # S;. But eyy = ell for all m 2 1, so

179

6. Twisted group algebras over fields

4

T,. Thus Ti # A;, so from dirnF(A/T;) 5 1, we deduce that dirn~(A;/T;) = 1, as desired. ell

In what follows, if p = 0 then by definition all elements of G are p'-elements. 6.13. Theorem. (Asano-Osima-Takahasi (1937)). Let G be a finite group, let F be an algebraically closed field of characteristic p 2 0 and let a E Z2(G,F * ) . Then the number of nonisomorphic irreducible F"G-modules is equal to the number ofa-regular classes of p'-elements of G.

Proof. Let r be the number of nonisomorphic irreducible F*Gmodules. If charF = 0 , then FaG is semisimple (Proposition S.ll), hence r = ( Z ( F " G ): F ) This proves the case charF = 0, by applying Lemma 6.8. Now assume that charF = p > 0 and put S = [F*G,F"G] and

T = { x E F"G1xPm E S

for some integer

rn 2 1)

Choose an F-basis a l , . . . , a t of T . We may find the integer m so large that afm E S for each i and pm is at least as large as the order of a Sylow p-subgroup of G. Owing to Lemma 6.9, we have

for all A; E F . Hence x = equivalently

xET

Cxgg E T if and only if xPm E S or,

if and only if

c

x;mXSpmgpmE

s

where, by our choise of rn, each gPm is a $-element and for each g E G, A, E F* satisfies g P m = A:mgpm. Let C1, Cz, . . . ,C, denote the a-regular classes of p'-elements of G, and Put Cz*= (9 E GlgPmE C;} (1 5 i 5 s )

Induction from normal subgroups

180

By Lemma 6.7, we may assume that CY is a standard cocycle. Applying (8) and Lemma 6.10, we therefore deduce that x = C xga E T if and only if /

\

(c

g€C?

pm

= XgXg)

c

(XgXg)Prn

=0

S€CY

for each i E (1,. . . ,s}. Thus xgg E T

if and only if

c

xgXg = 0

(9)

gECf

for all i E (1,. . . ,s}. Now define 1c, : F"G + F x the F-linear map given by

x F (s times ) as

with ,8; = C g E C f zgXg. Then 1c, is clearly surjective and, by (9), T is the kernel of +. Thus, by Lemma 6.12, r = dirnF(F"G/T) = s

and the result follows.

7.

Total and absolute indecomposability of induced modules

The notion of indecomposability of a module can be strengthened in two different ways. The first is absolute indecomposability in the sense of Huppert and the second is absolute indecomposability in the sense of Green. Since we shall use both notions, it will be convenient to rename the stronger one (namely absolute indecomposability in the sense of Huppert) as total indecomposability. Throughout this section, N denotes a normal subgroup of a finite group G, R a commutative local ring and A an R-algebra. Let V be an A-module and let S = EndA(V). We say that V is totally indecomposable if = R * 1s J ( S )

s

+

7.1. Lemma. The following conditions are equivalent: (i) V is totally indecomposable.

7. Total and absolute indecomposability of induced modules

181

(ii) J ( R ) S E J ( S ) and S / J ( S ) is ofR/J(R)-dimension I . (iii) J(R)S C J ( S ) and S / J ( S ) E R / J ( R ) as R/J(R)-algebras. (Note that if R is noetherian and V finitely generated, then J ( R ) S J ( S ) b y Lemmas f.7and 1.2).

+

Proof. (i)+(ii): By hypothesis, S = R 1s J ( S ) ,hence

S / J ( S ) 2 R . i s / ( R . is n J ( S ) )

-

Since R is local, so is its homomorphic image R 1s and hence Re 1s n J ( S ) J ( R . l s ) . Thus, by Corollary 1.4.4(i), R . l s n J ( S ) = J ( R - I s ) . SInce J ( R 1s) = J ( R ) ls, it follows that J ( R ) S C J ( S ) . Thus S / J ( S ) is an algebra over the field R / J ( R ) having dimension 1 since = R 1s -t J ( S ) . (ii)=+(iii): Obvious. (iii)+(i): Obvious.

-

-

s

It follows from Lemma 6.1 that if V is totally indecomposable, then

V is strongly indecomposable and hence indecomposable. Moreover, as a partial converse, we have

7.2. Lemma. Let V be finitely generated, let R be a complete noetherian local ring and let the jield R/ J ( R ) be algebraically closed. If V is indecomposable, then V is totally indecomposable.

Proof. Let V be indecomposable. By the proof of CaroDary V is strongly indecomposable, hence S / J ( S )is a division ring. By Lemma 1.7 and 1.2, J ( R ) S J ( S ) and so S / J ( S )is an R/J(R)-algebra. Furthermore, by Lemma 1.7, S is a finitely generated R-module. Hen= S / J ( S ) is a finite-dimensional algebra over the field R/J(R). Since R / J ( R ) is aIgebraically closed, it follows that S / J ( S ) is of R / J ( R ) dimension 1. Hence, by Lemma 1.7, V is totally indecomposable. W The next observation will justify our concentration on totally indecomposable modules.

7.3. Lemma. Let V be a totally indecomposable G-invariant R N module, let S = EndRN(V) and let E = EndRG(VG). Then

182

Induction from normal subgroups

(i) E - J ( S ) J ( E ) and A = E / E - J ( S ) is a twisted group algebra of G / N over the field R / J ( R ) . (ii) V G is strongly indecomposable if and only if A is local. (iii) V G is totally indecomposable if and only i f A/ J ( A ) is of R/ J ( R ) dimension 1. Proof. (i) This is a direct consequence of Theorem 4.2(iv), (v) and Lemma 7.l(iii). (ii)and (iii) Since E - J ( S ) 2 J ( E ) , we have E / J ( E ) E A / J ( A ) as R/ J ( R)-algebras. Hence E is local (respectively, E / J ( E ) is of R/ J ( R)dimension 1)if and only if A is local (respectively, A / J ( A ) is of R / J ( R ) dimension 1). Invoking some general facts on twisted group algebras, we now record the following consequences of the results so far obtained.

7.4. Theorem. Let V be a totally indecomposable G-invariant RN-module, let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. Then V G is strongly indecomposable. Proof. Let A be as in Lemma 7.3(i). Then, by Proposition 6.4, A is local. Hence, by Lemma 7.3(ii), V Gis strongly indecomposable.

7.5. Theorem. Let V be a totally indecomposable G-invariant RN-module. If V G is totally indecomposable, then either G = N or c h a r R / J ( R ) = p > 0 and G / N is a p-group. The converse is true if the field R/ J ( R ) is perfect. Proof. We may harmlessly assume that G # N . Let A be as in Lemma 7.3(i), let F = R / J ( R ) and let H = G / N . Then A F"H for some a E Z 2 ( H , F * )and, by Lemma 7.3(iii), V G is totally indecomposable if and only if

dimFF"H/J(F"H) = 1

(1)

Assume that (1) holds. Then, by Lemma 6.1(ii), F"H Z F H and hence F H is local. Since H # 1, we conclude from Proposition 2.10.11 that charF = p > 0 and H is a pgroup.

7. Total and absolute indecomposability of induced modules

183

Conversely, assume that F is perfect, charF = p > 0 and H is a F H . Hence (1) holds and pgroup. Then, by Lemma 6.2(ii), F a H the result follows.

7.6. Theorem. Let R be a complete noetherian local ring, let V be a finitely generated totally indecomposable RN-module and let H be the inertia group of V . If charR/ J ( R ) = p > 0 and H / N is a p-group, then V G is indecomposable.

Proof. Owing to Theorem 2.1, we may assume that G = H , in which case the result follows by virtue of Theorem 7.4. H The result above can be strengthened under the additional assumption on R. This is the content of the following theorem.

7.7. Theorem. Let R be a complete noetherian local ring such that the field R/ J ( R ) is algebraically closed. If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then the following conditions are equivalent: (i) V G is indecomposable. (ii) V G is totally indecomposable. (iii) H = N or c h a r R J J ( R )= p > 0 and HIN is a p-group. Proof. b y Lemma 7.2,(i) and (ii) are equivalent. Moreover, by Theorem 2.1, V G is indecomposable if and only if V H is indecomposable. Hence we may assume that G = H , i.e. that V is G-invariant. By Lemma 7.2, V is totally indecomposable. Moreover, since R is a complete noetherian local ring, V G is indecomposable if and only if V G is strongly indecomposable (see proof of Corollary 1.8). Hence, by Lemma 7.3(ii), V G is indecomposable if and only if the twisted group algebra A of that lemma is local. The desired conclusion is therefore a consequence of Proposition 6.4. Assume that V is a finitely generated totally indecomposable Ginvariant RN-module, where R is a complete noetherian local ring. Put F = R / J ( R ) ,E = EndRG(VG)and S = EndRN(V). Then, by

Induction from normal subgroups

184

Lemma 7.3(i),

E / E . J ( S )E F " ( G / N ) and

for some

CY

E Z 2 ( G / N ,F*)

c

E * J(S) J(E) By Theorem 1.5, any decomposition F"(G/N) = L1@

@ L,

into indecomposable right ideals of F " ( G / N ) can be lifted to a decomposition E = II@*..@I, into indecomposable right ideals of E. We are now ready to prove the following result essentially due to Conlon (1964). 7.8. Theorem. Let V be a finitely generated totally indecomposable G-invariant RN-module, where R is a complete noetherian local ring, Then, in the notation above, the following properties hold: (i) V G = @r=lI;V is a decomposition into indecomposable RGmodules such that

( I i v ) 2~ e;V

with e; = ranks(Ii) = dimFL;

(ii) I;V E IjV as RG-modules if and only i f L; % Lj as right F"(G/N)-modules. In particular, the number of nonisomorphic indecomposable direct summands ofVG is equal to the number of nonisomorphic irreducible Fa(GIN)-modules. (iii) If V is R-free, then each I;V is R-free and rankR(I;V) = (dimFL;)(rankR(V))

(1 5 i 5 n )

Proof. All of the stated properties will follow from Theorems 5.6, 1.5 and 2.12.4 provided we show that ranks(I;) = dimFL;

(1 5 i 2 n )

-

Because F a ( G / N )is identifiable with E / E J ( S ) ,the latter will follow provided we show that for e; = ranks(1;)

+

dimF(1; E J ( S ) ) / E- J ( S ) )= e;

(2)

7. Total and absolute indecomposability of induced modules

185

To prove (2), fix i E { 1,. . . ,n } and put ei = e. Then choose an S-basis { d l , . . . ,d e } of Ij and extend it to an S-basis

(this is always possible because all direct summands of E are free Smodules by Theorem 5.6(i)). Then we have

which implies that

Ij f l E J ( S ) = $;=,d;J(S)

+

-

Thus each element in (I; E - J ( S ) ) / E J ( S ) is a unique S / J ( S ) linear combination of the images of d l , . . . ,d , in ( I ; E J ( S ) / E J ( S ) . Invoking Lemma 7.1, we conclude that (2) holds and the result is established.

+ -

*

7.9. Corollary. Let R be a complete noetherian local ring such that the field F = R / J ( R ) is algebraically closed of characteristicp > 0 . If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then f o r

E = EndRH(VH)

and S = EndRN(V)

the following properties hold: (i) E / E . J ( S ) 2 F " ( H / N ) for some LY E Z 2 ( H / N , F * ) . (ii) The number of nonisomorphic indecomposable direct summands o f V G is equal to the number of a-regular classes ofp'-elements of H / N . Proof. (i) By Lemma 7.2, V is totally indecomposable. Now apply Lemma 7.3(i). (ii) By Theorem 2.1(i), we may assume that H = G, i.e. that V is G-invariant . The desired conclusion now follows by applying Theorems 7.8(ii) and 6.13.

For the rest of this section, we assume that R satisfies the following two conditions:

186

Induction from normal subgroups

(i) R is a complete local ring. (ii) R is a principal ideal domain. (for example, R can be a field, or the ring of P-adic integers, P a prime ideal of an algebraic number field). If R is not a field and R satisfies (i) and (ii), then R is said to be a complete discrete valuation ring. An integral domain S containing R is an extension of R, written SIR, if the following hold: (i) S is a principal ideal domain and a local ring. (ii) S is R-free. (iii) J ( S ) e= J ( R ) S for some integer e 2 1. We refer to e as the ramification index of SIR. In case e = 1, we say that SIR is unramified. Note that if SIR is an extension of R, then S / J ( S ) is a field extension of R / J ( R ) .Of course, if R is a field, then an extension SIR is simply an arbitrary field extension. Also, if R is a field, then all extensions S/R are unramified. We say that the extension SIR is finite if S is a finitely generated R-module. In this case, if R is a complete discrete valuation ring, then so is S. This is a consequence of a classical theorem of Cohen (1946). Let A be an R-algebra and let SIR be an extension of R. Then we Put As = S @ R A It is clear that A s is an S-algebra. Now assume that A is R-free of finite rank and let { q ,. . . ,a , } be an R-basis of A. Then the map

is an injective homomorphism of R-algebras. Identifying A with its image in As, it follows that each element of A s can be uniquely written in the form EKl X;a;, X i E S. If V is an A-module, then V . = S @ R V becomes an As-module under a module action

We say that V is absolutely indecomposable if for every finite extension SIR, Vs is an indecomposable As-module.

7. Total and absolute indecomposability of induced modules

187

7.10. Proposition. Let A be an R-algebra which is R-free of jinite rank, let V , W be finitely generated A-modules and let SIR be an extension of R. Then

If V = W , then the two sides of this are isomorphic S-algebras. Proof. Let { s;} be an R-basis of S . Each element in HornA (V,W ) , can be written uniquely in the form Csj 8 fi with fj E H o ~ A ( V , W ) and with finitely many f; # 0. Consider the map given by

+

HomA(V,w)S

HOmAs(VS,WS)

Then clearly 1c, is an S-homomorphism and, in case V = W , 1c, is also a ring homomorphism. If $(Cs; @I f j ) = 0, then Cs; 8 f;(v) = 0 for all v E V . Thus all f; = 0 and therefore 1c, is injective. Now fix X E H o ~ A ~ ( Ws). V S , For each i, define f; by

Because V is finitely generated, there are only finitely many nonzero fi. Clearly, f; E HornA( V,W ) for each i and $(C s; 8 fi) = A. Thus 1c, is surjective and the result follows. H 7.11. Lemma. Let A be an R-algebm which is R-free of finite rank and let SIR be a finite extension of R. Then, for E = EndA(V), (i) J ( E ) s J ( E s ) . (ii) E s / J ( E ) s ( E / J ( E ) ) sas S-algebras.

c

Proof. (i) By Lemmas 1.7 and 1.2, E is a finitely generated Rmodule with J ( R ) E 5 J ( E ) ,so E / J ( R ) Eis a finite-dimensional algebra over the field R / J ( R ) with J ( E / J ( R ) E )= J ( E ) / J ( R ) E .Hence J(E)" 5 J ( R ) E for some rn >_ 1. If sl,. . . ,,s E S and x1).. . ,x, E J ( E ) ,then we have

Induction from normal subgroups

188

and thus J ( E ) s / J ( R ) E sis a nilpotent ideal of Es/J(R)Es. Hence

and therefore J ( E ) s E J(Es). (ii) We may write S = Rsl @ Then

... @ Rs, for some sl,. . . ,s,

E S.

which implies that

as required.

7.12. Lemma. Let D be a division ring and let F be a subfield of Z ( D ) such that dimFD < 00 and D # F . If Xo E D - F , let f ( X ) be the rnonic irreducible polynomial over F with root Xo) and let K be a field extension of F by a root of f(X). Then K @ F D is not a division ring. Proof. By hypothesis, deg f ( X ) = n > 1 and

Since f ( X ) is reducible over K 8 1 %! K , we may write

and g and h are of degree at least 1. Owing to (3), we have

Since

7. Total and absolute indecomposability of induced modules

189

it follows that K 693 D has zero divisors. 7.13. Proposition. Let A be an R-algebra which is R-free of finite rank, let V be a finitely generated indecomposable A-module and

let E = EndA(V). (i) If V is totally indecomposable, then for a n y finite extension S I R , Vs is totally indecomposable. (ii) I f V is totally indecomposable, then V is absolutely indecomposable. (iii) There exists a finite unmmified extension S I R such that

where each Proof. 7.11,

I$ is a totally indecomposable As-module. (i) By hypothesis E / J ( E ) Z R / J ( R )and so, by Lemma

E s / J ( E ) sz S 6 9 R~/ J ( R )2 S / J ( R ) S Hence

Es/J(&)

=S/J(S),

proving (i) by applying Proposition 7.10. (ii) Direct consequence of (i). (iii) Given a finite unramified extension S I R , it follows that Vs is a direct sum of, say n ( S ) ,nonzero indecomposable As-modules, where 1 5 n ( S ) 5 dirnR,JcRl(V/J(R)V).We may therefore choose S so that n ( S ) is maximum. It is obvious that every component of Vs remains indecomposable when tensored with any finite unramified extension of S. Replacing R by S, we may thus assume that Vs is indecomposable for any finite unramified extension SIR. For any finite unramified extension S I R , put

Then D ( S ) is a division ring and a finite-dimensional S/J(S)-algebra. If T / S is a finite unramified extension, then T / R is a finite unramified extension. We now show that D ( T ) is a homomorphic image of T @s

Induction from normal subgroups

190

D W

Owing to Lemma 7.11(i), we have

VT

= ( t l @V S ) €B

..-€B ( t , €3 VS)

Consider the map X : T x EndA,( Vs) + EndA, (V T )given by

Then X obviously induces a well defined S-bilinear map

T x D ( S )+ D ( T ) and a homomorphism

Since by Proposition 7.10,

it follows that

is surjective. Thus D ( T ) is a homomorphic image of

T 63s D ( S ) . Choose a finite unramified extension K / R such that d i m ~ , ~ ( q D ( I < ) is minimum. If D ( K ) # l " / J ( K ) , then there is a monic irreducible polynomial f ( X ) E K / J ( K ) [ X ]of degree at least two having a root in D ( K ) . Choose fO(X) E K [ X ] such that the image of f o ( X ) in K / J ( K ) [ X ]is f ( X ) . Let S = K[b]where b is a root of f o ( X ) . It is easily verified that S / K is a finite unramified extension. By Lemma 7.12, S @K D ( K ) is not a division ring, hence D ( S ) is a proper homomorphic image of S € 3 D~ ( K ) and dirnsp(s)D(S)< d i r n ~ l ~ ( ~ ) D ( K ) , a contradiction. Thus D ( K ) = K / J ( K ) and the result follows. N

Lemma. For any two finite extensions S I / R and SZ/R, there is a finite extension T / R such that T is an extension of both S1 and S2. 7.14.

8. Crossed products over prime rings

191

Proof. This is an easy consequence of basic results in valuation theory. For a self-contained proof refer to Dornhoff (1972, p.328). W We are now ready to prove the following fundamental result.

7.15. Theorem. (Green (1959)) Feit (1962)). Let R be a complete local ring and a principal ideal domain. Let V be a finitely generated absolutely indecomposable R N - m o d u l e , let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. T h e n V Gis absolutely indecomposable.

Proof. Owing to Proposition 7.13(iii), we may choose a finite extension S I R such that every indecomposable component of (VG)s% ( Vs)" is totally indecomposable. Again, by Proposition 7.13(iii), there is a finite extension T / S such that VT is totally indecomposable. By Proposition 7.13(i), every indecomposable component of ( V G )=~( ( V G ) s ) , is totally indecomposable. But, by Theorem 7.6, (V"), is indecomposable. Hence ( V"), is totally indecomposable. If the theorem is false, then there is a finite extension K / R such that ( V G )is~decomposable. Invoking Lemma 7.14, we may choose a common finite extension L of T and K . By Proposition 7.13(i), (VG), is totally indecomposable. But K L , so (VG)Lis decomposable, a contradiction. 8.

Crossed products over prime rings

Throughout this section, G denotes a finite group, S a ring and S * G a crossed product of G over S. &call, from Proposition 3.4, that if (G, S,0,a ) is a crossed system of S * G, then S * G is a free (left and right) S-module freely generated by the elements ij,g E G, with multiplication given by

~ " s ~ ~ ~ , Y )( z~ Y1 ~ E3 2S, Z , Y E G) where a ( z , y ) E S and Is2 = a(z)(sz) E S. We now introduce some termonology. First of all, recall that an ideal f of S is said to be G-invariant if ( S I ~ : ) ( S ~= Y )s

gig-1 = I

for all g E G

Induction from normal subgroups

192

A G-invariant ideal I of S is said to be G-prime if III2 C I for Ginvariant ideals I; of S imply 11 E I or I2 E I. The ring S is said to be prime if for all nonzero ideals A, B of S we have AB # 0. An ideal P of S is called a prime ideal if SIP is a prime ring. A minimal prime ideal of S is a prime ideal P of S such that P is not properly contained in any other prime ideal. Our aim is to prove some purely ring-theoretic results concerning the structure of S * G, where S is a prime ring. These results will be applied in the next section to investigate the homogeneity of induced modules. 8.1. Lemma. If P is a prime ideal of S * G, then P n S is a G-prime ideal of S . Conversely) if A is a G-prime ideal of S ) then there exists a prime ideal P o f S * G such that P n S = A.

Proof. Observe that if I is an ideal of S * G, then since I is Ginvariant so is I n S. Conversely, if A is a G-invariant ideal of S , then by Theorem 3.6(ii), A . (S * G ) = ( S * G ) A = A * G is a two sided ideal of S * G with (A * G) n S = A. Assume that P is a prime ideal of S * G. Then, by the above, P n S is at least G-invariant. Now suppose that A and B are G-invariant ideals of S with AB P n S. Then (A * G ) ( B* G) = ( S * G)AB(S* G)

c c

( S * G)(P n S)(S * G) p,

c

so the primeness of P ensures that A * G P or B * G P. Thus A P n S or B C_ P n S, proving that P n S is G-prime. Conversely, let A be a G-prime ideal of S. Then A is a G-invariant ideal of S and so A*G is an ideal of S*G with (A*G)nS = A. Applying Zorn’s lemma, we may thus choose an ideal P of S * G maximal with respect to P nS = A. Let I , J be ideals of S* G properly containing P. Then I n S and J n S are G-invariant ideals of S properly containing A. Because A is G-prime, this impiies that

( I nS)(Jn S ) g A

193

8. Crossed products over prime rings

and hence I J

9 P.

Thus

P is prime as required.

In what follows, given a subset T of G, we write S * T for the set of elements of S * G with support in T . 8.2. Lemma. Let I be a nonzero left and right S-submodule of S* G and let T = {gl,g2,. . . , g n } be a subset of G with g1 = 1. Assume

that and I n ( S * T ' ) = O forall T ' c T

Iil(S*T)#O

For each i E { 1,. . . ,n } , define B; b y n

B; = { s E SI them exists p = C s j i j j E I

with s = sj}

j=1

Then (i) Each B; is a nonzem ideal of S . (ii) There exists a natural bijection f; : B1 4 B; such that for all s,t E S,a E B1 (a) f;(sat) = sfj(u)git (b) fl is the identity map. (iii) The elements of I n ( S * T ) are precisely the elements of the form:

c n

Q

=

f;(.)ji

with u E B1

i=l

Proof. Since I is a left and right S-submodule of S*G, it is clear that each B; is an ideal of S. Furthermore, the minimality of T implies that each B; is nonzero. Observe also that for each b; E B;, the minimality of T ensures that there exists a unique ,O = C s j i j j E I with si = b;. This is so since if p and p' are two such elements, then p - p' E I is an element of smaller support and so /? - p' = 0. By the foregoing, for each a E B1, there exists a unique CY = C;=,b& E I with bl = a . We may thus define fi : B1 4 Bi by setting f , ( a ) = b;. By the definition of f;,fl is the identity map and

Induction from normal subgroups

194

Conversely, it is clear that each element of I n (S * 2’) is of this form. Furthermore, each fi is clearly an additive bijection. Finally, let a E B1 and cy be as above and let s , t E S. Then

and since S; = 1, this implies that fj(sat) = sb;”t = s f ; ( a ) ” t

thus completing the proof. H For the rest of this section, S denotes a prime ring. In order to make further progress, we briefly discuss a certain ring of quotients M = Qo(S) which is defined in Martindale (1969). Consider the set X of all left S-module homomorphisms

f:A+S where A ranges over all nonzero ideals of S. We refer to two such homomorphisms f : A + S,g : B 3 S as equivalent if there exists a nonzero ideal C of S with C C A n B and with f(c) = g(c)

for all c E C

This relation is obviously reflexive and symmetric. It is also transitive, because for all nonzero ideals A, B of S, A n B is a nonzero ideal of S. We have thus defined an equivalence relation on the set X . For each f E X , let [f,A] (or f if no confusion can arise) denote the class of the function f above. We can define an arithmetic on the set &o(S) of these classes by

if,A1

[9, BI = [fs,BAI where fg is the composition first f then g . In this way we obtain a ring, called the Martindale ring of quotients of S. *

195

8. Crossed products over prime rings

The following five lemmas provide some basic properties of the ring Qo(S)* 8.3. Lemma. For any f : A + S in X , f = 0 if and only if f = 0. In particular, i f g : B + S is another element o f X , then f = ij

ifand o n l y i f f = g o n A n B . Proof. It is obvious that f = 0 implies f = 0. Conversely, suppose that f = 0. Then there exists a nonzero ideal B of S contained in A such that f ( B )= 0. Note that B A G B and so 0 = f ( B A )= B f ( A ) Because S is prime, it follows that f ( A ) = 0 and therefore f = 0. The second assertion follows from the first and the fact that f = g if and only if f - g = 0.

For every a E S, let

a, E

Homs(S,S) be defined by

a,(z) = sa.

8.4. Lemma. (i) The map S + Qo(S) d efined b y a H a, is an injective ring homomorphism. (ii) Iff : A + S is an element of X , then iirf = f ( a ) , .

+

Proof. (i) If a , b E S, then T, = 1, arb+- = (ab), and a, br = ( a b),. Hence the given map is a ring homomorphism. If a, = 0, then a, = 0, by Lemma 8.3. Thus a = 0 and (i) follows. (ii) For every a E A, Sa G A, hence f is defined on Sa. Therefore, for every x E S ,

+

proving that

- = f ( a ) r *W

~ , =f a,f

Thanks to Lemma 8.4(i), we may identify S with its image in Qo(S). With this identification, Lemma 8.4(ii) tells us that for f : A + S in

x,

af = f (4

for all a E A, f E QO(S)

(1)

196

Induction from normal subgroups

8.5. Lemma. Let M = Qo(S) be as above. (i) If m E M and Am = 0 for some nonzero ideal A of S , then

m=0. (ii) If ml,m2,...,mk E M , then there exists a nonzem ideal A of S with each Am; in S . (iii) M is a prime ring. (iv) If Q is an automorphism of S , then 0 extends uniquely t o an automorphism of M . (v) Z ( M ) = CM(S)and Z ( M ) is a field. Proof. (i) Let m E M with Am = 0. If m = f,then (1) shows that ! ( a ) = 0 for all a E A. Thus f vanishes on an ideal in its domain and therefore m = f = 0. (ii) Let ml,m2,. . . ,mk E M with m; = &. We may harmlessly assume that all the fi are defined on the common domain A. Then, by (11,

-

am; = afi = f;(a) E S

for all a E A. (iii) Let I be a nonzero ideal of M . If 0 # m E I , then by (ii) and (i), 0 # Am E I n S for some nonzero ideal A of S. Hence every nonzero ideal of A4 meets S nontrivially. Because S is prime, so is M . (iv) Let Q be an automorphism of S and let f : A + S. Then the map f":A"+S given by

mu) = f (4"

is clearly a left S-module homomorphism. From this we easily deduce that the map f H f" gives rise to an automorphism of M extending 0. To establish uniqueness of extension, it suffices to verify that if T is an automorphism of M fixing S elementwise, then T = 1. To this end, let m E M and let A be a nonzero ideal of S with Am C S. Then, for all a E A, we have

am = (am). = armT= am7 Thus A(m - m') = 0 and (i) implies that m7 = m. (v) Suppose that m E CM(S) and that m # 0. Because m cen-

197

8. Crossed products over prime rings

tralizes S, it is clear that T = {t E Sltm = 0) is a two-sided ideal of S and so we must have T = 0 by (i). Owing to (ii), we may choose a nonzero ideal A of S with Am C S. Then the map f : A -+ S defined by f(a) = am is an injective homomorphism of left S-modules. If B = f ( A ) ,then B = Am = mA since m E CM(S).Now there exists an inverse map g : B + S so that f g = 1 A . Since B is an ideal of S , ij E M and g is an inverse of f = rn. Moreover, because rn centralizes S, we must have 7n-l E C M ( S )and therefore C M ( S )is at least a division ring. Finally, conjugation by m induces an automorphism of M which is trivial on S. Hence, by (iv), the automorphism must also be trivial on M . Thus m E Z ( M ) and the result follows. H The preceding results indicate that the structure of Qo(S) is very close to that of S. Our next lemma will indicate that Qo(S) is large enough to contain certain needed additional units. 8.6. Lemma. Let CT be an automorphism o f S and let a , b E S be fixed nonzero elements. If for all s E S

as6 = bs'au

then there exists a unit m E M = Qo(S) such that b = am such that m-'sm = so

for all s E S

Proof. Put A = SaS,B = SbS and define the maps f : A + B and g:B+Aby

To prove that f is well-defined, it suffices to verify that C z i a y i = 0 implies Cx;by,O= 0. To this end, suppose that C x i a y i = 0. Then, for all s E S, the formula asb = bs'au yields

198

Induction from normal subgroups

and so C xiby: = 0 since a" # 0 and S is prime. Similarly, if C xiby; = 0, then for all s E S, we have

and we conclude that Cx;ayp-' = 0. Thus both f and g are welldefined and since they are clearly homomorphisms of left S-modules, we have f = m E M and ij E M . Furthermore, fg = 1~and g f = lg, so g = rn-l and m is a unit of M . Note that a,f is defined on S and for all x E S we have

Thus ii,f = b,, or equivalently, am = b. Finally, let c E S. Then gcrf is defined on B and for all xby E B we have

Thus gC,f =

and so rn-lcrn = c", as required.

8.7. Lemma. Let M = &O(S), let c be an automorphism of S and let A,B be nonzero ideals of S . Suppose that f : A + B is an additive bijection which satisfies f ( s a t )= s f ( a ) t " for all s,t E S and a E A . Then m = f is a unit an M , m-lsm = s" for all s E S and f ( a ) = am for all a E A .

Proof. Because f is an additive bijection, its inverse g : B + A is also an additive bijection. Furthermore, for all s , t E S and b E B , g(sbt) = sg(b)t"-' Observe also that f and g are left S-module homomorphisms so that m = f and g are elements of M . Moreover, because g is an inverse cif

8. Crossed products over prime rings

199

f,we have g = m-l. Let q E S. Then g q r f is defined on B and for all b E B , we have

Thus g q r f = q: and this yields rn-lqrn = q' for all q E S. Finally, because m = f we have f(a) = a m for all a E A. Let CT be an automorphism of S. Following Kharchenko (1975), we say that is X-inner if it is induced by conjugation by a unit of M = Qo(S). In other words, these automorphisms arise from those units m E M with m-lSm = S.If ml and m2 are two such units, then clearly so is mlm2. Applying Lemma 8.5(iv), we deduce that the set of all X-inner automorphisms of S is a normal subgroup of A u t ( S ) . Now let S * G be given and let G* be the group of graded units of S * G. Then G* acts on S by conjugation and the elements of U ( S ) surely act as X-inner automorphisms. Taking into account that G ' / U ( S ) 2 G, we conclude that

Ginn = { g E GIs I+

is an X-inner automorphism of

gs

S}

is a normal subgroup of G.

By Lemma 8.5(iv), the automorphism s H g s of S extends to a unique automorphism of M which we denote by the same symbol. We now define M*G to be the free M-module freely generated by (31s E G } and with multiplication given by (az)(@) = a"ba(x, y)zy

for all a , b E M and x , y E G. Here of course a : G x G + U ( S ) is given by

a ( x , y ) = zyzy- 1

The first part of the lemma below demonstrates that M G over M extending S * G.

* G is in fact

a unique crossed product of

Let M = Qo(S) and let F = Z ( M ) . (i) There exists a unique crossed product M * G extending S (ii) If E = C M * G ( M ) , then E C M * G;,,, M * Ginn = M

8.8. Lemma.

* G. @F

E

Induction from normal subgroups

200

and E = FtG;,,, some twisted group algebra of Gin,, over the field F . (iii) If H is a subgroup of Gin,, then

M

*H

=

( E n M * H ) ) and E n ( M * H ) = F t H

(iv) If L is a G-invariant ideal o f E , then L ( M * G ) = ( M * G ) L is an ideal ofM*G. Furthermore, (M*G)L considered as a left M-submodule of M * G is a direct summand of M * G . Also, (M*G)Ln(M*Gi,,) = M L and if L # E , then ( M * G ) L n M = 0. Proof. (i) Owing to Lemma 8.5(iv), the automorphisms of S extend uniquely to those of M . Thus, the above definition of M * G is the only possible extension of S * G. We are therefore left to verify the associativity of the multiplication. Let G* be the group of graded units of S * G. observe that G* acts on S and therefore the uniqueness of extension ensures that we obtain a group action of G* on M . This fact will be used implicitly in the computations below. Given a , b, c E M and x, y, z E G, we have

-

[ ( a i ) ( b i j ) ] ( c .= ~ )mlxyz

and

( a i )[ (bij)(cz)] = m a w for some ml,m2E M . An easy computation shows that ml = a"ba(x, y)"Yca(x, y)-lsl

and m2

= a"b"( YC)SZ

where s1 and s2 are elements of S independent of a , b, and c. Because

the a , b,and c terms in the two expressions are equal. But in the special case when a = b = c = 1 the above products belong to S * G and so s1 = s2. Hence ml = m2 and M * G is associative. (ii) Let y E E and let x E Suppy, say y = rnI .. By Lemma

+-

201

8. Crossed products over prime rings

8.5(i),(ii), there exists a E S with ay E S Because y commutes with sa, we have asay = aysa

* G and with x E S u p p a y .

for all s E S

We deduce therefore that

asamii = amiisa = am"s"aii Since b = am is a nonzero element of S, we see that the identity

asb = bxsxa holds for all s E S. Applying Lemma 8.6, we deduce that s H "s is an X-inner automorphism of S and thus x E Ginn. This proves that E C M * Gin,. For each g E G;,,, choose a unit m, E M inducing the automorphism s H g s on S and put ij = mi's. Then the elements 6 for all g E Gin, form an M-basis for M * G;,,. The elements # , g E Ginn are obviously F-linearly independent and we claim that they form an F-basis for E. To this end, observe that each ij is a unit in M * G which acts by conjugation on M centralizing all of S. Invoking Lemma 8.5(iv), we conclude that j must centralize all of M , i.e. j E E . Suppose that y E E C M * Gin,. Then we can write y = C-y,ij with y, E M and it is clear that each ygij centralizes M . Because ij is a unit of E , we have y, E M r ) E = F and we infer that the elements form an F-basis for E . Furthermore, because F C Z ( E ) we must have M * Gin, = M @= E . To establish the Iast assertion, note that E is an associative Falgebra with basis (519 E Gin,}. Furthermore, for s,y E G;,,, itg E E and itij = m ( z , y ) @ for some m ( z , y ) E M . It follows that E is isomorphic to FtG;,,, some twisted group algebra of Gin, over the field F . (iii) This is a direct consequence of the way the algebra FtG;,, is constructed. (iv) Let L be a G-invariant ideal of E. Then, by definition of E , we have LM = M L and the G-invariance yields L j = gL for all g E G.

Induction from normal subgroups

202

We deduce therefore that L(M * G) = ( M * G)L is an ideal of M

* G.

Let Y be a transversal for Gin, in G and let L' be the F-complement for L in E. Because M * Ginn = M @gF E = M L @ ML', we have

M

*G=CM YEY

L @~

CML'~ YEY

Moreover, since L is G-invariant,

( M * G)L = C

( M * G;,,)Ly

=

Y€Y

C MLij Y €Y

which implies that ( M * G)L is an M-direct summand of M thermore, C y e yMLjj is clearly a direct sum and so

* G.

Fur-

( M * G)L n ( M * Gin,) = M L If L # E , then we can choose L' above to contain the identity element 1. But then CyEY ML'Y contains M and hence ( M * G)L n M = 0, as required. From now on, we shall employ the following notation:

M = &o(S),E= CM*G(M) and F = Z ( M ) We wish to apply the foregoing results in order to establish a bijective correspondence between the prime ideals P of S * G satisfying P n S = 0 and the G-prime ideals of E. Recall that the group G acts on E by the rule: g x = gxg-1 for all x E E , g E G and that, by definition, a G-invariant ideal A of E is G-prime if A1 A2 A for G-invariant ideals A; of E implies that A1 C A or A2 E A. If L is a G-invariant ideal of E , then we put L" = L ( M

* G) n S * G

so that L" is an ideal of S * G by Lemma 8.8(iv). For any ideal I of S * G we put

I d = { 7 E El Ay 5 I

for some nofizero ideal A of S}

203

8. Crossed products over prime rings

We claim that I d is a G-invariant ideal of E . Indeed, assume that y1,y2 E I d with Alyl I , A272 I and let t E E. Owing to Lemma 8.5(ii), there exists a nonzero ideal B of S with B t C S*G. Since y 1 , ~ 2 and t centralize S, we have

AIB(7lt) = (AlTl)(Bt)c I and

BAl(tT1) = (Bt)(Al71)c I

+

Thus 71 y 2 , y l t , t ~ 1E I d so I d is an ideal of E which is clearly Ginvariant, as claimed. Let I be an ideal of S * G. Following Lorenz and Passman (1979), we say that I is S-cancelable if for every x E S * G and any nonzero G-invariant ideal A of S , Ax I implies that x E I . It is obvious that any prime ideal P of S * G satisfying P n S = 0 is S-cancelable.

c

Let L, Ll,and L2 be G-invariant ideals of E . Then (i) L" is S-cancelable. (ii) L;"Ly (L1L,)".

8.9. Lemma.

Proof. (i) Suppose that A z C L" for some x E S * G and some nonzero G-invariant ideal A of S. By definition of L", we have Ax C ( M * G)L. By Lemma 8.8.(iv), ( M * G)L is a left M-module direct summand of M * G and hence

M

* G = ( M * G ) L@ K

for some M-submodule K of M ( M * G ) L ,x 2 E I<, we have

* G.

Writing x =

21

A ( x - X I ) = Ax2 E ( M * G ) L f l I< = 0 Applying Lemma 8.5(i), we see that x = z1 and thus

z E ( M * G)L n S * G = L",

+ x2 with

XI

E

Induction from normal subgroups

204

proving (i). (ii) Owing to Lemma 8.8(iv), we have ( M * G)Li = Li(M * G) and so

LyL; C L i ( M * G)L2(M * G ) = L1L2(M * G ) Moreover, LyLi C S * G and thus

LyLy

c L1L2(M * G ) n S * G = (LiL2)",

as desired. As a final preparatory result, we now record the following technical lemma. 8.10. Lemma. With the notation above, we have (i) For any G-invariant ideal L of E , L = LUd. (ii) If I is an ideal of S * G, then I Id".

c

Proof. (i) Given y E L , it follows from Lemma 8.5(ii) that By C S * G for some nonzero ideal B of S. Then By = L ( M * G ) n S * G = L", so y E LUdand therefore L 5 LUd.Conversely, if y E Lud,then A y C L" for a nonzero ideal A of S. Thus

by Lemma 8.8(iv). Let L' be an F-complement for L in E and write y = y1 t 7 2 with y1 E L and 7 2 E L'. Then

and Lemma 8.5(i) implies that y = y1 E L. (ii) Fix x E I. The case )SuppxJ= 0 being trivial, we show that x E Id" by induction on ISuppxl. Suppose that x # 0 and that the result is established for all elements y E I of smaller support size. Choose T C Suppx minimal with respect to the property that I n S * T # 0. We may assume that 1 E .'2 Indeed, if y E T , then suppxy-' = (Suppx)y

-' 3- Ty-l

8. Crossed products over prime rings

205

Now Ty-* also has this minimal property and 1 E Ty-'. Because it suffices to verify that xy-' E Id", we can replace x by sj-l and T by

Ty-'. If T = {gl = 1,g2,. . . ,g,}, then Lemma 8.2 applies and we employ its notation. There exist nonzero ideals A = B1, B2,. . . ,B, of S and additive bijections fi : A + Bi satisfying f ( s a t ) = sfi(a)git

for all a E A , s , t E S

Thus, by Lemma 8.7, there exist units ml, m2,. . . ,m, of M such that m-lsm = g r s for all s E S and f;(u) = am* for all a E A. Moreover, since fl is the identity function, we have ml = f 1 = 1. Let ,f3 = Crnig; E M * G. Because gls = gjsg;' for all s E S, we see that migi centralizes S. But m;gi then yields an automorphism of M acting trivially on S. Applying Lemma 8.5(iv), we deduce that mig; E E = CM*G(M) and thus p E E. Furthermore, since f i ( a ) = ami for all a E A, it follows from Lemma 8.2(iii) that for all a E A

Thus, by definition, we have ,f3 E I d . Let s be the identity coefficient of x and let a E A . Put 7 = ax - aps. Since Suppp = T C_ Suppx, 1 E T and ml = 1, we see that ISuppyl < ISuppzl. Moreover, y E I since u p E I , so by induction we have 7 E Id". Observe that /3 E I d and ups E S * G so ups E Id" and thus ax E Id" for all a E A. Thus, if

then D is a nonzero G-invariant ideal of S with Dz C_ Id". But, by Lemma 8.9(i), Id" is S-cancelable and therefore x E Id". The desired conclusion now follows by induction. We are now ready to prove the main result of this section. 8.11. Theorem. (Lorenz and Passman (1979)). Let S * G be a crossed product of the finite group G ouer the prime ring S and let E = FtGinn be the centratizer of M = Qo(S) in M * G where F = Z ( M ) .

Induction from normal subgroups

206

Then the map P I-+ Pd yields a bijective correspondence between the prime ideals P of S * G with P n S = 0 and the G-prime ideals of E . The inverse of this map is given b y L H L". Proof. Let P be a prime ideal of S * G with P n S = 0, and let L be a G-prime ideal of E. We must show that (i) Pd is a G-prime ideal of E and P = Pdu. (ii) L" is a prime ideal of S * G with L" n S = 0 and L = Lud. We first demonstrate that P = Pdu. Owing to Lemma 8.10(ii), we have P C Pd". Suppose that z E Pdu. Then z E S * G and z E (A4* G)Pd so

By definition of P d , there exist nonzero ideals D;of S with Di& = G;Di C P . Setting D = r)gEGnr=l gD;, it follows that D is a nonzero Ginvariant ideal of the prime ring S with 6;D P for all i. Furthermore, by Lemma 8.5(ii), there exists a nonzero ideal B of S such that BP; C S * G for all i and by the foregoing argument we may also assume that B is G-invariant. The conclusion is that

( B * G ) z ( D* G) = ( S * G ) B z D ( S* G) 5 C ( S * G)BPiSjD(S * G ) i

But P is prime and P does not contain either B * G or D * G since P n S = 0. Hence z E P , so Pdu 5 P and thus P = Pd". To show that P d is G-prime, suppose that MIA42 C_ P d for some G-invariant ideals MI, M 2of E . Then, by Lemma 8.9(ii),

M,"M," C_

(MlM2)"

2 PdU= P

Since P is prime, this implies Mi" C P for some i and, by Lemma 8.10(i), Mi = MTd 5 P d . Thus Pdis G-prime and (i) is established. Because E is a finite-dimensional algebra over the field F , we see that L is in fact G-maximal. Moreover, it is an immediate consequence

8. Crossed products over prime rings

207

of Lemma 8.8(iv), that L" n S = 0 and, by Lemma 8.10(i), we have L = Lad. Suppose that L" c I for some ideal I of S * G. Then I d 2 LUd= L and I d # L since otherwise Lemma 8.10(ii) would yield I Id" = L". Thus the G-maximality of L implies that I d = E and therefore I n S # 0. To prove that L" is prime, assume that L" C 11, L" C I2 for some ideals I1,12 of S * G. Then, by the above, 1 1 n S # 0 and 1 2 n S # 0, hence 0 # ( I , n S ) ( I zn S ) 1112n S

c

Since L" n S = 0, we deduce that result is established. 4

1112

L". Thus L" is prime and the

As an easy application of the results so far obtained, we finally record

8.12. Theorem. (Lorenz and Passman (1979)). Let S * G be a crossed product of the finite group G over the prime ring S . (i) A prime ideal P of S * G is minimal if and only if P n S = 0 . (ii) There are only jnitely many such minimal prime ideals, say Pl, .. . ,P,. Furthermore, if chars = p is a prime and G is a p-group, then n = 1. (iii) J = n;=, Pi is a unique largest nilpotent ideal of S * G. Proof. Because E = FtG;,, is a finite-dimensional algebra over the field F , there are only finitely many, say L1,. . . , Ln, of G-prime ideals of E . Furthermore, if chars = p is a prime and G is a p-group, then since charF = chars, it follows from Proposition 6.5 that E has a unique prime ideal. Thus in this case E has a unique G-prime ideal and therefore n = 1. Returning to the general case, we obviously have J ( E ) = n;=lLi. If P; = Ly , then by Theorem 8.11, PI, . . . , P, are the unique prime ideals of S * G which are disjoint from S - (0). Put J = n?.=,P;.Then J C Ly so J d C Lyd = L; by Theorem 8.11 and therefore

J d C n;==,Li =J(E)

208

Induction from normal subgroups

Applying Lemma 8.10(ii), we see that J C Jdu J(E)". Because J ( E ) is nilpotent, we deduce, from Lemma 8.9(ii), that both J ( E ) " and J are nilpotent. On the other hand, each Pi surely contains all nilpotent ideals of S * G. Thus J 2 J(E)" and hence J = J(E)" is clearly the largest nilpotent ideal of S * G. Finally, let P be any prime ideal of S * G. Because J is nilpotent, P 2 J and so P 2 P; for some i. Thus the minimal ideals of S * G are the minimal members of the set {PI,. . . ,P,}. But P; 2 Pj implies that L; = pi" 3 - pd3 = Lj and thus, since Lj is G-maximal, we must have i = j . This shows that P I , .. . ,P, are precisely the minimal prime ideals of S * G, thus completing the proof. 9.

Homogeneity of induced modules

Throughout this section, N denotes a normal subgroup of a finite group G and R a commutative ring which is either artinian or a complete noetherian local ring. Let V be a finitely generated indecomposable RN-module. Our aim is to discover circumstances under which the induced module V Gis homogeneous, i.e. V Gis a direct sum of isomorphic indecomposable modules. The following ring- theoretic result will clear our path. 9.1. Theorem. Let p be a prime and let S*G be a crossed product of a finite p-group G over a simple artinian ring S of characteristic p . Then S * G is an artinian ring such that S * G / J ( S * G ) is simple.

Proof. Because S * G is a finitely generated S-module and S is artinian, S * G is certainly artinian. Since S is simple, S is also prime. Applying Theorem 8.12(ii), we deduce that S * G has a unique minimal prime ideal, say P . Again, by Theorem 8.12(iii), P is a unique largest nilpotent ideal of S * G. But S * G is artinian, hence P = J ( S * G). Because P is prime, we deduce that S * G / J ( S * G) is a prime ring. But S * G / J ( S * G) is semisimple artinian, hence a direct product of, say n, simple rings. If n > 1, then there are two nonzero ideals of

9. Homogeneity of induced modules

209

S * G / J ( S * G) whose product is zero, which is impossible. Hence S * G/J(S * G) is simple and the result follows. The following result in the special case where R is a perfect field of characteristic p > 0 was established by Harris (1986). The general case is due t o Karpilovsky (1989).

Let R be a commutative ring which is either artinian or a complete noetherian focal ring and assume that R / J ( R ) is of prime characteristic p . If N is a normal subgroup of a finite group G and V a finitely generated indecomposable RN-module such that H/N is a p-group, where H is the inertia group of V , then 9.2. Theorem.

for some positive integer e and some indecomposable RG-module U . Proof. Owing to Theorem 2.1, we may harmlessly assume that H = G, in which case GIN is a p-group and V is G-invariant. Put

E = E ~ ~ RVG G )( , S = E n d ~ ~ ( vand ) A =E/E J(S) Because V is G-invariant, A is a crossed product of GIN over S / J ( S ) and E J ( S ) C J ( E ) , by Theorem 4.2(iv). Observe also that, by Lemmas 2.11.2 and 1.7, and Corollary 1.6, S is local. Hence S / J ( S ) is a division ring. Furthermore, J ( R ) S C J ( S ) by Lemmas 1.2 and 1.7. Thus S / J ( S ) is of characteristic p and therefore, by Theorem 9.1, A / J ( A ) is simple. Moreover, A / J ( A ) is artinian since it is a finitely generated R/J(R)-module. But

J ( A ) = J ( E ) / E- J ( S ) since E . J ( S ) C J ( E ) . Thus A / J ( A ) E E / J ( E )and therefore E / J ( E ) is simple artinian.

. . ,(p7 be the primitive idempotents of E corresponding to Let yl,. the decomposition of V G into indecomposable direct summands. By Lemma 5.2(iii), it suffices to show that p;E S pjE for all z , j E { 1 , . . . , T } . Let pi be the image of (pi in E = E / J ( E ) . By Proposition 1.4.17 and Theorem 1.5, it suffices to verify that &; E p j for

Induction from normal subgroups

210

all i , j . But each (pi is a primitive idempotent of the simple artinian algebra El hence the result. W

Remark. Assume that R is the field of two elements, G the symmetric group of degree 3 and N the alternating group of degree 3. Then there exists a unique (up to isomorphism) irreducible RN-module of dimension 2. It is well known (see Huppert and Blackburn (1982, Ex.70)) that V G2 2U where U is the unique irreducible RG-module of dimension 2. Consequently, the positive integer e of Theorem 9.2 need not be equal to 1. 10.

F’robenius and symmetric algebras

Our intent here is to establish a basic knowledge about Frobenius and symmetric algebras. All the information recorded will be applied in the next section. Throughout this section, we set A* = l $ o r n ~ ( A , F )where , A denotes a finite-dimensional algebra over a field F . The left (right) Amodule means the regular left (right) A-module. 10.1. Lemma. Let V be a finite-dimensional space over a field F and let f : V x V + F be a bilinear form. Then the following conditions

are equivalent: (i) f (z, V ) = 0 implies z = 0. (ii) f (V,z) = 0 implies x = 0. Proof. Let el, e 2 , . . . ,en be a basis of V and let b;j = f (ei, e j ) . Then it is clear from the bilinearity that f ( z , V ) = 0 if and only if f(z,e i ) = 0 , 1 _< i 5 n. Similarly, f ( V ,z) = 0 if and only if f ( e ; ,z) = 0, 1 5 i 5 n. Now write z = C A j e j with A j E F . Then

Hence f ( x , V ) = 0 if and only if (A1, A 2 , . . . , A,) system of homogeneous linear equations

is a solution of the

211

10. Frobenius and symmetric algebras

Similarly, f (V, z) = 0 if and only if the A's satisfy bilA1

+

+ + binA,

= 0,

15 i 5 n

(2)

We know from linear algebra that (1) or (2) has only a zero solution if and only if det(b;j)# 0. Thus the lemma is proved. We refer to a bilinear form f : V x V + F as being nonsingular if it satisfies the equivalent conditions of Lemma 10.1. Let f : A x A 4 F be a bilinear form. Then f is said to be associative if f ( z y , z ) = f (5,y z )

for all z, y , z E A

while f is called symmetric if f(Z,Y)

=f ( Y 4 )

for all 3 , Y E A

10.2. Corollary. Let $ E A* and let f : A x A 3 F be defined by f(z,y) = $(zy). Then f is an associative bilinear form on A which is symmetric if and only if $ ( z y ) = $(yz) for all z, y E A . Furthermore, the following conditions are equivalent: (i) KerlC, contains no nonzero right ideals of A. (ii) Ker$ contains no nonzero left ideals of A . (iii) f is nonsingular.

Proof. It is obvious that f is a bilinear form on A. Given z, y, z E A , we have f ( z y 7 z ) = Itt((zy)z) = Itt(+Z))

=f

(Z7 Y Z ) ,

proving that f is associative. By definition, f(z, y) = f ( y , 5) if and only if $ ( z y ) = $(yz). Hence f is symmetric if and only if $ ( z y ) = $(yz) for all z , y E A . Note that f ( z , A ) = 0 if and only if $ ( z A ) = 0, while f ( A , z ) = 0 if and only if $ ( A x ) = 0. The desired conclusion is therefore a consequence of Lemma 10.1. 10.3. Lemma. The following conditions are equivalent: (i) There exists $ E A* such that Ker$ contains no nonzero right

212

Induction from normal subgroups

ideals of A. (ii) There exists $ E A" such that Ker$ contains no nonzero left ideals of A. (iii) There exists a bilinear form f : A x A + F which is both associative and nonsingular.

Proof. (i)+(ii): Apply Corollary 10.2. (ii)+(iii): Apply Corollary 10.2. (iii)+(i): Let ll, : A + F be defined by $(z) = f(z,1). It is obvious that ll, E A*. Let a E A be such that $ ( a A ) = 0. Then f ( a b , 1) = 0 for all b E A. But f(ab, 1) = f ( a , b), so the nonsingularity of f implies that a = 0. H We say that A is a Frobenius algebra if it satisfies the equivalent conditions of Lemma 10.3.

The following conditions are equivalent: (i) There exists $ E A* such that Kerll, contains no nonzero right ideals of A and ll,(xy) = +(yz) for all z , y E A . (ii) There exists E A* such that ker$ contains no nonzero left ideals of A and $ ( x y ) = ll,(yz)for all z, y E A . (iii) There exists a bilinear form f : A x A * F which is associative, nonsingular and symmetric. 10.4. Lemma.

+

Proof. (i) +(ii): Apply Corollary 10.2. (ii) +(iii): Apply Corollary 10.2. (iii) +(i): Define II, : A + F by $(z) = f ( z , l ) . Then, by Lemma 10.3, $ E A* and ICerII, contains no nonzero right ideals of A. Furthermore, since f is symmetric, we have $(ZY)

= f b y , 1) = f (z, Y ) = f ( Y , 2) = f ( y z ,1) = $ ( Y 4

as required. We say that A is a symmetric algebra if it satisfies the equivalent conditions of Lemma 10.4. In order to avoid exessive verbosity, it will be convenient to say that ( A , $ ) is a Frobenius (or symmetric) algebra whenever $ satisfies conditions of Lemma 10.3(or Lemma 10.4).

10. Frobenius and symmetric algebras

213

As a preliminary to the next observation, note that A* is an ( A ,A)bimodule via

for all z , a E A and f E A*. Lemma. (i) Suppose that ( A , $ ) is a Frobenius algebra. Then the map f : A + A* given b y f ( a ) ( z )= $ ( z a ) for all z, a E A is an isomorphism of left A-modules. Moreover, if ( A , $ ) is a symmetric algebra, then f is an isomorphism of (A,A)-bimodules. (ii) Suppose that f : A + A* is an isomorphism of left A-modules and set $ ( a ) = f ( l ) ( a ) for all a E A. Then (A,$) is a Frobenius algebra. Moreover, i f f is an isomorphism of (A,A)-bimodules, then ( A , + ) is a symmetric algebra. 10.5.

Proof. (i) It is clear that for all a E A , f ( a ) E A*, and that f (a: t y ) = f (a:) f ( y ) for all a:, y E A. Given r, z E A, we have

+

proving that f is a homomorphism of left A-modules. By the definition of +, f is injective, and since dirnFA = dimFA*, f is a bijection. If ( A , $ ) is a symmetric algebra, then for all r,z E A, we have

as required.

(iii) It is obvious that 1c, E A*. Suppose that $(aA) = 0. Then, for all x E A , 0 = f ( l > ( a a : )= f ( z ) ( a ) and hence a = 0. Thus ( A , $ ) is a Frobenius algebra. Assume that f is an isomorphism of (A, A)-bimodules. Then, for all x , y E A , we have

as desired. W

Induction fkom normal subgroups

214

10.6. Corollary. The following conditions are equivalent: (i) A is a Frobenius (respectively, symmetric) algebra. (ii) A E A* as left A-modules (respectively, A E A* as ( A , A ) -

bimodules). Proof. Direct consequence of Lemma 10.5. W

n:==,

10.7. Lemma. Let Al,Az,. . . ,A, be F-algebras. Then Ai is a Frobenius (respectively, symmetric) algebra i f and only i f each A; is a Frobenius (respectively, symmetric) algebra.

Proof. Assume that $; E A* is such that (A;,$;)is a Frobenius A; and define : A + F by algebra, 1 5 i 5 n. Set A =

n?=,

+

Then obviously $ E A*. Put a = ( a l , . . . , a , ) and assume that +(Aa)= 0. Let z = ( 2 1 , . . . ,z,) E A be such that zj = 0 for j # i. Then $ ( x u ) = $;(zja;) = 0 for all zi E A;. Hence each ai = 0 and therefore a = 0, proving that ( A ,$) is a Frobenius algebra. Conversely, if ( A ,$) is a Frobenius algebra, put $; = $ o X i where X i : Ai + A is the canonical injection. Then (Ai,$;) is obviously a Frobenius algebra. If each ( A ,$i) is a symmetric algebra, then for a = ( a l , . . . ,an),b = ( b l , . . . ,b,) we have

proving that ( A ,$) is a symmetric algebra. Conversely, if ( A ,+) is a symmetric algebra, then for ai, b; E A, we have

Thus each (Ai,$;)is symmetric and the result follows.

10.8. Lemma. If A is a Frobenius (symmetric) algebra, then for a n y positive integer n, M,(A) is a Frobenius (symmetric) algebra.

10. Frobenius and symmetric algebras

215

Proof. Let $ E A* be such that ( A , $ ) is a Frobenius algebra. Define the map X : Mn( A ) + F by

Then obviously X E Mn(A)*.Assume that

(ajj) E

Mn(A) is such that

We must show that ( a i j ) = 0. To this end, fix i , j E (1,. . . ,n} and denote by e;j the matrix with (i,j)-th entry 1 and 0 elsewhere. Then, for any a E A , X[(a;j)ae;j]= $(uj;u) = O Since ( A , $ ) is a Frobenius algebra, we deduce that aj; = 0. Thus ( a ; j ) = 0 and therefore M n ( A )is a Frobenius algebra. If ( A ,$) is a symmetric algebra, then n

n

/ n

n

n

\

n

n

n

proving that M,(A) is also symmetric. W 10.9. Proposition. symmetric.

I f A is a semisimple algebra, then A is

Proof. By Wedderburn’s theorem, A is isomorphic to a finite direct product of full matrix rings over division rings. Hence, by Lemmas 10.7 and 10.8, we may assume that A is a division algebra over F . Then, for any nonzero $ in A*, ( A , $ ) is a Frobenius algebra. Denote by [ A , A ]the additive subgroup of A generated by all elements of the form ab - ba, with a , b E A . It is clear that [A,A] is an F-subspace of A . To prove that ( A ,$) is a symmetric algebra, it suffices to verify that $ ( [ A ,A ] )= 0 for some nonzero $ in A*. The latter will follow provided

Induction from normal subgroups

216

we show that [A,A] # A. This last statement being independent of the ground field, we may assume that F is the centre of A . Let E / F be a field extension such that AE E M n ( E ) for some n 2 1. Because each matrix in [ M , ( E ) , M , ( E ) ]has trace 0 , we have [AE,AE] # A E . But [AE,A E ]= [A,A ] E ,so [ A ,A] # A , as required. 10.10. Proposition. If A is a Frobenius dgebra and G is a finite group, then any crossed product F-algebra A * G is Frobenius.

Proof. Let $ E A* be such that ( A , $ ) is a Frobenius algebra. Define the map X : A * G + F by

Then it is clear that X is an F-linear map. Let x = C x g g E A * G be such that X ( x y ) = 0 for all y E A * G. For any a E A , g E G, we have zag-l = C x t i a g - 1 = c x t t atg ---I tEG

t€G

Thus

0 = X(zag-') = $ ( x g g a )

for all a E A

and so $ ( x g A ) = 0 . Hence each x g = 0 and therefore x = 0, as we wished to show. Again, let A be a finite-dimensional algebra over a field F . For each subset X of A, we denote by I ( X ) and r ( X ) the left and right annihilators of X defined by

I ( X ) = { a E AlaX = 0 } , r ( X ) = { a E AlXa = 0} It is clear that l ( X ) and r ( X ) are left and right ideals of A , respectively. Moreover, if X is a left (respectively, right) ideal of A , then I ( X ) (respectively, r ( X ) )is a two-sided ideal of A. Suppose that ( A ,$) is a Frobenius algebra and that X is a subset of A. Define the subsets I X and X I of A by

' X = { a E AI$(aX) = 0 } X I = {a €Al$(Xa)=O}

217

10. F'robenius and symmetric algebras

It is clear that if X is a subspace of A, then so are ' X and X I . 10.11. Theorem. Let ( A , $ ) be a Frobenius algebra and let X be an F-subspace of A. (i) ('X)* = ' ( X ' ) = X and dimFX' = dirnF'X = dzrnFA -

dimFX. (ii) If X is a left (respectively, right) ideal of A, then r ( X ) = X I (respectively, l ( X ) = ' X ) . In particular, for every idempotent e of A (Ae)' = (1 - e ) A A/(Ae)' E eA as right A-modules d i m F e A = dimFAe

(iii) If X is a left ideal of A, then dimFX

+ d i m F r ( X ) = dimFA

and I ( r ( X ) )= x

If X is a right ideal of A, then dimFX

+ dimFl(X) = dimFA

and r ( l ( x ) )= x

(iv) The mapping X H r ( X ) is a duality of the lattice of left ideals of A onto the lattice of right ideals of A, that is X H r ( X ) is a bijection of the set of left ideals of A onto the set of right ideals of A such that

for all left ideals X l , X 2 of A . Similarly, X H i ( X ) is a duality of the lattice of right ideals of A onto the lattice of left ideals of A. (v) If ( A ,$) is symmetric, then r ( X ) = l ( X )for any two-sided ideal X ofA.

Proof. (i) Let { a l , .. . ,a , } be an F-basis of A such that { a l ,. . . ,a,} is an F-basis of X ( m 5 n ) . The map

218

Induction from normal subgroups

is an isomorphism of F-spaces such that the image of X' consists of all ( A 1 , . . . ,A,) with A1 = . . . = A, = 0. Hence dimFX' = dimFA dimFX and, by a similar argument, d i m F ' X = dimFA - dimFX. We conclude that d i r n F ( 'X)' = dirnFX Because ('X)' 2 X , it follows that ('X)' = X , as desired. The equality ' ( X ' ) = X is proved similarly. (ii) Suppose that X is a right ideal of A. Then a E ' X if and only if $ ( a x ) = 0 or if and only if $ ( a X A ) = 0. SInce the latter is equivalent to aX = 0, we have 1 ( X ) = ' X . The equality r ( X ) = X' for a left ideal X of A follows by the same argument. Aez = 0 ++ ex = 0 ++ Given z E A, we have z E (Ae)' z E (1 - e)A, so (Ae)' = (1 - e)A and A/(Ae)' G eA. Finally, dimFAe = dimFeA, by applying (i). (iii) This follows directly from (i) and (ii). (iv) That the given map is a bijection follows from (iii). Clearly we have

Applying (iii), it follows that

The corresponding assertion for left ideals follows similarly. (v) Assume that ( A , $ ) is a symmetric algebra. Then I X = X' for any subset X of A. The desired conclusion follows from (ii). Let V be a left A-module. Then the socZe of V , written SocV, is defined as the sum of irreducible submodules of V . The socle of a right A-module is defined similarly. The Zeft and right socles of A are defined as the socles of A A and AA, respectively. In case the left and right socles of A coincide, we shall write SocA to denote both of them. 10.12. Theorem. Suppose that A is a Frobenius aZgebra.

(i) The left and right socles of A coincide. Moreover, dimFSocA =

10. Fkobenius and symmetric algebras

219

dirnF(A/J(A)). (ii) For each primitive idempotent e E A, (SocA)e is the unique irreducible submodule of Ae. In particular, Soc(Ae) = (SocA)e. (iii) If A as symmetric, then for any primitive idempotent e E A, Soc(Ae) E A e / J ( A ) e Proof. Setting J = J ( A ) , it follows that r ( J ) and l ( J ) are the left and right socles of A, respectively. Let e be a primitive idempotent of A. Then J e is the unique maximal submodule of Ae and A e / J e is irreducible, by Theorem 2.12.4. By Theorem 10.11, the map X H X' is an inclusion-reversing bijection between the set of left ideals X of A contained in Ae and the set of right ideals of A containing (Ae)'. We between thus obtain an inclusion-reversing bijection X H X'/(Ae)' the set of submodules of A e and the set of submodules of A/(Ae)*. Now (Ae)' = (1 - e ) A and A/(Ae)' 2 eA by Theorem 10.11. Thus we have an inclusion-reversing bijection X H eX' between the set of submodules X of A e and submodules of eA. In particular, e(Je)' is the unique irreducible submodule of eA. Further, for y E A, we have y E ( J e ) l H Jey =

o w ey E r ( J )

which implies that e(Je)* = e r ( J ) . Thus e r ( J ) is the unique irreducible submodule of eA. Because J annihilates all irreducible A-modules, we have e r ( J ) J = 0. Thus r ( J ) J is annihilated by every primitive idempotent e E A. The latter implies r ( J ) J = 0 and so r ( J ) l ( J ) . The reverse inclusion holds by symmetry. The moreover assertion being a consequence of Theorem 10.11, (i) is established. The same argument (reversing left and right) also proves (ii). To prove (iii), let $ E A* be such that ( A ,$) is a symmetric algebra. By the foregoing, r ( J ) e is the unique irreducible submodule of Ae and J r ( J ) e = 0. Thus every f E H o m A ( A e , r ( J ) e ) induces a map X E HornA(Ae/Je,r(J)e). Hence we need only show that HornA( Ae, r ( J ) e ) # 0. Since there is an isomorphism of additive groups HornA(Ae,r ( J ) e ) e r ( J ) e given by f H f ( e ) , we are left to verify that e r ( J ) e # 0. Assume by way of contradiction that e r ( J ) e = 0. Then

Induction from normal subgroups

220

and so r ( J ) e = 0, a contradiction. This completes the proof of the theorem. 11.

Symmetric crossed products

Our aim is to prove that if S is a finite-dimensional semisimple algebra over a field F and G a finite group, then any F-algebra which is a crossed product of G over S is symmetric. Let S be an arbitrary ring, let G be a finite group and let S * G be a crossed product of G over S. As usual, for any s E S,g E G, we put

Then z H g z , z E Z(S), provides an action of G on Z ( S ) . Note that if S is an algebra over a field F, then S * G is an F-algebra if and only if G acts trivially on F . We now fix the following data: E is a G-invariant field contained in Z ( S ) I( = EG = { a E Elga = a for all g E G} F is a subfield of IC such that K / F is a finite field extension and dimES C 00 t = trE/K E + I(. Note that

t ( E )= I(

(1)

and t(ga)= t(u)

for all a E E,g E G

(2)

(see Lang (1984, Chapter VIII)). Let II, E H o ~ E ( SE,) . We say that 1c, is G-invariant if cp( gs) = gcp(s)

for all g E G, s E S

11.1. Lemma. I f ( S ,cp) is a Frobenius E-algebra, 0 and f : S * G + F is defined b y

then (i) ( S * G , f ) is a Frobenius F-algebra.

#X

(3) E Homf;(K,F )

11. Symmetric crossed products

221

(ii) If (S,cp) is a symmetric E-algebra and cp is G-invariant, then ( S * G , f) is a symmetric F-algebra. Proof. (i) If 1c, = X 0 t cp, then 1c, E H o r n ~ ( Sand, ) by (l),( A o t ) ( E ) = F . Assume that +(US) = 0 for some a E S. If cp(aS) # 0, then cp(aS) = E and $(US) = ( A 0 t ) ( E )= F = 0, a contradiction. Thus cp(aS) = 0 and therefore a = 0, proving that (S,T)) is a Frobenius F-algebra. The desired assertion now follows from Proposition 10.10. (ii) Assume the additional hypotheses of (ii) and let z = C zgg, y= C ygg be two elements of S * G , zg,yg E S. Then 0

c

f (4 = Wcp(

zggyg-lg-')))

g€G

=

-

c c Nt(cp(sY,-ls-'4>>

A(t(cp(~,(sv,-ls-'>>>>

SEG

=

s€G

-

since jyg-1g-l E S and cp(ab) = cp(h) for all a , b E S. Hence

f (4 =

c Wcp [ c c

g€G

=

g(yS-lF%s)]

-

wgcp(Yg-1g-lzgg)))

(bY(3))

gEG

=

A(t(cp(Yr19-'%g)))

g€G

=

(bY(2))

-

Wcp(CY g - 1 g - l z g 3 ) )

= f ( Y 4

as required.

11.2. Lemma. Let R be a commutative ring and let a G-graded R-algebra A be a crossed product of G over Al. Let e l , . . . ,en be orthogonal idempotents of Z ( A ) n A1 with 1 = el + - - . e,. Then

+

A = $y.=le;A is a direct sum decomposition of A into ideals e;A where each e;A is a G-graded R-algebra which is a crossed product of G over e;Al such that

Induction from normal subgroups

222

(i) e; is the identity element of e;A. (ii) ( e ; A ) , = e;A, for all g E G . (iiif e i j E U(e;A)n ((eiA),) for all g E G . (iu) U ( A )= @ L 1 U ( e ; A ) .

Proof. The proof is straightforward and so will be omitted. In order to make further progress, we need to introduce reduced characteristic polynomials. Let A be a finite-dimensional algebra over a field F . Then A is said to be separable if for every field extension E of F , AE = E @ F A is a semisimple E-algebra. For each a E A , let char.pol.AIF(a)be the characteristic polynomial of the F-linear map x H ax, x E A . The trace map TA/Fand norm map N A I Fare then defined by

where m = dimFA. Now assume that A is a separable F-algebra. Then there exists a field extension E / F and an isomorphism

of E-algebras. For each a E A , let

We now define the reduced characteristic polynomial of a E A by S

red.char.pol.A,F(a) =

char.pol.cpi(a), i=l

where char.pol.cp;(a) is the characteristic polynomial of the matrix cpi(a). It can be shown (see Curtis and Reiner (1981, p.159)) that this reduced characteristic polynomial is independent of the choice of E and of the isomorphism f . Further, its coefficients all lie in the ground field

11. Symmetric crossed products

223

F. Now put m = E n ; and write

for a E A. Then trA/F is called the reduced trace, and nrA/F is called the reduced norm. It is easily verified that trA/F : A -+ F is an F-linear map and

trA/F(ab)= t T A / F ( b U ) for all a , b E A

(4)

We now quote the following two standard facts. 11.3. Proposition. Let A be a separable F-algebra. Then the reduced trace trA/F gives a symmetric bilinear associative nonsingular form, namely

(a,b) ++ Proof.

~ T A I F ( ~ ~ (a,b ) E

A)

See Curtis and Reiner (1981, p.165).

11.4. Proposition. Let A be a semisimple F-algebra. Then A is separable if and only if the centres of the division algebras associated with the Wedderburn components of A are separable field extensions of F. Proof.

See Curtis and Reiner (1981, p.145).

Let A, B be isomorphic rings and let u : A -+ B be a ring isomorphism. Let F be a field contained in Z ( A ) such that A is a finitedimensional separable F-algebra. If L = a(F),then L is a field contained in Z ( B ) ,B is a finite-dimensional separable L-algebra and

Let a E A and let

where n

1 1 is an integer and A; E F for all i E ( 1 , . . . ,n } .

Induction from normal subgroups

224

11.5. Lemma.

With the notation above,

red.char.pol.slL(a(a))= X "

+ Q ( X ~ ) X " - ' + + a ( X , - l ) ~+ a(X,) * *

Proof. Let F and L be algebraic closures of F and L , respectively. It is well known (see Lang (1984, VII, Theorem 2.8) that Q : F + L can be extended to a field isomorphism i? : F + 1. Thus there is a ring isomorphism CY

:F 8 F

A

+L 8 L

B

such that cY(X 8 a ) = e(X) 8 a ( a )

for all X E F and all a E A. Because P 8~ A is a finite-dimensional semisimple F-algebra, there exists an F-algebra isomorphism

for some positive integer rn and some positive integers r; for all 1 5 i rn. The isomorphism 8 : F + 1 induces the ring isomorphism m

M

i=l

r=l

Setting 7 = p o p o a-', we obtain an 1-algebra isomorphism m

Also let

5

11. Symmetric crossed products

225

m

red.char.pol.B,L(a(a)) =

char.pol.$i(~(u)) i=l

We have

~i[char.pol.(cpi(a))] = char.pol.$;(a(a))

(1 5 i 5 rn)

thus completing the proof. W We have now come to the demonstration for which this section has been developed. 11.6. Theorem. (Harris (1989)). Let G be afinite group and let S be a finite-dimensional semisimpte algebra over a field F. Then any F-algebra A which is a crossed product ofG over S is symmetric.

Proof. Let el, . . . ,en be pairwise orthogonal primitive idempotents of Z(S) with 1 = el ... en. Setting Bi = eiS, we then have S = B1 @ - - - @ B, where each Bi is a simple F-algebra. The map ei I-+ gei = jeig-l, g E G, provides an action of G on {el,. .. ,en}. If f 1 , . . . ,f k , k 5 n, are the G-orbit sums, then f l , . .. ,f k are mutually orthogonal idempotents of Z ( A ) n S. Since a finite direct product of symmetric F-algebras is a symmetric F-algebra (Lemma 10.7), it follows from Lemma 11.2 that it suffices to assume that G acts transitively on { e l , . . . ,en}. Put e = el, f = 1 - e , B = B1 = eS, E = Z ( B ) and denote by H the stabilizer of e. It is obvious that E is a field, F E F e = eF 5 E , AH = {C shhlsh E S, h E H } is an F-algebra which is a crossed product of H over S, e , f are orthogonal idempotents of Z ( A H )f l S with 1 = e -t f . By Lemma 11.2, eAH is a crossed product of H over

+ +

Induction from normal subgroups

226

B. For each h E H , put k = eh. Then k is a graded unit of eAH of degree h and the map p H kpk-', p E E , h E H , provides an action of H as a group of automorphisms of E. Let I< = E H be the H-fixed subfield of E , so that F E2 Fe = eF C I<. Put t = t r E / K : E + K and fix 0 # X E H o ~ F ( K , F )Note . that B is a finite-dimensional simple E-algebra, hence the reduced trace ~ T B I EE H o r n ~ ( BE, ) is defined by Proposition 11.4. Furthermore, by (4), t r B ~ E ( a b= ) trB/E(ba)

for all a,b E B

and, by Proposition 11.3, the kernel of t r B / E contains no nonzero right ideals of B. Also t r B / E is H-invariant by Lemma 11.5. Define f : eAH + F as in Lemma 11.1 with t r B / E playing the role of cp. Then, by Lemma 11.1 (ii), ( e A H , f ) is a symmetric F-algebra. Hence, for any graded unit u of AH and any b E B ,

f(eubeu-') = f(b)

(5)

Choose gl,. . . ,gn in G such that e; = g;e;Z-l. Then B; = zBl~-' for a l l i E { l , . . . , n } . Definecp:S=@Z1B; + E by : if y = Cr=lyi E S for yi E Bi, put

c f(G-'!/;G) n

cp(d =

i=l

Then obviously cp E H o ~ F ( SF ,) , cp(ab) = cp(ba) for all a , b E S and Kercp contains no nonzero right ideals of S. Applying Lemma l.l(ii), with F playing the role of E , we are left to verify that cp( g x ) = cp(x) for all x E S,g E G. The latter will follow provided we show that for any fixed j E (1,. . .,n}, g E G, and z E Bj, cp( g z ) = ~ ( 2 ) . We have gBj3-l = ijijjB1G-'ij-' = B k for some k E { 1,. . . , n}. Hence ijE = u a where u is a graded unit of AH. SInce z-'zz E B1 = B , we have u z z g j u = e u z z s - eu

-'

and hence, by ( 5 ) ) f(uz-'z~u-')= f ( j 7 ' Z i j j ) . cp(

gz)

But then

= cp(gzg-') = f(g&Mj-li&) = f(ug;'Zijju-') = f(g371zgi) =

44,

227

11. Symmetric crossed products

thus completing the proof. H

11.7. Corollary. Let G be a finite group and let S be a finitedimensional algebra over a field F . If an F-algebra A is a crossed product of G over S, then A / A - J ( S ) is a symmetric F-algebra.

-

Proof. By Theorem 3.7(ii), A / A J ( S ) is a crossed product of G over S / J ( S ) . Since S / J ( S ) is a finite-dimensional semisimple Falgebra, the result follows by virtue of Theorem 11.6. H

To provide another application of Theorem 11.6, we need the following simple observation. 11.8. Lemma. Let R be a commutative ring and let A be a G graded R-algebra. If N is a normal subgroup of G , then A can be regarded as an GIN-graded R-algebra by means of

AgN = @xEgNAz Moreover, if A is a crossed product of G over Al, then A can also be regarded as z crossed product of G I N over

AN = @xENAx Proof. Let T be a transversal for N in G. It is clear that

A = @tETAtN Given t l , tz E T , we have

AtlNAtzN =

NAr) (@y€t2NAy) @xEtlN,yEtzNAxy= @zEtitzNAz = AtItzN (@x€tt

Assume that A is a crossed product of G over A l . Then, for all g E G, there exists a unit g of A in Ag in which case j E A g ~as, required. H 11.9. Corollary. Let F * G be a crossed product of a finite group G over a field F . If N is a normal subgroup of G , then

F

* G / ( F * G ) ( J ( F* N ) )

Induction from normal subgroups

228

is a symmetric FG-algebra, where FG is the fixed field of G. Proof. We first note that A = F * G is an Fc-algebra. Since G is finite, F/FGis a finite Galois extension. In particular, F is a finitedimensional algebra over the field FG. Hence S = F * N is a finite dimensional algebra over FG.By Lemma 11.8, A can be regarded as a crossed product of GIN over S. Thus, by Corollary 11.7, A / A . J ( S ) is a symmetric FG-algebra, as required. H The following result for ordinary group algebras is due to Harris (1989).

11.10. Corollary. Let F"G be a twisted group algebra o f G over

F. Then FaG/FaG- J(F"N)

is a symmetric F-algebra. Proof. Since G acts trivially on F, we have F = FG. Now apply Corollary 11.9. R As a preliminary to our final result, we next record

11.11. Lemma. Let n be a positive integer and let A = F[X]/(X"). Then A is a symmetric F-algebra.

-+

Proof. Put s = X ( X " ) . Then { l,s,s2,. . . ,sn-l} is an F-basis of the commutative F-algebra A. Since any ideal of F[X] containing ( X " ) is of the form ( X ' ) , 0 5 i 5 n , ( X o ) = F [ X ] , it follows that are all nonzero ideals of A. The map A , A s , ... ,

given by

is obviously F-linear. Since for all i E {0,1,. , . ,n - l}, si # Ker+,

11. Symmetric crossed products

229

it follows that kerll, contains no nonzero ideals of A. Thus ( A , $ ) is a symmetric algebra, as required. Assume that an F-algebra A is a crossed product of G over an Falgebra S. It is natural to consider the following problem. Problem. What are necessary and sufficient conditions for A to be a symmetric F-algebra? An important sufficient condition is given by Theorem 11.6. The following example illustrates that the above problem in full generality is extremely complicated.

11.12. Example. (Dude). There exists a finite-dimensional algebra A over a field F such that A is a crossed product of a group G over an F-algebra S such that fi) A is a symmetric F-algebra.. (ii) S is not a symmetric F-algebra. Proof. Put B = F [ X ] / ( X 2 )so , that B = F [ x ]with x = X + ( X 2 ) and B is a symmetric F-algebra by Lemma 11.11. Let A = M z ( B ) and let eij, 1 2 i,j 2 2, be the usual matrix “units)) of A. Then A is a symmetric F-algebra, by virtue of Lemma 10.7. The elements eij, xejj, 1 5 z,j 5 2 constitute an F-basis of A and xejj = eijx, Let G =< g

x2 = o

(1 5 i , j

5 2)

> be a cyclic group of order 2 and put

A, = Felz

+ Fe21 + F s e l l + Fxez2

Then one immediately verifies that A becomes a G-graded algebra. Since e21 e12 E A, and

+

it follows that A is a crossed product of G over S. By the foregoing, we are left to verify that S is not a symmetric F-algebra. To this end, observe that J ( S ) = Fze21 Fxe12 and S =

+

Induction from normal subgroups

230

Sell @ Sezz (direct sum of S-modules). Put P = Sell. Then P = Fell Fzezl and J ( P ) = J ( S ) P = Fse21. Thus P is a projective indecomposable S-module such that SocP = J ( P ) . Since ellJ(P)= 0 and e,l(P/J(P))# 0, we have

+

SOCP = J ( P ) y P / J ( P ) Thus S is not a symmetric F-algebra, by virtue of Theorem 10.12(iii). The proof is therefore complete. 12.

EndFG(VG)is symmetric

All modules below are assumed to be finitely generated over their ground fields. Throughout, N denotes a normal subgroup of a finite group G, F an arbitrary field and V an FN-module. Our aim is to prove that if V is completely reducible, then the F-algebra EndFG(VG) is symmetric. We begin by recording the following two elementary results. 12.1. Lemma. Let V be an irreducible FN-module and let H be the inertia group o j V . Then any FH-homomorphism 8 : V H + V H extends to a unique FG-homomorphism 0' : V" t V G and the map EndFH( V H ) + EndFc( V") 6 H 8' is an isomorphism of F-algebras.

Proof. Let gl,gZ,. . . ,gs be a transversal for N in H and gl,. . . ,gk a transversal for N in G ( k 2 s). Then

O'(9i 8 v) = g i q 1 €3 v)

(v E V )

Then it is immediate that 8' is a unique element of EndFG(v") extending 6. Thus the map 0 H 8' is an injective homomorphism of

12. E + G ( V ~ )

is symmetric

231

F-algebras. Note that ( V H )is~the sum of all submodules of ( V G ) , isomorphic to V . Since V is assumed to be irreducible, we conclude that for any t,b E EndFG(VG),we have @ ( V H ) V H .This shows that the given map is surjective and the result follows.

12.2. Lemma. Let V be a completely reducible FN-module. Then there exist a positive integer s , irreducible nonconjugate FN-modules V,, . . . ,V , and positive integers n1, n2,. . . ,n, such that

n 8

EndFG( V G )

i= 1

Mni(EndFH,(qHi)

where H, is the inertia group of K .

Proof. Write V = f&m;V; where each is irreducible and mi 2 1 is an integer. We may assume that V,,. .. ,V , ( s 5 k) are such that Vl, . . . ,V, are mutually nonconjugate and each Q , j E (1,. . . ,k} is conjugate to some K, i E ( 1,. . . ,s} (hence yGg KG). It follows that there exist positive integers n l , 1 2 2 , . . . ,n, such that

vG2 $:=ln;KG

($tlniK)"

Thus we may harmlessly assume that V = $:=lniV;. Let T be a transversal for N in G containing 1. Assume that i , j E (1,. . . ,s} and i # j. Then, by Proposition 2.2.4(i), HOmFG

( n j & ) G ) HomFN ( n i K , ( n j q G ) N )

Thus HomFG

( ( n i K ) G(njl$)G) , g ninj ( $ t , g H O m F N ( V , , t C3 4) = 0,

since V , and t 8 V, are irreducible and nonisomorphic FN-modules for all t E G. This fact immediately

EndFG(VG)

g

N

N

Induction from normal subgroups

232

where the last isomorphism follows from Lemma 12.1. It is now an easy matter to prove the following important result. 12.3. Theorem. (Harris (1989)). Let F be an arbitraryfield and let V be a finitely generated completely reducible FN-module. Then the F-algebm EndFc(VG) is symmetric.

Proof. Owing to Lemmas 12.2, 10.8 and 10.7, we may harmlessly assume that V is irreducible and G-invariant. Then, by Theorem 4.2(ii), the F-algebra EndFc(VG) is a crossed product of GIN over S = EndFN(V). But V is irreducible, hence S is a division ring by Schur's lemma. Since V is finitely generated, S is of finite F-dimension. Now apply Theorem 11.6. Another application of Theorem 11.6 is given by 12.4. Theorem. Let F be an arbitrary field and let V be a finitely generated G-invariant FN-module. Then the F-algebra

is symmetric.

Proof. Put E = EndFc(VG) and S = End&V). Since V is Ginvariant it follows from Theorem 4.2(iv) that E / E - J ( S ) is a crossed product of G I N over S / J ( S ) . Bearing in mind that S / J ( S ) is a finitedimensional semisimple F-algebra, the result follows by applying Theorem 11.6. m 13.

Graded modules

Throughout this section, G denotes a group, R a commutative ring and A a G-graded R-algebra. While we shall have occasion to use right modules, any unspecified module will be understood to be left and unitary. All the information recorded here will be utilized in the next section.

233

13. Graded modules

We say that an A-module M is G-graded (or simply graded) provided there exists a family

P g I 9 E GI of Al-submodules of M indexed by G such that the following conditions hold:

M = $gEGMg

(direct sum of Al-modules)

AxMyG Mzy

for all z,y E G

(1) (2)

The above definition certainly implies that the regular module A A is graded (with M, = A, for all g E G ) . We refer to M, as the 9 component of M . By the g-component of rn E M , we understand the m,. unique rn, E M, defined by rn = CgEG A submodule N of a graded module M is said to be a graded submodule if for all rn E N and all g E G, rn, E N . Expressed otherwise, N is graded if

N = @gEG(N fl Mg) Thus if N is a graded submodule of M , then N is a graded module with Ng = N fl Mg for all g E G. A graded left A-submodule 1 of A is called a graded left ideal of A. We call (1) a G-grading of M and refer to M, as the g-component of M . When (2) can be replaced by the stronger condition

AxMy = Mzy

for all z,y E G

(3)

we say that M is a strongly graded A-module. Let M and N be two graded A-modules. An A-homomorphism

is called graded if

f(M,)

Ng

for all g E G

(4)

In case f is an isomorphism, we say that M and N are isomorphic as graded A-modules.

234

Induction from normal subgroups

13.1. Lemma. Let f : M + N be a graded homomorphism of two graded A-modules M and N . Then (i) I - e rf is a graded submodule of M . (ii) f ( M ) is a graded submodule of N . (iii) M / K e r f E f ( M ) as graded A-modules, where the g-component of M / K e r f is defined b y

+

( M l K e r f ) , = (Mg K e r f ) / K e r f

(9 E G )

Proof. For any given rn E M , f ( m )= f(Erng)= cf(rng), where each f ( r n g ) is in N g . This shows that f(m),= f ( r n g )

for all g E G , m E M

(5)

If rn E K e r f , then f ( r n ) = 0 and so, by ( 5 ) , f ( r n g ) = 0. Thus m, E K e r f , proving (i). Property (ii) is a direct consequence of (5). To prove (iii), let f’ : M / K e r f -+ f ( M ) be the isomorphism induced by f. Then

for all rn E M . Thus f * is a graded isomorphism, as required. Let V be any Al-module. Then the tensor product

is naturally an A-module with

for all z, y E A and o E V . We shall refer to V Aas the induced module. Bearing in mind that

A = @gEGAg it follows from Lemma 3.l(ii) that we may identify the Al-module A, 6 3 V~ with ~ its image in V A ,for any g E G. With this identification, we have the following direct decomposition of Al-modules:

13. Graded modules

235

Setting

( V A ) ,= A,

@Al

V

for all g E G

it follows from (6) and the inclusion

A,A, C A,,

for all s , y E G

that V A is a graded A-module. Let M be a graded A-module. For each g E G, put

Mg = M and define

for all s E G. Then Mg is obviously a graded A-module which differs from M only by grading. We shall refer to Mg as a conjugate of M . 13.2. Theorem.

(Dude (1980)). The following properties of a G-graded algebra A are equivalent to each other: (i) A is strongly G-graded. (ii) Every graded A-module is strongly graded. (iii) For any graded A-module M , the natuml map

is a graded isomorphism. (iv) For any graded A-module M , there exists an Al-module V such that M s VA as graded A-modules.

Proof. (i)

=+

(ii): Let M be a graded A-module. Because A is strongly graded, A,A, = A,, for all z , y E G , so by Lemma 3.1(i), we have Mxy= AIM,, = A,A,-1 MxyC A x M y C M,,

236

Induction from normal subgroups

where we have used (2) twice. Thus A,M, = ME,, as required. (ii) =+ (iii): Fix g E G. Then

so $ is a graded epimorphism. The kernel N of $ is a graded A-module (Lemma 13.l(i)) whose g-component N, is given by

Ng = N n (A,

@ A ~MI)

By assumption, N is strongly graded so that for all g E G, Ng = A,N1. Now N1 is the kernel of the natural isomorphism of A1 8~~Ml onto Ml, so Nl = 0. Thus N = 0 and therefore $ is a graded isomorphism. (iii) + (iv): Put V = MI and apply (iii). (iv) + (i): Invoking Lemma 3.1(i), (ii), we obtain for all g E G

A,Al = A, Hence, for any Al-module V ,

Applying the hypothesis, we conclude therefore that

for any graded A-module M . The regular A-module A is a graded A-module with g-component As for all g E G. Letting M to be the conjugate module A" for some z E G, we define

A,A, = Ag(A")l = (A"), = A,, by (7). Thus A is strongly G-graded, as asserted.

To provide a useful application of the above result, we need to record the following simple observation. 13.3. Lemma. If V is projective A-module.

a

projective Al-module, then V A is

a

237

13. Graded modules

Proof. By hypothesis, A1 .$A1 = V$ 6 for some A1-module V,. Since A @ I A ~A1 2 A, it follows that A @ - * @ A VA @ FA. Thus V Ais a direct summand of a free A-module and therefore is projective, as required. $ a

+

13.4. Corollary. Let A be a strongly G-graded algebra such that Al is semisimple artinian. Then every graded A-module is projective.

Proof. Let M be a graded A-module. Then, by Theorem 13.2(iii), there is a graded isomorphism M + M t = A @ I A ~ MI. Since A1 is semisimple artinian, it follows from Corollary 1.5.11 that every Almodule is projective. Hence Ml is a projective Al-module. Thus, by Lemma 13.3, A4 is projective. As a preliminary to our final result, let us recall the following piece of information. Let A be a strongly G-graded R-algebra and let I be a G-invariant ideal of Al. Then, by Theorem 3.6, I A = A I is a graded ideal of A. Hence, by Lemma 3.5(i), A/IA is a G-graded R-algebra with

( A I I A ) , = (A,

+ IA)/IA

for all g E G

13.5. Proposition. Let A be a strongly G-graded R-algebra, let I be a G-invariant ideal ofA1 and let V be a G-graded A-module. Then, for any n 2 0, I"V/I"+lV is a G-graded A/IA-module (here I' = A1).

Proof. Since I is G-invariant, so is I" for all n 3 1. Hence each I"V is a submodule of V. Since I A annihilates I"V/I"+'V, it follows that I"V/I"+'V can be regarded as an AIIA-module. Because V is G-graded, there exists a family {Vglg f G} of Alsubmodules of V such that

for all z, y E G. Setting

+

(I"V/I"+'V), = (I"T/, I"+lV)/I"t'V

(g E G)

Induction from normal subgroups

238

we immediately deduce that I"V/I"+'V is a direct sum of all (I"V/In+lV)g, g E G. Finally, for any z,y E G, we have

+ +

(A / I A), (I" V/In+'V ) , = (A,I"V, In+' V)/I"+lV = (I"A,V, I"+'V)/I"+'V (PV,, -I- I"+'V)/I"+'V = (I"V/In+lV),y,

c

thus completing the proof. 14.

Induction from irreducible modules and their projective covers

Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. All vector space algebras have finite dimension over the stipulated field. If n is a positive integer and V is a module, then nV denotes the direct sum of n copies of V . Let P ( V ) be a projective cover of an irreducible FN-module V . Our aim is to tie together the decompositions of V Gand P(V)" and to describe indecomposable direct summands. Among other properties, it will be demonstrated that if W is an indecomposable direct summand of V G ,then Soc(W) and W / J ( W )are isomorphic irreducible FG-mod ules. Let S be a ring and let M be a (left) S-module. For any subset X of S, we put A ~ ~ M (=X{)m E M l X m = 0} and

X M = (C2imjIxi E X,mi E M } If X is a right ideal of S , then AnnM(X) is a submodule of M . For convenience of reference, we first record the following elementary observat ion. 14.1. Lemma.

L e t S be a ring, let M be a n S - m o d u l e a n d let

14. Induction from irreducible modules and their projective covers 239

be a direct sum decomposition of M . Then, for any left ideal I of S and for any subset X o f S we have (i) I M and IMj are submodules o f M and I M = $j,jIMj. (ii) M / I M 2 $jEj(Mj/IMj) as S-modules. (iii) A n n ~ ( x=) $jE~AnnM3 (X).

Proof. The proof is straightforward and so will be omitted. H 14.2. Lemma. Let H be a subgroup ofG containing N and let W be an FH-module. Then (i) w ~ / J ( F N ) W E (~w / J ( F N ) w ) ~ . (ii) [ Annw ( J (F N ) ) I G2 A n n w (~J (F N ) ). Proof. (i) Since J(FN)FG = F G . J ( F N ) , J ( F N ) W is obviously a submodule of W . Consider the natural exact sequence 0 + J ( F N ) W -%

w -5W / J ( F N ) W -+ 0

By Theorem 2.2.9(i), the corresponding sequence of FG-modules 0 + ( J ( F N ) W ) Gl% W GlY ( W / J ( F N ) W ) " -+o is exact. Since

the desired conclusion follows. (ii) Since g J ( F N ) = J ( F N ) g for all g E G, A n n w ( J ( F N ) )is a submodule of W . Consider the natural exact sequence

0

+ A n n v ( J ( F N ) )5 W

5W/Annv(J(FN))

-+

0

As above, the corresponding sequence of FG-modules

is exact. Since fm(1 8 a ) = A ~ ~ ~ G ( J ( F(ii) N is ) )established. , H

240

Induction from normal subgroups

14.3. Lemma. Let V be an N-projective FG-module and let n be a positive integer such that J ( F N ) " V = 0 . Then in the chain

V = J(FN)OV 2 J ( F N ) V 2

* * .

2 J(FN)'+'V 2 J ( F N ) n V = 0

of FG-submodules of V , each factor

v;. = J ( F N ) ' v / J ( F N ) ' + ' v

(0

5 i 5 n - 1)

is a projective module over the symmetric F-algebra F G / F G . J ( F N ) . Proof. Since V is N-projective, it follows from Theorem 2.10.3(ii) Thus, by Lemma that V is isomorphic to a direct summand of 14.1, V;. is a direct summand of W; = J(FN>'(VN)"/J(FN)'+'(V~)~. By Corollary 11.10, F G / F G - J ( F N ) is a symmetric F-algebra. It therefore suffices to verify that W; is a projective F G / F G J ( F N ) module. We know, from Lemma 11.8, that F G is a crossed product of GIN over F N , where

-

( F G ) g=~ gFN

for all g E G

Hence, by Theorem 3.7(ii), F G / F G . J ( F N ) is a crossed product of GIN over F N / J ( F N ) . Now F N / J ( F N ) is semisimple artinian, hence by Corollary 13.4, it suffices to verify that W; is a (GIN)-graded F G / F G . J(FN)-module. Since the latter is a consequence of the fact that (VN)' is a GIN-graded FG-module and Proposition 13.5, the result is established.

We have now come to the demonstration for which this section has been developed.

(Harris (1989)). Let N be a normal subgroup of a finite group G , let F be an arbitrary field and let V be an irreducible FN-module. Let V1,V2,. . . , V, be all nonisomorphic irreducible FG-modules for which V is a direct summand of (K)', let r; be the multiplicity of as a composition factor of VG/J ( V G ) , and let W; = P ( V ; . ) / J ( F N ) P ( V , )Then . 14.4. Theorem.

14. Induction from irreducible modules and their projective covers 241

and each W; satisfies the following properties: (i) W; is indecomposable and W; E Wj implies i = j . (ii) W; 2 A n n p ( v ) ( J ( F N ) ) . (iii) Wj/J(Wj) socwj 2 K.

=

Proof. By Theorem 2.12.3, P ( V ) / J ( F N ) P ( VS ) V . Hence, by Lemma 14.2(i), P ( V ) G / J ( F N ) P ( V )s GVG (1) and therefore P(V)"/J(P(V)G)2 VG/J(VG) (2) Now, by Theorem 2.2.14, V1,.. . ,V, are all nonisomorphic composition factors of V G / J ( V G )hence ,

v ~ / J ( 2v er=lr;K ~)

(3)

Because P ( V ) Gis projective and, by (2) and (3),

P ( v ) ~ / J ( P ( v ) ~ey=lrjx, ) it follows from Theorem 2.12.3(iii) that

P(V)"

2

@y="=,jP(v,)

(4)

Applying (1) and Lemma 14.1(ii), we also deduce that

V G 2 @r=lrr;W;

(5)

Now SocP(V)= A n n p ( v ) ( J ( F N )E ) V by Theorem 10.12(iii). Hence, by Lemma 14.2(ii) applied to W = P ( V ) ,we have G

V G2 [Annp(v)( J (F N ) ) ] 2 A n n p p q ~( J (F N ) ) Thus, by (4) and Lemma 14.1(iii),

V G $:=lr;Annp(v)( J (F N ) )

(6)

Now fix i E { 1,. . . ,n } . Then

Wa/J(Wi)

=

P(K)/J(P(K)) 2 v, E Soc(P(v;:)) =

SOC ( A n n P ( v ) ( J ( F N ) ) )

(7)

Induction from normal subgroups

242

Hence W; and Annp(v,)(J ( F N ) ) are indecomposable FG-modules. Note that P ( v ) is a projective (hence N-projective) FG-module. Therefore, by Lemma 14.3, W;is a projective module over the symmetit follows ric F-algebra F G / F G - J ( F N ) . Since, by ( 7 ) , W;/J(Wi) from Theorem 10.12(iii) and (7) that

x

By ( 5 ) , (6) and the Krull-Schmidt theorem, A n n p ( ~ l ( J ( F i Vis) )a projective module over F G / F G J ( F N ) . Hence, by (8) and Theorem 10.12(iii), Wi AnnP(v,)(J(FN)),

-

=

thus completing the proof. H

(Harris (1989)). Let N be a normal subgroup of a finite group G, let F be an arbitrary field and let W be an irreducible FG-module. Then W is N-projective if and only if 14.5. Corollary.

J ( F N ) P ( W )= J ( F G ) P ( W )

(9)

Proof. Assume that W is N-projective. Then, by Theorem 2.10.3(ii), W is isomorphic to a direct summand of (WN)'. Hence there exists an irreducible FN-module V such that V is a direct summand of WN and W is a direct summand of VG. Since W is a direct summand of V G , we have W Z W; for some i, where W; is as in Theorem 14.4. Now W; is irreducible, hence W E V , (by Theorem 14.4(iii)) and J ( F G ) P ( V , )= J ( F N ) P ( V , ) ,proving (9). Conversely, assume that (9) holds. Let V be an irreducible direct summand of W N . By Theorem 14.4, P ( W ) / J ( F N ) P ( W )is a direct summand of V G . Since by (9), P ( W ) / J ( F N ) P ( W ) W , it follows that W is a direct summand of VG.Therefore W is a direct summand ~ so W is N-projective, by applying Theorem 2.10.3(ii). of ( W N )and

15. Inflated modules over twisted group algebras Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. Given an absolutely irreducible FN-module

15. Inflated modules over twisted group algebras

243

V , our future aim is to provide a detailed description of the decomposi-

tion of VG.Since twisted group algebras and projective representations will play the main role in the ensuing discusion, it will be convenient to relegate all the necessary background into a separate section. Let V be a finite-dimensional vector space over F . A mapping p : G -+

GL(V)

is called a projective representation of G over F if there exists a mapping a : G x G t F* such that

p(1) = 1v

(2)

Thus an ordinary representation is a projective representation with a ( x , y ) = 1 for all z,y E G. If we identify G L ( V ) with G L ( n , F ) , n = d i r n ~ Vthe , resulting map is called a projective matrix representation of G over F . As in the case of ordinary representations, we shall treat the terms “projective representation” and “projective matrix representation” as interchangeable. Applying the associativity of the multiplication in G, it follows from (1) and (2) that a E Z2(G,F*). To stress the dependence of p on V and 0, we shall often refer to p as an a-representation on the space V . Two projective representations p; : G t GL(V;),i= 1,2, are said to be linearly equivalent if there exists a vector space isomorphism f : t V . such that

for all g E G. The projective representation p on the space V is called irreducible if 0 and V are the only subspaces which are sent into themselves by all the transformations p(g),g E G. An a-representation p on the space V is called completely reducible if for any subspace W invariant under all transformations p ( g ) , g f G, there exists another such subspace W’ with V = W @I W‘. The following result shows that the study of a-representations is equivalent to the study of F”G-modules.

244

Induction from normal subgroups

15.1. Lemma. There is a bijective correspondence between arepresentations of G and F"G-modules. This correspondence preserves sums and maps bijectively linearly equivalent (irreducible, completely reducible) representations into isomorphic (irreducible, completely reducible) modules.

Proof. Let p be an a-representation of G on the space V . We may clearly define a homomorphism f : F"G + E n d ~ ( vby ) setting f ( i j ) = p(g) and extending by linearity. Thus V becomes an F"Gmodule by setting

Conversely, given an F"G-module V, and hence a homomorphism f : FOG + E n d ~ ( v )put , p ( g ) = f ( i j ) . Then p ( g ) E GL(V) since i j is a unit of F"G. Furthermore, p(1) = 1 and for all 2,y E G ,

so that p is an a-representation on V. This sets up a bijective corre-

spondence between a-representations and PG-modules. A subspace W of V is invariant under all p ( g ) if and only if W is an F"G-submodule of V . Consequently, the correspondence preserves sums and maps bijectively irreducible (completely reducible) representations into irreducible (completely reducible) modules. We next note that an F-isomorphism f : V, + V, of F"G-modules is an F"G-isomorphism if and only if Jf(v) = f(ijv) for all g E G, E V , Suppose now that p; : G + GL(V,),i = 1 , 2 are two arepresentations. Then p1 is linearly equivalent to p2 if and only if there is an F-isomorphism f : V i + & such that p2(g)f = f p l ( g ) for all g E G. The latter is equivalent to p z ( g ) f ( v ) = f p l ( g ) v or to gf(v) = f ( g v ) , for all g E G , v E V,. Thus two a-representations are linearly equivalent if and only if the corresponding modules are isomorphic. W Given a E Z 2 ( G / N ,F * ) ,we denote by infa the element of Z2(G,P ) defined by (infa)(z, Y )=a(xK

Y W

15. Inflated modules over twisted group algebras

for all x , y E G. Then the map a

H

H 2 ( G / N ,F')

245

inf Q induces a homomorphism

+ H2(G,F*)

called the inflation map. Observe that since (infa)(z,y)= 1 for all x , y E N , we may embed F N in FpG by n H fi for all n E N , where p = i n f a or p = (infa)-'.

Theorem. (Cliford (1937)). Let V be an absolutely irreducible G-invariant FN-module. Then there exists w = WG(V)E Z 2 ( G / N ,F') such that for 7 = infw, V can be extended to an FrGmodule, i.e. there exists an FYG-module W such that WN E V . Furthermore, if w is a coboundary, then V can be extended to an FGmodule. 15.2.

Proof. Let r be the representation of N afforded by V . To prove the first statement, it suffices (by Lemma 15.1) to exhibit -y E Z 2 ( G , F * ) ,whose values are constant on the cosets of N and a 7representation p of G such that p(n) = r ( n )for all n E N . By assumption, for each g E G , there exists a nonsingular matrix L ( g ) such that r(9-'n9) = L ( g ) - ' W L ( g )

We may, of course, assume that L(1) = 1. Let T be a transversal for N in G containing 1 and put

Then, for all g E G , n E N ,

From Schur's lemma, we readily verify that p(x)p(y)differs from p(xy) only by a scalar

Induction from normal subgroups

246

Assume that w is a coboundary, say

w ( z N ,y N ) = X(sN)t(yN)X(syN)-' for some X : G/N + F* with X(N) = 1. Then, setting p*(g) = X(gN)-'p(g), it is immediate that p* is an ordinary representation of G such that p*(n) = p ( n ) = r ( n )for all n E N . From now on, we shall refer to 2u = wG(V) as the obstruction cocycle

of

v.

We now return to our study of F"G-modules. Assume that

(i = 1,2)

p ; : G --+ GL(V,)

is an a;-representation. Consider the map pi €3 p2 : G

--t

GL(Vi 8 V2)

given by €3 P2)(9) = Pl(S) €3 P 2 b ) for all g E G. One immediately verifies that p1 8 p2 is an q a 2 representation. We shall refer to p1 €3 p2 as the (inner) tensor product of the projective representations p1 and p2. It will convenient to reformulate the above in terms of modules. Let V and W be F"G and FPG-modules, respectively. Then the vector space V @ F W is an F"PG-module, where the action of elements g, g E G, is defined by (P1

g(v@w)=gv€3gw

(v E v,wE W )

15. Inflated modules over twisted group algebras

247

and then extended to V@FW and FaPG by F-linearity. It is clear that if pv and pw are cy and P-representations of G afforded by V and W , respectively, then pv @ pw is afforded by the F"PG-module V € 3 W ~ .

15.3. Lemma. Let V and W be FaG and FPG-modules, respectively, and let EIF be a field extension. Then

(V @ F W ) E2 VE@E WE

as EaPG-modules

Let ( ~ 1 , . . . ,v,} and ( ~ 1 , . . . ,w,} be F-bases of V and Proof. W , respectively. Then (1 @ (v; @ w j ) ) and ((1 €3 v ; ) 8 (1 8 wj)}, 1 5 i 5 n, 1 5 j I m, are E-bases of (V 8~W ) Eand V , @ E WE, respectively. Consequently, the map

extends to an E-isomorphism of (V 8~ W ) Eonto VE@ E WE. It is obvious that for all v E V , w E W,$(l @ (v @ w)) = (1@ v) €3 (1 @ w). Since, for all g E G,

and

We now come to the main topic of this section. Let a E Z2(G/N, F*), where F is an arbitrary field and GIN acts trivially on F*. Given an P(G/N)-module V , we can form an F'"f"G-module inf(V)whose underlying space is V and on which elements g , g E G, act according to the rule: gv = g N v (v E V ) We shall refer to z n f ( V ) as being inflated from V .

248

Induction from normal subgroups

15.4. Lemma. Let CY E Z2(G/N,F*)and let V be an F"(G/N)module. (i) V is irreducible (absolutely irreducible, indecomposable, totally indecomposable) if and only if inf(V) is irreducible (absolutely irreducible, indecomposable, totally indecomposable). (ii) For any F"(G/N)-modules Vl and %,& E v2 if and only if inf( &) Z inf (V2). (iii) If W is an F'"f*G-module such that fiw = w for all n E N, w E W , then W = inf(U) for some F*(G/N)-module U. (iv) IfE/F as a field extension, then inf(VE) (infv),.

Proof. The proof is straightforward and so will be omitted. Let H be a subgroup of G. Then, for any CY E Z2(G,F*),the restriction of CY to H x H is an element of Z2(H,F*).To prevent our expressions from becoming too cumbersome, we shall use the same symbol for an element of Z2(G,F*) and its restriction to Z2(H,F*).With this agreement, we will identify F*H with the subalgebra of F*G consisting of all F-linear combinations of the elements &, h € H . Given an F*G-module V, we denote by VH the F"H-module obtained by the restriction of algebra; thus as an F-module, V H equals V, but only the action of F*H is defined on VH. Given an F"H-module W , we form the F"G-module

and refer to W Gas the induced module. 15.5. Lemma. F"S-module. Then

Let S be a subgroup of Q = G/N and let Vbe an inf(VQ) z inj(V)'

Proof. Let H be the subgroup of G with S = H/N. If {gl,. . . ,gn} is a left transversal for H in G, then {glN, . . . ,gnN} is a left transversal for H / N in GIN. Consequently, the map

15. Inflated modules over twisted group algebras

249

and

Consequently, putting X = a ( g N ,g;N)a-'(gjN, h N ) , we obtain

as required.

15.6. Lemma. Let a E Z 2 ( G , F * ) ,let V be an FOG-module and, f o r any given cp E EndFaN(VN),g E G, define 'pg by

Then 'pg E E n d p N ( V N ) and the map

is an F-algebra automorphism. Proof. have

Given g E G, n E N , v E V and cp E E n d p N (V N ), we @ ( n v ) = gcpg-'(nv) = gcp[(g-'ng)(g-'v)]

= gg-'ngcp(g-'v) = f i c p g ( v ) ,

Induction from normal subgroups

250

proving that cpg E EndpN(VN). It is obvious that cp H cpg is F-linear and injective. Bearing in mind that -- 1 g-I = a(g,g-')g-' and 9-1 = a-1 ( g , g-'>ij, we obtain (cpg-l)g(w)

=

gcpg

-'--I

g

-

--

v-gg-'cpg-'

1

g

w

= g a ( g , g-l)g-lcpa-'(gl g-l)gij-lv

=

cp(4,

which shows that (&)g = cp. Thus cp H cpg is a bijection. Finally, if cp1,cpz E EndpN(VN), then (cpM)(v)

=

c p ~ g c p 2 s - 1= ~ scpls-19cp2s-1v

= s ( ' P ~ ' P & - ~=v ( c p 1 ~ 2 ) ~ ( v )

Hence

(cpI'p2)g

= cpfcp; and the result follows.

In what follows, we write Am for the F-algebra of all m x m matrices with entries in an F-algebra A.

15.7. Lemma. Let V be an F*G-module and let & and & be FPQ-modules, Q = G I N , ofequal dimension m with F-bases ( ~ 1 , .. . ,w m ) and { w:, . ., ,w;}) respectively. Put y = ( i n f p ) a and W;= inf (K), i = 1,2. (i) There is an injective F-linear map

j=1

then B(cp) = ( c p j ; ) . (ii) I f & = V2 and wi = w;)1 5 i 5 m, then

15. Inflated modules over twisted group algebras

251

is an injective homomorphism of F-algebras. (iii) Suppose that gw; =

m

m

j=1

j=1

C ajj(g)wj,gwI = C bj;(g)wi

If Q E H o ~ F ~ G ( W@ FI V,W2 @ F V ) and

ifO((p)

(ajj, bjj E F ) = (Fji), then

Proof. (i) We may write any element in W28 V uniquely in the form w(i 8 vj for some v j E V . Hence, given and v E V,

cp E H o r n ~ ( W 1 8V,W2 8 V ) there exists a unique

cpj;

E E n d F ( V ) for which m

P(Wj

@v) =

1w;8

(1 5 i

Cpjjv

5 rn)

j=1

It therefore suffices to verify that if Q is an FYN-homomorphism, then each Pjj E E n d F a N ( V N ) . To this end, assume that cp is an FYN-homomorphism and fix n in N . Then m

C

j=1

W;

8 v j j ( f i v > = v(wj 8 f i v ) = ~

( f i8~n ui ) = ~ ( i i i ( w8 j v)) m

= ncp(w; @I v) =

fi(Cw; 8 Qj;v) j=1

m

=

c w ;8 f i Q j j ( V ) j= 1

and thus 'pjj(fiv)= ncpj;(v),as required. (ii) Let Q , $ E E ~ ~ F - ~ N ( (@IWVIN))N and let O(Q) = ( $ j i ) . Then

(vji),O($) =

Induction from normal subgroups

252

which shows that O(cp$) = O(cp)O($). (iii) Fixing g E G, cp E H o ~ F ~ G@(FW V,~W, @ F V ) ,we have

m

m

j = 1 k=l

m

/ m

and

k=l

whence

m

m

aji(9)vkjgv j=1

=

bkj(g)&'jiv j=1

Substituting v for ij-lv, we therefore derive m bkj(g)gyji(g-'v)

253

15. Inflated modules over twisted group algebras

thus completing the proof. We have now accumulated all the information necessary to prove the following result essentailly due to Huppert and Willems (1975, Satz. 2.)

15.8. Theorem. Let N be a normal subgroup of a finite group G, let F be an arbitrary field and let V be an FaG-module. Then for Q = GIN and for FPQ-modules U, U1 and U2 the following properties hold: (i) If VN is totally indecomposable and U is indecomposable, then

i n f ( U )@ F V is an indecomposable FrG-module, where y = (inf pja. (ii) If V N is absolutel~irreducible and U is irreducible, then inf (Uj @ F V is an irreducible FYG-module. (iii) If V , is totally indecomposable, then inf ( & ) @ F V = inf ( U ~ ) @ F V implies U, U2.

Proof. We put W = inf(U) and W; = inf(V;),i = 1,2. (i) Suppose that cp is an idempotent of EndF7G(W @ V ) . It suffices to verify that cp = 0 or cp = 1. Choose an F-basis ( ~ 1 , .. . ,wn}of W and write m

Swi = C a j i ( g ) w j

(9 E G,aji(g) E F )

(6)

j=1

In the notation of Lemma 15.7(iii) with 1 5 i 5 rn, we have O(p) = ( c p j ; ) , where m

U1

=

U2 = U and w;

= wi,

m

C aji(g)Ykj = jC akj(9)v:; =1

(7)

j=l

Applying Lemma 15.7(iii)) ( c p j ; ) is an idempotent of (EndFaN(VN)),. Bearing in mind that VN is totally indecomposable, we have

EndpN(VN) = F - l v

+ J(EndpN(VN))

Induction from normal subgroups

254

Thus we may write

+

= X j i IV

vji

with

Xji

E

(1 5 i , j

xji

5 m)

E J ( E ~ ~ F ~ N ( VByN Lemma )). 15.6, it follows that

F , ~ j i

~ 3' 9

=. X j i * l v

+

for all 9 E G

~ ; i

where x;i f J ( E ~ ~ F ~ N ( VComparing N)). the coefficients of Iv in (7), we derive m m j=1

j=1

Applying (6), it now follows easily that the map cp' : W -+W given 1 5 i 5 n, is an F'"fflG-homomorphism. by cp'(w;) = & X j i w j , Note that, by Lemma 15.4, W is indecomposable. Observe also that the matrix ( X j i ) of cp' is obtained from ( c p j i ) by reduction modulo J(Endp,( V N ) ) .Since ( c p j i ) is an idempotent, so is 'p' and hence cp' = 0 or cp' = 1, by the indecomposability of W . Replacing cp by 1 - cpl if necessary, we may therefore assume that cp' = 0, in which case Qji

EJ(E~~F~N(VN for) )all

;,j

Since J(EndpN(V,)) is a nilpotent ideal, there is a positive integer s such that ( c p j i ) s = 0. Thus cp = pa = 0 and i n f ( V )8~V is indecomposable. (ii) Let M # 0 be an FYG-submodule of W @ F V and let u be a nonzero element of M . Then we may write k

=CS~@V (Vi ~ E V , SE~ W ) i= 1 s1 # 0 and 211,. . . ,vk linearly independent over F . u

with Since V , is absolutely irreducible, E n d p , ( V ~ ) = F . Hence, by the Jacobson's density lemma, given v E V ,we may find a = CxEN uzZ in F*N, ax E F such that avl = v and a V i = 0 , 2 5 i 5 k. Since the elements 5 , s E N , operate.trivial1y on W , we obtain a(si

@ Vi) =

c c

axZ(si

@ Vi)

ZEN

=

a,(& @ Z V J

SEN

= s; @ a v i

15. Inflated modules over twisted group algebras

255

and therefore k

We conclude that

s1

k

@

V 2 M and hence, for all g E G,

However, FaGsl is a nonzero submodule of W and W is irreducible, by Lemma 15.4(i). Thus M = W @F V and therefore W @ F V is irreducible. (iii) Assume that cp : Wl @ F V -+ W2 @ F V is an FYG-isomorphism. Then we have dirnFWl = dirn~W2and so we may apply Lemma 15.7(iii). If 6(cp) = ( c p j ; ) then in the notation of Lemma 15.7(iii), we have m m aji(g)Pkj

=

j=1

C

bkj(g)vTi

j=1

By repeating the argument of (i), we derive m

m

j=1

j=1

for some Xj; E F . The latter implies that the map cp' : W1 defined by

c

+

Wz

cp-l,

we

m

cp'(W1)

=

xjiw;

j=1

is an F'"fPG-homomorphism. Applying the same procedure to deduce that cp' has an inverse and therefore

Thus, by Lemma 15.4(ii), U 1 2 U2 as FPQ-modules. This completes the proof of the theorem.

In the notation of Theorem 15.8, assume that is absolutely irreducible. Then inf (U)@ F V is absolutely irreducible if and only if so is U. 15.9. Corollary.

VN

Induction from normal subgroups

256

Proof. Let E / F be a field extension. Applying Lemmas 15.4(iv) and 15.3, we have

( i n f ( U )@F

V)Eg

inf(U)E @ E V , g inf(UE)8, VE

(8)

Assume that i n f ( U ) @F V is absolutely irreducible. Then, by (8)) inf(UE) is irreducible. Hence UE is irreducible, by Lemma 15.4, and therefore U is absolutely irreducible. Conversely, assume that U is absolutely irreducible. Then UE is irreducible and since ( V E ) N = ( V N ) E , (V')N is absolutely irreducible. Applying (8) and Theorem 15.8(ii), we conclude that ( i n f ( U )@F V ) , is irreducible. Thus inf(U )@F V is absolutely irreducible, as we wished to show.

16. Induction of absolutely irreducible modules Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. Given an absolutely irreducible FN-module V , our aim is to provide a detailed analysis of the decomposition of VG.

16.1. Lemma. Let A be a finite-dimensional algebra over F , let V be an absolutely irreducible A-module, and let W = U @ F V , where U is some finite-dimensional vector space over F . For each a E A, let cp, E End=(V) be defined b y cp,(v) = uv,v E V . (i) I f 6 E E n d ~ ( wis) such that

6(l@ 9,) = (1 @ ya)6

for all a E A

then 6 = II, @ 1 for some II, E EndF(U). (ii) If6 E GL(W) and T E GL(V) are such that 8-l(1 @ cpa)6 = 1 @ T - ' ~ , T

for all a E A

then 0 = $ 8 T for some II, E GL(U).

Proof. (i) Let {ul,. . . ,u m } be an F-basis for U . Then, for all v E V , we have m

o(ui 8 v) =

C uj 8 Oji(v) j=1

(1)

16. Induction of absolutely irreducible modules

257

for some 8ji E E n d ~ ( v )Given . a E A, we have

m

j=1

Because the {uj} are linearly independent, we infer that

or, equivalently, that 0,; E E n d ~ ( v ) . Since V is absolutely irreducible, it follows that Bj; = X j i - l v for some Xi; E F . Now define T,!J E EndF(U) by

Then II, E GL(U)and, by (l),8 = $ 8 1, proving (i). (ii) Setting 7 = 1 87,we have

and therefore, by (i), 87-' = II, 8 1 for some II, E G L ( U ) . Hence

as required.

We are now ready to prove the following result partially due to Clifford (1937). 16.2. Theorem. Let F be an arbitrayfield, let V be an absolutely irreducible G-invariant FN-module, and let w = wo(V) E Z2(&,F*),

Induction from normal subgroups

258

Q = GIN,be an obstruction cocycle of V . Let ezt(V) denote any extension of V to an FOG-module, where a = infw ( b y Theorem 15.2,such an extension always exists). '1 {Ul,Uz,.. . ,Un)is a f i l l set of nonisomorphic irreducible (rspectively, absolutely irreducible) F"-' &-modules, then {inf(Ui)@F ezt(V)I 1 5 i 5 n} is a full set of nonisomorphic irreducible (respectively, absolutely irreducible) FG-modules whose restriction to N is a direct sum of copies of v.

Proof. By Corollary 15.9, it suffices to treat the case where all

U;are irreducible. Let M denote an irreducible FG-module such that MN is a direct sum of copies of V and put S = ezt(V), Si = inf (Ui),1 5 i 5 n. We may clearly assume that M = W @ V , where W is an F-space such that for all n E N,w E W and v E V ,

the

If 7 : N

G + G L ( M ) and T : G + G L ( S ) are the representations afforded by V,M and S, respectively, then by (2) and by the equality SN = V, we obtain + GL(V),p :

p(n) = 1 @ 7 ( n ) and ~ ( n=r)(n) ) for all n E N

(3)

Also if ng = gnl for some n,nl E N and g E G, then T(n)T(g) = T(g)T(nl) because a(n,g)= a(g,n)= 1 for all n E N,g E G. Thus

and hence, by (3), we have

16. Induction of absolutely irreducible modules

259

Invoking Lemma 16.1(ii), for each g E G there exists $ ( g ) E GL(W) such that

P ( d = $(s>@ 4 7 ) Setting gw = $(g)w,w E W , it follows that W is an Fa-lG-module such that M g W ezt(V) Moreover, since M is irreducible, so is W . Finally, since $(n) = 1 for all n E N, we conclude that W i n f ( U ) for some irreducible F"-'Qmodule U . Thus

M

inf(Vi)@ ezt(V)

.

for some i E {I,. . ,n }

Conversely, assume that U is an irreducible F"-'Q-module and let {ul,. . . ,uk} be an F-basis of U . By assumption, V is absolutely irreducible. Because V = e z t ( V ) N , it follows from Theorem 15.8(ii) that

inf(U ) 8 ezt ( V ) is irreducible. Finally, write

i n f ( U )8 ezt(V)= u 1 @V

$ - a .

$uk @ V

where each ui @ V is an FN-module, by the definition of i n f ( U ) . Since the map v H u, @ o is an FN-isomorphism of V onto u; 8 V , we deduce that ( i n f ( U )@ ezt(V))N is a direct sum of copies of V . Since, by Theorem 15.8(iii),

the result is established.

16.3. Corollary. Let N be a normal subgroup ofG and let W be an irreducible FG-module such that some irreducible direct summand V of WN is absolutely irreducible. Let H be the inertia group ofV and let w = W H ( VE) Z 2 ( H / N ,F*)be an obstruction cocycle of V . Then (i) There exists an irreducible F"-'(H/N)-module U such that

w 2 ( i n f ( U )8 e z t ( V ) ) G

Induction from normal subgroups

260

(ii) I f U is absolutely irreducible, then so is W .

Proof. (i) This follows from Theorems 2.4.2(iii) and 16.2 and transitivity of the induction. (ii) By Theorem 16.2, i n f ( U ) @ e z t ( V ) is absolutely irreducible. Now apply Corollary 2.4.5(iv). We now come to the main result of this section. 16.4. Theorem. Let F be an algebraically closed field, let V be an irreducible FN-module, let H be the inertia group of V and let charF not divide ( H : N ) . Denote by w = O H ( V E ) Z 2 ( H / N ,F') an obstruction cocycle o f V and by e x t ( V ) an extension ofV to an F"Hmodule with Q = infw. Let U1,. . . ,U,, be all nonisomorphic irreducible F"-'(HIN)-modules and, for each U;,let inf (U;) be the Fa-' H-module

injlatedfim U;. Then the

are nonisomorphic irreducible FG-modules such that for m; = dimFU;,

Proof. By Theorem 2.1 and transitivity of the induction, it suffices to prove that the inf(U;) @ e z t ( V ) are nonisomorphic irreducible FHmodules such that

To this end, we first invoke Theorem 2.10.14 to deduce that V H is completely reducible. Thus there exist integers t l , t Z , . . . ,t , and nonisomorphic irreducble FH-modules W1,W2,.. . ,W, such that V H2 @&W; By Theorem 2.2.14, ti equals the multiplicity of V as an irreducible constituent of (W;)N.Invoking Theorem 16.2, we may thus assume that W; % i n f ( U j ) e z t ( V ) (1 5 i 5 s )

17. Applications

261

Note that the multiplicity of V as an irreducible constituent of ( i n f ( U i ) @ e s t ( V ) ) N is dimFU; = m;. Accordingly,

By Proposition 6.11, F"-'(H/N) is semisimple, because charF does not divide ( H : N ) . Taking into account that F is algebraically closed, we have n

Thus

dimFVH = ( g m : ) dimFV, proving that n = s and hence the result.

17.

Applications

In this section, we provide a number of applications of the preceding results. Throughout, F denotes a field and G a finite group.

17.1. Lemma. Let F be an algebraically closed field such that charF does not divide [GI, and let V be an irreducible module over a twisted group algebra F"G. Then dimFV divides the order of G.

Proof. By making a diagonal change of basis {glg E G} of FOG, we may replace cu by any cohomologous cocycle. Hence, we may assume that cy is of finite order, say m. Let E be a primitive m-th root of 1 in F and let G' =< e'glg E G, 1 5 i 5 m >. Then the map f : G' + G given by f ( E i g ) = g is a surjective homomorphism whose kernel is a central subgroup < E >. Moreover, if p : G --t G L ( V ) is an arepresentation of G afforded by V , then p* : G + G L ( V ) defined by p*(ciij) = Eip(g) is easily seen to be an irreducible representation of G'. Because charF does not divide ]GI and G*/ < E >%G, the result

Induction from normal subgroups

262

follows by applying the following standard fact (see Curtis and Reiner (1981)): If F is an algebraically closed field, charF does not divide IGl and 2 is a central subgroup of G , then the dimensions of irreducible FGmodules divide ( G : 2). Let p be a prime. A group G is said to be p-solvable if the composition factors of G are either pgroups or p’-groups. It is an immediate consequence of the definition that (i) Any extension of a psolvable group by a psolvable group is psolvable. (ii) Subgroups and homomorphic images of psolvable groups are psolvable. We are now ready to establish the following important result. 17.2. Theorem. Let N be a normal subgroup of G , let F be an algebraically closed field and let V be an irreducible FG-module. If charF = p > 0 divides ( G : N ) , assume that GIN is p-solvable. Then dimFV divides ( G : N ) d , where d is the dimension of an irreducible constituent of VN.

Proof. We first demonstrate that the result holds under either of the following hypotheses: (i) charF does not divide ( G : N ) . (ii) charF = p > 0 and G / N is a p-group. Denote by W an irreducible constituent of VN. Let H be the inertia group of W and let w E Z 2 ( H / N ,F*)be an obstruction cocycle of W . By Corollary 16.3(i), there exists an irreducible F”-’(H/N)-module S such that

v

( i n f ( S )C ~ ezt(W))G I

Because dimFS = dirnFinf ( S ) and dirnFW = dirnFezt(W), we have

dimFV = (dirnFS)(dirnFW)(G: H ) In case (i), dimFS divides ( H : N ) , by Lemma 17.1, so dirnFV divides

( G : N)dimFW. In case (ii), F”-l(H/N) is a local ring by Proposition 6.4. Hence

17. Applications

263

dimFS = 1 and so dimFV again divides ( G : N)dimFW. Turning to the general case, we use induction on /GI. Suppose that the result is true for groups of lower order than G. By the above, we may assume that charF = p > 0 divides ( G : N ) , in which case GIN is psolvable, by hypothesis. Because GIN is p-solvable, there exists a proper normal subgroup M of G containing N and such that G / M is either a p or p'-group. Denote by W an irreducible constituent of VM and let S be an irreducible constituent of WN. Then S is clearly an irreducible constituent of V N . By the foregoing, dimFV

divides (G : M)dimFW

and, by induction hypothesis,

dimFW

divides

( M : N)dimFS

Thus dimFV divides ( G : M ) ( M : N)dirnFS = (G : N ) d i m ~ S as , required. 17.3. Corollary. (Dude (1968), Swan (1965')). Let F be an algebraically closed field of characteristic p > 0 and let A be a normal abelian subgroup of a p-solvable group G. Then the dimensions of irreducible FG-modules divide ( G : A).

Proof. Note that all irreducible FA-modules are of dimension 1. The desired conclusion is therefore a consequence of Theorem 17.2.

As a prepararion for the proof of our next result, we now record the following properties. 17.4. Proposition. Let N be a normal subgroup ofG such that (G : N ) = pn, p prime, and let F be an algebraically closed field of characteristic p. If V is an irreducible FN-module, then

where U is an irreducible FG-module such that V is a constituent of UN.

Induction from normal subgroups

264

Proof. We know, from Theorem 2.12.4, that P ( V ) is an indecomposable FN-module. Hence, by Theorem 7.5, P(V)" is indecomposable. It follows from Proposition 2.12.5(i) that P(VG)E P(V)". By Corollary 2.2.5(i), there is an FG-epimorphism V G + U. Thus, by Lemma 2.12.2(ii), P ( U ) is isomorphic to a direct summand of P(V"). But, by the above, P(V") is indecomposable, hence the result. H Let N be a normal subgroup ofG and let V be an irreducible FG-module. If(G : N ) # 0 in F , then 17.5. Proposition.

Proof. By Theorem 2.10.14, J(FG) = F G J ( F N ) and so

where

with nonisomorphic irreducible FG-modules 2.12.3(iii)) F G = $idiP(x)

x. Then, by Theorem

and so (FG)N = $ i d ; P ( V i ) ~ (3) Bearing in mind that F G is a free (hence, projective) FN-module, we also have (FG)N = P((FG)N) = P ( (FG)N/ J ( FN)(FG)N) = ~ ( ( F G / J ( F G ) ) N ) (by (1)) = $;d;(P((K)N)) (by (2) and Lemma 2.12.l(iii)) Thus the desired conclusion follows by virtue of (3) and Proposition 2.12.5(ii). H Given a natural number n and a prime p , we denote by np the highest power of p dividing n and put np:= n/np.

17. Applications

265

Let G be a p-solvable group, let F be an algebraically closed field of characteristic p , and let V be an irreducible FG-module. Then (i) (Fong (1961)) dirnFP( V ) = /GIp(dirnFV),I. (ii) (Willems (1980)). If V = W Gfor some FH-module W , H is a subgrozlp of G, then P( V ) s P( W ) G 17.6. Theorem.

Proof. (i) We argue by induction on IGl. Let N be a maximal normal subgroup of G. Because G is psolvable, either [ G I N [= p" for some n 2 1 or G / N is a p'-group. Owing to Clifford's theorem, there is a decomposition

with nonisomorphic irreducible FN-modules K of equal dimension. Furthermore, by Theorem 17.2, es divides IG/NI. Suppose first that IG/NI = p". Because dirnFV = esdirnFV1, we have

(dirnFV),t = (dirnF&),t Note also that, by Proposition 17.4,

(5)

Thus

as required.

Next assume that GIN is a p'-group. Owing to Proposition 17.5, P ( V , ) S P(V)N and so dirnFP(V) = dirnFP(VN). On the other hand, by (4>1 ~(VN = )e ( P ( l 4 )CB @ P(K))

Induction from normal subgroups

266

and thus

where the last equality follows from the fact that [NIP = ]GIp and esl(G : N ) . Hence (i) is established. (ii) Owing to Proposition 2.12.5(i), P ( V ) is isomorphic to a direct summand of P ( W ) G .On the other hand, by (i), we have

Hence P ( V ) "= P(W)Gand the result follows. H 18.

The Loewy length of induced modules

Throughout this section, A denotes a finite-dimensional algebra over a field F . Unless explicitly stated otherwise, all A-modules are assumed to be left and finitely generated. The same assumption is made on FG-modules, where G is a finite group. Let V # 0 be an A-module. Then the descending chain

v 2 J(A)V 2 J(A)2V2 . . . of submodules of V is called the (lower) Loewy series of V . Since J ( A ) is nilpotent, there is an integer k, called the Loewy length of V such that J(A)"-'V # 0 but J(A)"V = 0 Observe that if J(A)'V = J(A)'+lV for i < k, then

J(A)"'V = J(A)k-i-'J(A)'V= J(A)"-'-'.J(A)'+'V = J(A)kV= 0,

18. The Loewy length of induced modules

267

which is impossible. Thus, if k is the Loewy length of V , then

V

J(A)V 3

18.1. Lemma. Let e

#0

-

a

-

3 J(A)k-lV 3 0

be an idempotent of a ring R. Then

J(eRe) = eJ(R)e Proof. Let I be a primitive ideal in R and let V be a faithful irreducible R/I-module. Then eV is an eRe-module. If eV = 0, then J ( e R e ) eRe C 1. Assume that eV # 0. Let 0 c W = e W eV where W is an eRe-module. Then V = RW since V is irreducible. Thus eV = e R e W W and so eV = W . It follows that eV is an irreducible eRe-module, hence J ( e R e ) V = J ( e R e ) e V = 0. Consequently, J ( e R e ) 5 I for any primitive ideal I in R. Thus J ( e R e ) J ( R ) and J(eRe) = eJ(eRe)e eJ(R)e. Assume that a E J ( R ) . Then eae E J ( R ) and so, by Proposition 1.4.10, ( 1 - b ) ( l - eae) = 1 for some b E R. Thus beae = eae b and multiplying by e on both sides yields eae + ebe = ebeae. Therefore ( e - ebe)(e- eae) = e so that e - eae is a left unit. Because e J ( R ) e is an ideal of eRe, it follows from Proposition 1.4.10 that e J ( R ) e C J ( e R e ) , as required.

c

c

+

If W # 0 is an A-module and E = EndA(W),then W will also be regarded as an E-module via 'pw = 'p(w) for all 'p E E , w E W . 18.2. Lemma. Let e # 0 be an idempotent of A, let W = A e and let E = EndA( W ) . Then

for all i 2 1

J(E)'W = Ae(eJ(A)e)'

In particular, the Loewy lentgh of W as an E-module is equal to the nilpotency index of J ( E ) . Proof. For each x E eAe, let By Lemma 1.5.9, the map

E E be defined by f3:(w) = wx.

f3:

eAe + E

Induction from normal subgroups

268

is an anti-isomorphism of rings. Moreover, by Lemma 18.1, J(eAe) =

eJ(A)e and so fz E J(E)' if and only if x E (eJ(A)e)i.Thus

J(E)'W = Ae(eJ(A)e)' Finally, J(E)'W = 0 if and only if (eJ(A)e)' = 0. Thus the Loewy length of the E-module W is equal to the nilpotency index of J(eAe) = eJ(A)e. Because eAe Z E", the result follows. We have now accumulated all the information necessary to prove the following result in which L ( V G )denotes the Loewy length of the induced FG-module V G .

(Clarke (1972)). Let F be a field of charucteristic p > 0, let N be a normal p'-subgroup of G , let H be the inertia group of an irreducible FN-module V and let E = EndFG(VG). Then (i) J(FG)"VG = FG J(FH)"VH = J(E)"VG for all n 2 1. (ii) L ( V c ) is equal to the nilpotency index o f J ( E n d F H ( V H ) ) . 18.3. Theorem.

Proof. We first show that (ii) is a consequence of (i). Indeed, we may take V = F N e for some primitive idempotent e of F N , in which case V G = FGe. Hence, by Lemma 18.2, the Loewy length of V Gas an E-module is equal to the nilpotency index of J(E). Thus (ii) is a consequence of (i) and Lemma 12.1. To establish (i), we write

as a sum of primitive idempotents of F N with e = el. Then we have

J ( F G ) V G = J(FG)e = FG(J(FG)e) = FGe(J(FG)e)t - - - FGe,(J(FG)e)

+

(1)

as left FG-modules, where the sum is not necessarily direct. For each u E e;FGe, let f a E HOmFG(FGe;,FGe) be defined by fa(.) = xu for all x E FGei. It follows, from Lemma 1.5.8, that the map

1

e;FGe a

HomFG(FGei,FGe)

+ H

f

a

18. The Loewy length of induced modules

269

is an F-isomorphism. In particular, if FGe; E FGe, then there is an a E e;FGe such that fa : FGe; + FGe is an FG-isomorphism. For the sake of clarity, we divide the rest of the proof into three steps. Step I. Here we demonstrate that J(FG)e = FGeJ(FG)e. Because FGeJ(FG)e J ( F G ) e ,it follows from (1) that it suffices to show that

e;J(FG)e ejFGeJ(FG)e

(1 5 i 5 m )

(2)

Let f; be the centrally primitive idempotent of F N with e;fi = e;, 1 5 i 5 m, and let f: be the sum of G-conjugates of f;. Then f: is clearly a central idempotent of FG. Next observe that if and thus

f1

and f; are not G-conjugate, then

frf1

=0

e;FGe = e;f;f;*FGf,*flel= 0 We may therefore assume that f1 and f; are G-conjugate, say g - l f j g . Then ( g - l e ; g ) f l = g - l ( e i f i ) g = g-leig

f1

=

and therefore the irreducible FN-modules FN(g-'e;g) and F N e are in the same block F N f l . However N is a p'-group, so F N e Z FN(g-'e;g) and thus

FGe 2 FG(g-'e;g) E FGe; By the foregoing, we may find a E e;FGe such that

fa

: FGe; + FGe

is an FG-isomorphism. Consequently, there exists b E eFGe; with f;'(y) = yb for all y E FGe. Accordingly, zab = z for all z in FGe;. Thus e; = eiub = (e;a)b = ab

and therefore for any c E eiJ(FG)e, c = e;c = (ub)c= a(bc) E e;FGeJ(FG)e

This proves (2) and hence the required assertion. Step 2. Our aim is to prove that J(FG)"VG = J(E)"VG for all n 2 1. Owing to Lemma 18.2, we need only show that

J(FG)ne= (FGe)(eJ(FG)e)"

for all n 3 1

Induction from normal subgroups

270

The case n = 1 being a consequence of Step 1, we argue by induction on n. Thus assume that

J(FG)ke= (FGe)(eJ(FG)e)k for all k 5 n Multiplying (3) on the left by J ( F G ) gives

(3)

J(FG)"+'e = (J(FG)e)"+' for all k 5 n whereas multiplying (3) on the right by J(FG)e gives

(4)

( J ( F G ) k e ) ( J ( F G ) e=) (FGe)(eJ(FG)e)"'

for all

k 5n

(5)

We obtain therefore

J(FG)"+'e = (J(FG)e)"+l (using (4)with k = n ) = (J(FG)e)"(J ( F G ) e ) = (J(FG)"e)(J(FG)e) (using (4) with k = n - 1 ) = (FGe)(eJ(FG)e)"+' (using ( 5 ) with k = n )

+

which establishes (3) for k = n 1. Step 3. We now complete the proof by demonstrating that

J(FG)"VG= FG - J ( F H ) " V H

for all n

21

We keep the notation of Lemma 12.1 and put El = EndFH(VH).Then k

i=s+l

k

= J(FH)"VH

+C i=s+l

= FG * J ( F H ) " V H ,

gjJ(El)"(1 @ V )

(by Step 2)

18. The Loewy length of induced modules

271

as desired. Our next aim is to provide a sufficient condition under which L ( V G ) is equal to the nilpotency index of F P , where P is a Sylow psubgroup of the group H in Theorem 18.3. 18.4. Theorem. Let F be an algebraically closed field of characteristic p > 0 , let N be a normal p'-subgroup of G and let H be the inertia group of an irreducible FN-module V . If P is a Sylow p subgroup of H , assume that P N d H . Then (2) The Loewy length L(VG)o f V G is equal to the nilpotency index of J ( F P ) . (ii) n(p - 1) + 1 5 L(VG)5 p", where pn = IPI. (iii)L(VG)= n(p - 1) 1 i f and only if P is elementary abelian. (iu) L ( V G )= p" if and only if P is cyclic. (v) I f P is abelian, say P = l-If=l < gi > with I < g, > I = p";, then

+

L ( V G )= 1 - k t

k

Cpmi i=l

Proof. By Theorems 4.2 and 18.3(ii), L ( V G )is equal to the nilpotency index of J ( F * (H / N ) ) for some cy E Z 2 ( H / N ,P). Put = H / N and P = P N / N . By hypothesis, is is a normal Sylow p-subgroup of H and, by Lemma 6.2, F a p F P . Furthermore, since N is a p'-group, p E P and hence F a p 2 F P . We now use the following standard property of crossed products: if N is a normal subgroup of G of index n, S an arbitrary ring such that n is a unit of S and S * G is a crossed product of G over S, then

J ( S * G) = ( S * G ) J ( S * N )

(6)

(see Theorem 3.16, p.87 in Karpilovsky (1987a)). Since ( S * G ) J ( S* N ) = J ( S * N ) S * G by Theorem 3.6(iii), it follows from (6) that the nilpotency index of J ( S * G) is equal to that of J ( S * N ) .

Applying the general observation above to the twisted group algebra PI?, we conclude that the nilpotency index of J ( F " H ) is equal to that of J ( F P ) , proving (i). The rest follows from some general results on the Jacobson radical of group algebras, namely from Proposition

272

Induction from normal subgroups

1.10, p.313, Theorem 3.2, p.323 and Corollary 2.5, p.122 in Karpilovsky (1987b). H

273

Chapter 4 Projective summands of induced modules Let H be a subgroup of G, let F be a field and let V be an FH-module. It is natural to investigate circumstances under which V Ghas projective indecomposable summands. An important contribution to this problem was marked by an appearance of paper of Robinson (1989) to which this chapter is devoted. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G(if there are any) have relatively large dimension. Some of the results presented place restrictions on the size of Cartan invariants in terms of group-theoretic structure. Finally, we demonstrate that if there is only one (H,H)-double coset representative, say g, for which 1gHg-' n HI is not divisible by p , then ( l H ) G contains at most one indecomposable projective summand. Furthermore, if there exists such a summand, then it must be irreducible. 1.

The Reynolds ideal

Throughout this section, A denotes a finite-dimensional algebra over a field F . Recall that if $ E H o m ( A , F ) is such that Ker$ contains no nonzero right ideals (equivalently, Ker$ contains no nonzero left ideals), then the pair ( A , $ ) is said to be a Frobenius algebra. If, in addition, $(zy) = $(yz) for all z,y E A , then we refer to ( A , $ ) as a symmetric algebra. Some basic properties of Frobenius and symmetric

Projective summands of induced modules

274

algebras were established in Sec.10 of Chapter 3. We begin by recording certain additional properties of these algebras. For each subset X of A, we denote by I ( X ) and r ( X ) the left and right annihilators of X defined by

Z(X) = { a E AlaX = 0 } , r ( X ) = { a E AlXa

=0)

It is clear that 1(X) and r ( X ) are left and right ideals of A, respectively. Recall that if ( A ,+) is a Frobenius algebra and X is a subset of A , then the subsets ' X and X I of A are defined by

'X X'

= { a E A I $ ( u X )= 0 } = { a € A I $ ( X a >= 0 }

1.1. Lemma. (i) Suppose that ( A , $ ) is a Frobenius algebra. Then, for any subset X of A, l ( X ) is the largest left ideal of A contained in

LX. (ii) Let [ A , A ]be the commutator subspace of A. If ( A , $ ) is a symmetric algebra, then Z ( A ) = [A,A]*. Proof. (i) By the definition of ' X , we have Z(X) C ' X . Let L denote any left ideal of A contained in ' X . Then

+ ( A L X )= $ ( L X ) = 0 so L X = 0 and hence L Z(X), as asserted. (ii) Given z E A, z E [A,A]* if and only if for all 2,y E A +((EY - y+)

=

$(w) - $(yzz) = $(w> - $(w) =0

or if and only if

$(A(yz - zy)) = 0

for all y E A

Because the latter is equivalent to yz - zy = 0 for all y E A , the result follows. rn From now on, we assume that charF = p

T ( A )= { a E AlaPn E [A,A]

> 0 and put

for some n 2 1)

1. The Reynolds ideal

275

Then T ( A )is an F-subspace of A containing [A,A].

1.2. Theorem. (fr'iilshammer(1981)). Let A be a Jinite-dimensional algebra over a j e l d F of characteristic p > 0 . (i) T ( A ) = [A,A] J ( A ) and J ( A ) is the largest left ideal of A contained in J ( A ) . (ii) If A is a Frobenius algebra, then SocA = T(A)'A. (iii) If A is a symmetric dgebra, then T(A)' = Z ( A ) n SocA and, in particular T(A)' is an ideal of Z ( A ) .

+

Proof. (i) We first assume that A is a simple algebra. Because for any z E Z ( A ) and 2,y E A ,

both [A,A] and T ( A ) can be regarded as vector spaces over the field Z ( A ) . We may therefore assume that Z ( A ) = F , i.e. that A is a central simple F-algebra. Let E be the algebraic closure of F , so that

, and for some n 2 1. Then obviously T ( A E )= [ A EAE]

Taking into account that

we deduce that dzmF[A,A] = dzmFA - 1. If it were true that A = T ( A ) ,then since T ( A E )is an E-subspace of AE we would have AE = T ( A E ) ,a contradiction. Thus A # T ( A ) and hence, by the above, T ( A ) = [ A , A ] . Since A is simple, ( A , + ) is a symmetric algebra for some t,h E HomF(A,F) (Proposition 3.10.9). But then [ A , A )C Ker+ and therefore T ( A ) contains no nonzero left ideals of A. This proves the case where A is a simple algebra. Turning to the general case, we put A = A / J ( A ) and write

A = A1 $ * * * $ A ,

Projective summands of induced modules

276

where the Ai are pairwise orthogonal simple F-algebras. Then T ( A ; )= [A;,Ai],1 5 i _< rn, by the special case proved above. It follows that

+

Clearly, [ A , A ] J ( A ) T ( A ) and T ( A ) / J ( A )C_ T ( A ) . Thus T ( A )= T ( A ) / J ( A )and therefore T ( A )= [A,A] J ( A ) . Let L be a left ideal of A contained in T ( A ) . Then ( L + J ( A ) ) / J ( A ) is a left ideal of A contained in T ( A )and therefore in [A,A ] . Since A is semisimple, A is a symmetric algebra (Proposition 3.10.9). Thus, by the previous argument, [A,A]contains no nonzero left ideals of A. Hence L C J ( A ) and the required property follows. (ii) Assume that A is a Frobenius algebra. By Theorem 3.10.11, we have

+

' ( T ( A ) * )= T ( A )

(1)

r[Z(T(A)*A)]= T(A)'A (2) By (i) and (l),J ( A )is the largest left ideal of A contained in ' ( T ( A ) * ) . Applying Lemma 1.1(i), we deduce that J ( A ) = Z(T(A)') = Z(T(A)'A)

(3)

Thus, by (2) and (3), it follows that

SocA = r ( J ( A ) )= T(A)'A, as required.

(iii) Assume that A is a symmetric algebra. By (ii) and Lemma l.l(ii), it suffices to show that

T(A)' = [ A ,A]' n T(A)'A Because Z(T(A)'A) is a left ideal of A , we have

[Z (T(A )* A)]' = r [I( T (A)' A]

(4)

277

1. The Reynolds ideal

by Theorem 3.10.11. Consequently,

as we wished to show.

Let A be an F-algebra and let Soc(A) be the left socle of A . We shall refer to the ideal Z ( A ) n SocA of Z ( A ) as the Reynolds ideal of A . In what follows, we put

Rey(A) = Z ( A ) f l SocA The name is partially justified by the fact that Reynolds (1972) introduced this ideal for the special case where A is a group algebra. Our next aim is to determine Rey(FG) exclusively in terms of F and G which will give us the original ideal constructed by Reynolds. In what follows, G denotes a finite group, F a field, and Cl(G)the set of all conjugacy classes of G. 1.3. Lemma.

We have

[FG,FG] = { c z g g E FGI

c

xg = o for all C E CZ(G)}

g€C

Proof. This is a special case of Lemma 3.6.10. An easy proof can also be obtained from the following elementary argument. Fix C = { g l , . . .,gn} in CZ(G) and assume that z = Cy=lA;gi where A; E F and CZl A; = 0. Then z = Cyr: X;(g; - g n ) , and so in view of the identity g - tgt-' =

(gt-1)t - t(gt-1)

(t,g E G )

we have z E [FG,FG]. Since [FG,FG] is spanned by all z y - y z with z,y E G, the result follows by virtue of zy - yz = zy - z-l(xy)z. W

1.4. Lemma. Let tr : FG + F be defined b y tr

Then ( F G , t r ) is a symmetric algebra.

(C,,,

zgg)= 2 1 .

Projective summands of induced modules

278

Proof. It is an immediate consequence of the definition that tr is an F-linear map. Fix z = CgEGzggand y = CgEGyggin FG. Then we have t r ( z y )= zgyg-l = YgZg-1 = t r ( y 4

c

c

g€G

g€G

and tr(g-'z) = zg

for all g E G

In particular, if tr(FGz) = 0 then z = 0 as required. For the rest of this section, we assume that F is a field of characteristic p > 0. Given g E G, we write gp and gp1 for the p and p'-parts of 9 , respectively. We shall refer to C E C1(G) as being p-regular if all elements of C have order not divisible by p . For any subset X of G, we Put

x-'

= {z-llz E X }

and denote by F X the F-linear span of X . Let C1 = { l}, C2,.. . ,C, be all p-regular classes of G. Then the sets

are called p-regular sections of G. It is an immediate consequence of the definitions that (i) G = U;==,S; with S; n Sj = 0 for i # j and each Si is a union of conjugacy classes of G, S1 is a union of conjugacy classes of p-elements of G. (ii) Let X be a subset of G and let z = CSEG zgg E F G . Then

t r ( z X + )=

c

zg

g€X-1

and, in particular,

tr(zS?) =

c gES,-'

zg

1. The Reynolds ideal

279

1.5. Proposition. Let S1, Sz, . . . ,S, be all p-regular sections of G. Then (i) T ( F G ) = {x = C x g g E FGIC,,, xg = 0 for all i E { l , .. . , r } }

Proof. (i) We first claim that there exists n gpm = ypt

2 1 such that

for all rn = O(rnodn) and all g E G

(5)

Indeed, write (GI = p"k with ( p , k) = 1. Since (p,k) = 1, there exists a positive integer n such that pn 5 l(rnodk). Replacing n by its multiple, if necessary, we may assume that n 2 a . Let rn be divisible by n and m let g E G. Then gEm = 1 and, since p" l(rnodl),gg, = gpf. Hence P"

9

- P" -Yp

Pm

gp, = gp'7

proving ( 5 ) . Let x = C g E G x g g E FG. Then, by (5) there exists rn 2 1 such that gPm = gpt for all g E G and such that x E T ( F G ) if and only if xp" E [FG7FG].Owing to Lemma 3.6.9(i), xpm E

C x;"gpm

(mod[FG,F G ] )

9EG

T ( F G ) if and only if EgEGxgPmgpt E [FG,FG]. Applying Lemma 1.3, we deduce that x E T ( F G ) if and only if CgESi xi"' = 0 for all i E (1, . . . , r } . Because so 2 E

the required assertion follows. (ii) By Theorem 1.2(iij), it suffices to show that T(FG)' = Ci==, FS? By (i), we have

2 FS? C T(FG)' i=l

Projective summands of induced modules

280

Conversely, suppose that z = CgEGzgg E T(FG)* and let a , b E S;. Then and b-' belong to the p-regular section Sr' and so, by (i), a-' - b-' E T ( F G ) . But then tr(za-' - zb-l) = 0 which implies that

as required. 2.

Projective summands

Throughout, H denotes a subgroup of a finite group G and, unless explicitly stated otherwise, F is an algebraically closed field of characteristic p > 0. All modules are assumed to be finitely generated over their ground rings. Our main theme here is the study of projective summands of modules of the form V G ,where V is an FH-module. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G (if there are any) have relatively large dimension. Recall that a conjugacy class C of G is said to be p-regular if the order of the elements of C are not divisible by p. In what follows, we fix the following notation: C1, (72,. . . ,CT are all p-regular classes of G. Si = { g E Gig,, E Ci}, 1 5 i 5 r , are all p-regular sections of G.

si+ = CsES, 3.

Rey(FG) = Z ( F G )n Soc(FG) is the Reynolds ideal of FG. Note that, by Proposition 1.5(ii), T

Rey(FG) = C i=l

while by Theorem 1.2(i),(iii), Bey( F G ) = ( [ F G FG] ,

2.1. Lemma. Let

+ J ( FG))'

x be an F-character of G .

Then, for all g E G,

281

2. Projective summands

Proof. We may choose an integer m 2 1 such that g p m is p-regular. In a representation of G with character x, the linear transformation . say, in F and x(g) = corresponding to g has eigenvalues ~ 1 , ~ 2. ,. ,en, m E~ + . . + en. Similarly, x(gPm) = E; - - - &Ern. Since F has characteristic p , the binomial theorem shows that

+ +

x(g)Prn= &;m

+ + &Krn = x(gPrn) ' * *

Bearing in mind that gPrn= (gpt)pm,we have x(g)pm = x(gPrn)= x((gPdprn)= X(S,l)""

and therefore

Thus x ( g ) = x(gpt),as asserted. 2.2. Lemma. Let K , . . . , V , be all nonisomorphic irreducible F G modules and let x; be the character of G aforded b y V,, 1 5 i 5 r . Then 1< - i 5 r, form an F-basis (i) The elements z; = ,&Gx;(g-l)g, of Rey(FG). (ii) For any u E FG, i E (1,. . . ,r } , uz; = C g E G xi(ug-')g. (iii) An element u E F G annihilates V; if and only if uzi = 0,

l
Proof. (i) By Lemma 2.1, each xi is constant on p-regular sections. Hence each z; lies in Rey(FG), by virtue of (1). Note also that, by (I), dimFRey(FG) = r and so we need only verify that z1,. . . ,zT are F-linearly independent. If Ciz1A;z; = 0 for some A; E F , then

0=

2c

~;xi(g-')g =

i=l gEG

and therefore

c (2

AzXi(g-')) 9

g€G

i=l

T

Aix;(g) = 0 i=l

for all g E G

Projective summands of induced modules

282

But, by Proposition 1.6.3, X I , . . . ,xz are F-linearly independent, hence all X i = 0 and (i) is established. (ii) Write u = C u,x with u, E F, x E G. Then ug-' = C u,(xg-') and so

as required. (iii) By (ii), we must show that u annihilates if and only if xj(u9-l) = 0 for all g E G. If u annihilates E, then so does ug-' and hence xj(u9-l)= 0, for all g E G. Conversely, assume that xj(u9-I)= 0 for allg E G. Then xi(uFG) = 0. If p; is a matrix representation of FG which affords xi, then pi is a surjective homomorphism FG + M,,(F) of F-algebras for some n; 2 1. This follows from the assumption that F is algebraically closed and from Proposition 1.6.10. Thus

trace(p(u)Mni(F)) =0 which implies p ( u ) = 0. Hence u annihilates V , and the result is established. 2.3. Lemma. Let V be a n irreducible FG-module and let e be a primitive idempotent of FG. Then eV # 0 if and only if FGe 2 P ( V ) .

Proof. Assume that FGe 2 P ( V ) . Then V 2 FGe/J(FG)e by Theorem 2.12.3(ii). Hence eV # 0. Conversely, suppose that eV # 0, say ev # 0 for some v E V . Since V is irreducible, FGev = V . Hence the map

{

FGe

-+

V

X H X V

is a surjective FG-homomorphism. Thus V 2 FGe/J(FG)e and there-

fore, by Theorem 2.12.3(ii), P ( V ) E FGe. 2.4. Lemma. Let V be an FG-module. Then z;V

Soc(V).

283

2. Projective summands

Proof. Since zj E Z ( F G ) ,z;V is clearly a submodule of V . By Lemma 2.2(i), zi E Rey(FG) = Z ( F G ) n Soc(FG)

Hence J ( F G ) z j = 0 and therefore

J(FG)z;V = 0 Thus z;V

C Soc(V), as asserted.

W

We have now accumulated all the information necessary to prove the following result.

(Robinson (1989)). Let V,,V,,. . . ,V , be all nonisomorphic irreducible FG-modules, let V be an FG-module and let a; 2 0 be the multiplic2ty of P( as a direct summand of V . Then 2.5. Theorem.

x)

(1 5 i

z;V 2 n;V,

5 r)

In particular, z;V # 0 if and only if P ( K ) is isomorphic to a direct summand of V . Proof, We may write V = Wl @

- - @ W, where each W; is inde-

composable, in which case z j v = Z ; W 1 €9

* * *

@ z;W,

Hence it suffices to prove that if V is indecomposable, then (i) z;V = 0 if V P(V,). (ii) z;V E V , if V E P ( x ) . We first show that (i) and (ii) hold for principal indecomposable FGmodules. Then we show that if z;V # 0, then V is isomorphic to a principal indecomposable FG-module. This will obviously complete the proof. Let e be a primitive idempotent of FG and let V = FGe. Then t;V # 0 if and only if zje # 0. Hence, by Lemma 2.2(iii), z;V # 0 if and only if eV, # 0, which by Lemma 2.3 happens if and only if V 2 P ( K ) . Moreover, if V 3 P ( x ) ,then

SocV

P(V,)/J(FG)P(V,)

Projective summands of induced modules

284

by Theorems 3.10.12(iii) and 2.12.3(ii). But, by Lemma 2.4, ziV SocV and hence z;V = SocV. Thus, if V 2 P ( K ) , then z;V 2 K. Let V be an indecomposable FG-module. By the foregoing, we are left to verify that if ziV # 0, then V is isomorphic to a principal indecomposable FG-module. So assume that ziV # 0. Choose v E V with z p # 0. Define the homomorphism of left FG-modules cp : F G + V by cp(x) = m for all 5 E FG. Then z;FG g Kercp. Hence there exists a primitive idempotent e E F G with z;FGe g Kercp. This implies that q F G e # 0 and hence, by the foregoing, F G e 2 P ( K ) . Furthermore, since Soc(FGe) = z;FGe is irreducible, F G e n Kercp = 0. The latter implies that F G e is isomorphic to a submodule of V . But FGe is a projective FG-module, hence F G e is also injective, by Theorem 2.10.3. Thus F G e is isomorphic to a direct summand of V . Since V is indecomposable, we conclude that V %! F G e and the result follows. 2.6. Corollary. Let Vl, . . . ,V , be all nonisomorphic irreducible FG-modules and let ni = dimFV;, 1 5 i 5 r. Then

z;FG E n i x Proof. Since F is algebraically closed, we have

F G 2 $T=,njP(&) where P( q),. . . ,P (V , ) are pairwise nonisomorphic. The desired conclusion now follows by virtue of Theorem 2.5.

=

2.7. Lemma. Let A be a finite-dimensional algebra over a field. two primitive idempotents of A such that elA e2A, then there exists a unit u of A such that e2 = u-lelu.

If el and e2 are

Proof. This fact established in the course of the proof of Theorem 3, p.59 in Jacobson (1956). For future applications, we next record the following result.

(Robinson (1989)). Let V be an FG-module and let E = EndFG(V). Then J(E)V has no nonzero projective submodule. 2.8. Theorem.

2. Projective summands

285

Proof. It suffices to show that J ( E ) V has no projective indecomposable submodule. Assume by way of contradiction that M is a projective indecomposable submodule of V contained in J ( E ) V . By Theorem 2.10.3, M is also injective, hence M is a direct summand of V. Thus M = eV for some primitive idempotent e of E. Write 1 = el + * - . et where el = e, e 2 ) .. . , et are orthogonal primitive idempotenets of E. Since M is a projective indecomposable module, it follows from Theorem 2.5 that zjM = Soc(M) # 0 for some i E { 1,.. . r } . Now M = eV C J ( E ) V and

+

)

J(E)V = e J ( E ) V @ (1 - e)J(E)V which implies that M = eJ(E)V. It follows that 0

# z;M

= z;eJ(E)V

and therefore z;eJ(E)ejV

#0

(3)

for some j E { I , . . . ,t } . In particular, since

zieJ(E)ejV = eJ(E)ziejV, we have z,ejV

# 0.

Because ej is primitive, ejV is indecomposable. Since ziejV must have and ejV E M ziejV = Soc(ejV)

# 0 we

by virtue of Theorem 2.5. Hence eV ejV and, by Lemma 3.5.2(iii), e E 2 e j E as right E-modules. It follows from Lemma 2.7 that there exists a unit u of E such that ej = ueu-l. Hence '] eJ(E)ejV = e [ u ~ ( ~ ) u -(ueu-l)V

= eJ(E)eu-'V = eJ(E)eV

(4)

Now M = eV is eEe-invariant, hence so is ziM = Soc(M). But z;M is an irreducible FG-module and F is algebraically closed. Hence E n d ~ c ( z ; M=) F . 1 and thus there is a surjective homomorphism eEe

+ EndFG(z;M)

Projective summands of induced modules

286

of F-algebras. Since, by Lemma 3.18.1, J(eEe) = e J ( E ) e , it follows that zieJ(E)eV = eJ(E)ez;M = 0 and hence, by (4))that

zieJ(E)ejV = 0 Since the above contradicts (3), the result is established. We need an auxiliary result concerning projective covers of contragredient modules. The facts exhibited below are valid for an arbitrary field F . Let V be an FG-module and let V* be the contragredient of V . If W is also an FG-module and a E H o m ~ c ( VW , ) ,then the map

a* : w" --+

v*

defined by

is an FG-homomorphism. It is clear that if

is an exact sequence of FG-modules, then so is

s w * s v* + 0

0 + x* 2.9 Lemma.

IfV is an irreducible FG-module, then P ( V ) * P ( V * ) .

Proof. By Corollary 2.5.12, P(V)*is a projective indecomposable module. Since P ( V * ) is also a projective indecomposable module, it suffices to show that Soc(P(V)*)2 Soc(P(V*)).But Soc(P(V*))2 V * ,hence we must show that

v*= Soc(P(V)*) By definition of P ( V ) ,there is a surjective homomorphism f : P ( V ) + V . Hence f * : V* --+ P ( V ) *is an injective homomorphism. But P ( V ) *

2. Projective summands

287

is projective indecomposable, and V * is irreducible and Soc( P( V ) * ) is also irreducible. Thus f* gives the desired isomorphism V * + Soc(P(V)*)and the result follows.

We next establish the following useful reciprocity theorem. 2.10. Theorem. (Robinson (1989)). Let H be a subgroup of G and let V and W be irreducible F G and FH-modules, respectively. Then the multiplicity of P ( V ) as a direct summand of WG is equal to the multiplicity of P ( W ) as a direct summand of VH.

Proof. Let V,,V,, . . . ,V , be all nonisomorphic irreducible FGmodules and let Wl, . . . ,Wt be all nonisomorphic irreducible FH-modules. Let { X j l l 5 j 5 t } be the elements of R e y ( F H ) as the zi's, 1 5 i 5 r were for R e y ( F G ) . Denote by m;j the multiplicity of P(K)as a direct summand of WT and put kj = d i m F W j , 1; = d i r n F ( K ) , 1 5 j 5 t , 1 5 i 5 r. We have

FGXj

S

C

(FHXj)G (kjWj)"

(by Corollary 2.6)

kjWy

(5)

Note also that, by Theorem 2.5,

where nij is the multiplicity of P(V,) as a direct summand of FGXj. By (5), kjrnij = n,j and hence, by (6),

dimFz,FGXj = kjrnijl;

(1 5 i

5 r, 1 5 j 5 t )

(7)

For any x = Cz,g E F G , x , E F,g E G , put x* = Cz,g-'. Then the is the contragredient map x H x* is an anti-automorphism of F G . If of V , then V;, . . . ,Vr*are all nonisomorphic irreducible FG-modules. Similarly, W:, . . . ,W: are all nonisomorphic irreducible FH-modules. If xa is the character of G afforded by v,*,then x t ( g ) = xi(g-'), hence

Projective summands of induced modules

288

This shows that 2; performs the role of zj with respect to K*. Similarly, Xj' performs the role of X j with respect to W'. We see that dimF(ziFGXj) = dimr(Xj'FGz:), since x H x* is an anti-automorphism of FG. Again, by Theorem 2.5 and our foregoing observation on Xj', we have

dimF(XfFGzf) = dirnF( W')s;j =

Jcjs;j

where sij is the multiplicity of P(Wj.)as a direct summand of ( F G z f ) H . It follows from (7) that m..l. = s.. t3 8 '3 (8) On the other hand, by Corollary 2.6, FGzr ( d i m F V , * ) v = Z;V,.. Hence sij = ljpij where p j j is the multiplicity of P(W,') as a direct Thus, by (S), mij = p i j for all i E (1,. . . ,r } , summand of j E (1,. . . , t } . But by Lemma 2.9, p ; j is the same as the multiplicity of P ( W j ) as a direct summand of (&)H, hence the result.

(v)~.

We are now ready to examine the dimension of projective summands. The numbers C;k in the theorem below are known as Cartan invariants of FG.

2.11. Theorem. (Robinson (1989)). Let H be a subgroup of G and let V,, . . . , Vr and W1,.. . ,Wt be all nonisomorphic irreducible F G and FH-modules respectively. Let mij be the multiplicity of P ( x ) as a direct summand of and let c;k be the multiplicity of v k as a composition factor of P ( x ) . Then, for any given i E { 1, , . . ,r }

dimFP( K) 2

c c CjkmkjdimF(P(Wj)) t

r

j=1 k=l

with equality if is projective. Furthermore, if H is of p'-index and the equality holds, then is projective.

Proof. By hypothesis, v k occurs with multiplicity

sition factor of P(K),hence

c;k

as a compo-

2. Projective summands

289

By Theorem 2.10, ( V k ) H contains P(Wj) as a direct summand with multiplicity m k j . Since P ( W l ) ,. . . ,P ( W t ) are nonisomorphic, we see that @ & I m k j P ( w j ) is a direct summand of ( h ) hence ~, t

dimFh 2

mkjdimFP(Wj)

(10)

j=1

The desired inequality now follows from (9) and (10). Assume that is projective. Then so is by Theorem 2.10.8(i). Hence (K)H @;=,mijP(Wj)

v

(v)~,

and, since cjk = 0 for i # k,c;; = 1, it follows that the equality holds. Conversely, suppose that the equality holds and H is of #-index. Then there is certainly some k E (1,. . . ,r } and some j E (1,. . . ,t} such that C ; k m k j # 0. In particular, C;k # 0 and so v k is a composition factor of P ( q ) . Furthermore, since m k j # 0, cjk # 0 the equality in (10) must hold and thus

Hence ( V ~ )isHprojective and so, by Theorem 2.10.8(i), V k must also be projective. Since & is a composition factor of P ( K ) ,it follows that i = k, i.e. K is projective. This completes the proof of the theorem.

As a useful companion to the result above, we next prove the following theorem.

2.12. Theorem. (Robinson (1989)). Let H be a subgroup o f G and let &, . . . , V,. and W1,. . . , Wt be all nonisomorphic irreducible FG and FH-modules, respectively. Let m;j be the multiplicity of P ( K ) as a direct summand of W y and let cjk be the multiplicity of wk as a composition factor of P(W,), 1 5 i 5 r , 1 5 j,k 5 t . Then, for any j E {L--,t}, r

(G : H)dimFP(Wj) 2

t

C C cjkmjkdimFP(K) i=l k=l

with equality if and only if Wj is projective.

Projective summands of induced modules

290

Proof. For 1 5 k 5 t , wk occurs C j k times as a composition factor of P(W j ) ,hence

c CjkdimFWk t

dimFP(Wj) 2 Again, for 1 5 k Wf, hence

(11)

k=l

5 t , P ( X ) occurs mjk times as a direct summand of

c m;kdZmFP(K) r

(G : H)dimFWk 2

(12)

i=l

The desired inequality now follows from (11) and (12). Assume that Wj is projective. Then C j k = 0 for k # j and By Theorem 2.10.8(ii), W y is projective, hence

P ( W j ) G W:

cjj

= 1.

eI=lmjjP(K)

which implies the desired equality. Conversely, suppose that the equality holds. Then there is some i E { 1, . ..,r } and some k E (1,. . . ,t } such that cjkmjk # 0. Since cjk # 0,mik # 0, the equality in (12) must hold and thus

Hence W t is projective and therefore so is Wk, by Theorem 2.10.8(ii). Since c j k # 0, wk is a composition factor of P ( W j ) . Hence k = j and Wj is projective as required. W As a preliminary to our final result, we record the following property which holds without any restriction on F .

2.13. Lemma. Let V be a projective FG-module. Then E = E ~ F G ( Vis) a symmetric F-algebra.

Proof. Since V is projective, the trace map TrF : EndF(V) + E is surjective by Theorem 2.10.3. We define the bilinear form

f:ExE+F

291

2. Projective summands

as follows: if z = T r G, (z I) and y = T rG, ( y I),

X I , yl

E E n d ~ ( v )then ,

f ( z , y ) = truce(zyl) We have

)

truce ( ~ Gr( ,xI> y/

= truce

= truce ( x ~ r f ( y l ) )

Assume that Tr?(yI) = T r G, ( y11). Then truce(zy") = truce(z'Tr,G ( y11 )) = truce(x'Tr,G ( y I)) = truce(zy')

(by (13))

(by (13))

proving that f is well defined. Note also that f is symmetric since by (13), trace(zy') = truce(yz'). Since z y = T r f ( z y ' ) for all z E E , we have f ( z z , y ) = truce(zzy') = f ( x , z y ) and hence f is associative. Finally, assume that f ( z , E ) = 0. Then t r u c e ( z E n d F ( V ) ) = 0 which clearly implies z = 0. Thus f is nonsingular and the result follows. Let V be an FG-module and, for each z E FG, let be defined by Cp&)

=

(pz E

Encl~(V)

(v E V )

If z E Z ( F G ) , then each cpz is in Z ( E ~ ~ F G ( V and ) ) thus we have a homomorphism

Projective summands of induced modules

292

of F-algebras. We shall refer to this homomorphism as the natura2 ho-

momorphism. We have now accumulated all the information necessary to prove the following important result. 2.14. Theorem.

(Robinson (1989)). Let V be any FG-module and write V = P @ W where P is projective and W has no nonzero projective direct summands. (i) Under the natural homomorphism Z ( F G ) + Z(EndFG(V)),the image of Rey(FG) is Rey(EndFc(P)). (ii) The number of isomorphism classes of indecomposable projective direct summands of V is equal to dimFRey(EndFG(P)).

Proof. By Theorem 2.5, Rey(FG) annihilates W . Hence the image, say M , of Rey(FG) is in Z(EndFG(P)).Furthermore, by Theorems 2.5 and 2.8, Rey ( F G )[ J (EndFc(P ) )PI = 0

which shows that M . J(EndFG(P))= 0. Hence

by the fact that EndFG(P) is a symmetric algebra (Lemma 2.13) and Theorem 3.10.12( i). Let E = EndFc(P). Then, by Theorem 1.2(iii) and Lemma 2.13,

Rey(E) = T ( E ) * and therefore, by Theorem 3.10.11(i),

dirnF(Rey(E))= dimF(E)- dirnF(T(E)), which, by Lemma 3.6.12, is the number of nonisomorphic irreducible Emodules. Thus, by Lemma 3.5.2, dirnF(Rey(E))is equal to the number of isomorphism classes of indecomposable direct summands of P. Since the latter number is equal to the number of isomorphism classes of indecomposable projective direct summands of V , (ii) is established. By the foregoing, we are left to verify that dimFM is equal to the number of isomorphism classes of indecomposable direct summands of

293

2. Projective summands

P. To this end, let

. . ,z,

be the F-basis of Rey(FG) defined in Lemma 2.2(i). After renumbering the z;’s, if necessary, we may assume that 2 1 , . . . ,zk, k 5 T , are all the 2;’s which do not annihilate P. Their images in R e y ( E ) are linearly independent since z j annihilates any direct summand of P whose socle is not isomorphic to V, (Theorem 2.5). Thus, by Theorem 2.5 dirnFM is equal to the number of isomorphism classes of indecomposable direct summands of P. This completes the proof of the theorem. 21,.

It is worthwhile to mention that a special case of Theorem 2.14(ii), in which V = F G is a classical Brauer’s theorem which asserts that the number of nonisomorphic irreducible FG-modules is equal to the number of p-regular classes of G. It will next be shown (Corollary 2.17) that Lemma 2.13 holds under more general circumstances. We need the following two useful observations.

2.15. Lemma. Let S be an arbitrary ring and let V be an S module. If W is a direct summand of V and e : V + W is an idempotent projection, then

Ends(W)

eEnds(V)e

Proof. We have W = eV and V = W @ U , where U = (1 - e ) V . It is clear that the subring K of E n d s ( V ) defined by

is isomorphic to Ends(”). Now $ ( U ) = 0 if and only if $ = $e. Since $ ( e v ) = e$e(v) (1 - e)+ev,v E V , we have +(W)C W if and only if (1 - e)$e = 0, which happens if and only if = e$. Thus I< = e E n d s ( V ) e and the result follows.

+

+

2.16. Lemma. Let A be a symmetric algebra over an arbitrary field F . Then, for any idempotent e of A, eAe is a symmetric algebra.

+

Proof. Let E H o m F ( A , F ) be such that ( A , $ ) is a symmetric algebra, and let X E H o m ( e A e , F ) be the restriction of to eAe.

+

Projective summands of induced modules

294

Assume that X(eAes) = 0 for some z E eAe. Then z = eze = ze and

which implies z = 0. W

2.17. Corollary. Let A be a symmetric algebra over an arbitrary field F . If P is a projective A-module, then EndA(P) is a symmetric algebra. Proof. By hypothesis, P is a direct summand of a free A-module Q )M,(EndA(AA)) E hf,(Ao), it of rank n 2 1. Since E ~ ~ A ( 2 follows from Lemma 3.10.8 that EndA(Q) is a symmetric algebra. The desired conclusion now follows by Lemmas 2.14 and 2.15. W

The following result will enable us to provide an explicit description of Rey(EndA(P)), where P is a projective A-module. 2.18. Proposition. (Thkvenaz (1988)). Let A be a finite-dimension algebra over an arbitrary field, let P be a projective A-module and let S o c ( E n d ~ ( P )be ) the right socle o f E n d ~ ( P ) .Then

Proof. If P is A-free of rank 1, then EndA(P)" E A by right multiplication. If r, denotes the right multiplication by a E A, then r, E Soc(EndA(P)) if and only if r,rb = 0 for all b E J ( A ) , which happens if and only if Aa = Irn(r,) Soc(P). If P is A-free of rank n, then EndA(P)" E M,(A) and note that

Soc(M,(A)) = M,(Soc(A)) Applying the case n = 1 to each entry f E &?,(A), the result easily follows in this case. If P is arbitrary, then P is a direct summand of a free A-module Q. Let e E EndA(Q) be an idempotent projection onto P. Then, by Lemma 2.15, EndA(P) eEndA(Q)e

3. Applications

View

295

f E EndA(P) as an element of eEndA(Q)e. We have Soc( Endp( V ) ) 2 eSoc( end^ (Q)e = Soc(EndA(&))f l eEndA(&)e

Thus f E Soc(EndA(P))if and only if f E Soc(EndA(&)). On the other hand, Soc(P) = Soc(Q) n P , since P is a direct summand of Q. Thus Imf Soc(P) if and only if I m f C Soc(Q). The result now follows from the case where P is A-free. 2.19. Corollary. Let A be a finite-dimensional Frobenius algebra over an arbitrary field F, let P be a projective A-module and let f E EndA(P). Then f E Rey(EndA(P)) zf and only i f f E Z ( E n d A ( P ) )

and f ( P )

Soc(P).

Proof. Apply Proposition 2.17 and Theorem 3.10.12(i). 3.

Applications

Throughout this section, G denotes a finite group and F an algebraically closed field of characteristic p > 0. Given two subgroups H , L of G, we fix the following notation: V,, . . ,V , are all nonisomorphic irreducible FG-modules. Wl, . . . , W, are all nonisomorphic irreducible FH-modules. Ul, . . . ,U, are all nonisomorphic irreducible FL-modules. Cjk is the multiplicity of v k as a composition factor of P ( K ) . rnij is the multiplicity of P ( x ) as a direct summand of WG. n;j is the multiplicity of P ( K ) as a direct summand of UjG". 3.1. Theorem. (Robinson (1989)). Suppose that W = Wj and u = us are one-dimensional. Then there are at least &=1 c;kmkjn;, ( H ,L)-double coset representatives g for which 1gLg-l n HI is not divisible b y p and

( gU)gLg-'nH WgLg-IrlH Proof. By hypothesis, V k occurs with multiplicity C;k as a composition factor of P(Vk). On the other hand, by Theorem '2.10, ( V ~ )conH tains P ( W ) as a direct summand with multiplicity m k j . Since P ( W )

296

Projective summands of induced modules

is projective (hence injective), it follows that P ( l 4 ) contains ~ P ( W ) as a direct summand at least C j k m k j times. Since P ( K ) occurs njq times as a direct summand of UG,it follows that P ( W )occurs at least &=1 cjkmkjn;, times as a direct summand of (U"),. By Mackey Decomposition Theorem (Theorem 2.7.1), we have

where T is a full set of double coset representatives for ( H ,L ) in G. Applying Theorem 2.10 within H , we see that for any g E T the multiplicity of P ( W )as a direct summand of (( gU)gLg-lnH)H is the same as the multiplicity of the projective cover of (gU)g,cg-in~as a direct summand of WsLg-lnH. Because W is one-dimensional, this last multiplicity is equal to 1 if IgLg-' n HI is not divisible by p and ( g U ) g , c g - ~ n ~ W g ~ g and ~ is n~, 0 otherwise. This obviously yields the result. 3.2. Corollary. Let H be a subgroup of G such that there is only one ( H ,H)-double coset representative, say g , for which 1gHg-l n HI is ) ~ not divisible b y p . If 1~ is the trivial FH-modute, then ( 1 ~contains at most one indecomposable projective direct summand. Furthermore, if there exists such a summand, then it must be irreducible.

Proof. We may choose Wl to be the trivial FH-module. Applying Theorem 3.1 (with S = H,W = U = Wl), we have

i,k=l

Since c;i 2 1, it follows that there can be at most one value of i for which P ( K ) occurs as a direct summand of Wf. Furthermore, if there is such a value of i , then m;l = c;; = 1. The fact that c;; = 1 and FG is symmetric forces, by Theorem 3.10.12(iii), P ( q ) to be irreducible. So the corollary is true. 3.3. Theorem. (Robinson (1989)). Assume that H is a Sylow p-subgroup of G, and put

I = { i E { I , . . . ,r}IK

is not projective

}

297

3. Applications

Then the number of ( H ,H)-double coset representatives g , for which IgHg-' n HI is not divisible b y p , is at least

Proof. Since H is a p-group and charF = p , it follows that t = 1 and Wl is the trivial FH-module. As in the proof of Theorem 3.1, P ( K ) H contains P ( W l )as a direct summand at least EL=, C i k m k l times. Hence, if V , is not projective, then by Theorem 2.10, P ( ~ $ ) contains H P(W1) as a direct summand with multiplicity at least

Since WF contains P(K)as a direct summand mil times, we deduce that (W,")Hcontains P(W1)as a direct summand with multiplicity at least r xmtl a€

I

+

x

milcikmkl

i,k=l

The result now follows as in Theorem 3.1. H

This Page Intentionally Left Blank

299

Chapter 5 Green theory The aim of this chapter is twofold: first to present some basic results of Green theory with refinements and extensions achieved in scope of recent developments, and second to provide a number of applications. An analogy between the role of vertices in the Green correspondence and elementary subgroups in Brauer’s induction theorem has been first discovered by Green (1971). We present an Alperin’s theorem which shows that there is, in fact, a direct connection: the Green correspondence implies Brauer’s theorem. As is well known, the process of reduction modulo a prime p is one of the most important tools in group representation theory. A fundamental result, called the Brauer lift, asserts that this process, in an appropriate sense, is surjective. A separate section is devoted to a result of Alperin (1986) which demonstrates how Green’s theorem readily and directly implies the Brauer lift. In the final section, we establish a theorem of Alperin (1986) which relates the Green correspondence with normal subgroups.

1. Vertices and sources Throughout this section, G denotes a finite group and R a commutative ring. All modules are assumed to be finitely generated over their ground rings. It will also be assumed that, for every subgroup H of G, any nonzero RH-module has the unique decomposition property. For example, by Corollaries 2.11.6 and 3.1.8, as R one can take any ring of

300

Green theory

one of the following types: (i) R is artinian. (ii) R is a complete noetherian local ring. Let V and W be RG-modules. We say that V is a component of W if V is isomorphic to a direct summand of W . 1.1. Lemma. Let H be a subgroup of G. (i) If V is an indecomposable RG-module and V is a component of M G for some RH-module M , then V is a component of WGfor some indecomposable component Wof M . (ii) If W is an indecomposable RH-module and W is a component O f LH for some RG-module L , then W is a component of VHfor some indecomposable component V of L .

Proof. (i) Write M = MI @ posable. Then

M G = MF

-

+

t$j

@ M,

@

where each M; is indecom-

M,"

and, by the unique decomposition property, V is a component of MF for some i E { I , . . . , r } . (ii) Write L = L 1 @ . - - @ Lt where each Li is indecomposable. Then

and, by the unique decomposition property, W is a component of ( L ~ ) H for some i E {I, ..., t } . 1.2. Lemma. Let H and S be subgroups of G and let V be an RS-module. (i) If W is an indecomposable component of (V"),, then there H exists g E G such that W is a component of [( gV)gSg-lnH] . I n particular, b y Theorem 2.10.7, W is 9Sg-l n H-projective. (iz) If W is an H-projective indecomposable component of V G , then there exists g E G such that W is a component of [( gV)gSg-lnH]G. I n particular, W is 9Sg-l n H-projective.

1. Vertices and sources

301

Proof. (i) By Theorem 2.7.1,

where T is a full set of double coset representatives for ( H , S ) in G. Now apply the unique decomposition property. (ii) By Theorem 2.10.3, W is a component of ( W H ) ~ .Since WHis a component of ( V G ) ~ it, follows from (1) that ( W H ) is~ a component of

((vG)H)G

&=T

[( t V ) t S t - l " H ]

H

The required assertion follows by virtue of the unique decomposition property. 1.3. Lemma. Let H be a subgroup of G, let V be an RG-module and let g E G. Then

(i) ' ( V H ) V S H g - 1 . (ii) ( V H ) G ( v g H g - l ) G . (iii) V is H-projective if and only i f V is gHg-'-projective. Proof. (i) The map f : 5 H g - I ' ( V H ) defined by f ( v ) = g- 1v is clearly an R-isomorphism. Since, for any z E gHg-', ---f

f(m) = g-*zv = (g-lzg)(g-lv) = 2 * f ( v ) ,

f is an

R(gHg-*)-isomorphism.

(ii) This is a direct consequence of (i) and Lemma 2.1.5(ii). (iii) By Theorem 2.10.3, V is H-projective if and only if V is a component of ( V H ) G . Now apply (ii). Let V be an indecomposable RG-module. Then a subgroup Q of G is sa.id to be a vertex of V if the following two properties hold: (a) V is Q-projective. (b) V is not H-projective, for any proper subgroup H of Q. Suppose that Q is a vertex of V . By Theorem 2.10.3, V is a component of ( V Q ) ~ hence , by Lemma l.l(i), V is a component of U G for some indecomposable RQ-module U . Any such U is called a source of V .

302

Green theory

1.4. Theorem. Let V be an indecomposable RG-module with vert e x Q and let an RQ-module U be a source of V . (i) U is a component of VQ and Q is the vertex of U . (ii) An RQ-module W is a source of V if and only if W E gU for some g E NG(Q). (iii) A subgroup H of G is a vertex of V if and only i f H is Gconjugate to Q . (iv) For any given subgroup H of G , V is H-projective if and only i f H 2 9Qg-l for some g E G. (v) If R is a complete noetherian local ring and charR/ J ( R) = p > 0 , then Q is a p-group.

Proof. (i) If S is a vertex of U , then U is a component of (Us)Q. Hence UG is a component of (US)' and therefore V is a component of (Us)'. Thus, by Theorem 2.10.7, V is S-projective and so S = Q. Since V is a component of (VQ)G, it follows from Lemma l . l ( i ) that there exists an indecomposable component X of VQ such that V is a component of X'. In particular, X is a source of V and hence, by the above, Q is the vertex of X . Since V is a component of U G , VQ is a component of (U"),, hence X is a component of ( U G ) Q .Thus, by Lemma 1.2(i), X is a component this implies that of [( gl/)gQg-lnQ]Q. Since Q is the vertex of

x,

gQ9-l = Q

X Z gU

and

(2)

g-lX for some g E NG(&) and therefore, by Lemma 1.3(i), Thus U is a component of ~-'(vQ) 2 (ii) If W is a source of V , then by (2),

u

vQ.

xZ

g u

E

tW

for some g , t E NG(Q) and therefore W E t-'gU. Conversely, for any given g E NG(Q), gU is an indecomposable RQ-module such that V is a component of ( g ~ z) UG. ~ (iii) Assume that H = gQg-' for some g E G. By Lemma 1.3(iii), V is H-projective. Any subgroup of H is of the form gQ1g-l for some subgroup Q1 of Q. If V is gQlg-'-projective, then V is Q1-projective, by Lemma 1.3(iii). Since Q is a vertex of V , we have Q1 = Q and hence

1. Vertices and sources

303

gQ1g-l = H , proving that H is a vertex of V . Conversely, assume that H is a vertex of V . Then V is an H projective indecomposable component of U G , hence by Lemma l.Z(ii), V is gQg-lnH-projective for some g E G. Thus gQ9-l 2 H and, since 9Qg-l is a vertex of V , gQg-' H . Hence H = gQg-l, as required. (iv) If V is H-projective, we may choose a subgroup Q1 of H to be a vertex of V . Hence, by (iii), gQg-' = Q1 C H for some g E G. Conversely, if H 2 9Qg-l then V is H-projective, since V is 9Qg-lprojective. (v) Let P be a Sylow p-subgroup of G. By hypothesis, (G : P) is a unit of R, hence V is P-projective, by Corollary 2.10.4. It follows from (iv) that P 2 9Qg-l for some g E G. Thus Q is a p-group. 1.5. Corollary. Let H be a subgroup of G , let V and W be indecomposable RG and RH-modules, respectively, and let P and Q be vertices o f V and W , respectively. (i) If V is a component of W G ,then P C 9Qg-l for some g E G. (ii) If W i s a component of VH,then Q C gPg-' for some g E G.

Proof. Let U be an RP-module which is a source of V and let S be an RQ-module which is a source of W . (i) Since W is a component of S H ,W Gis a component of SG,hence V is a component of SG. Thus V is &-projective and, by Theorem 1.4(iv), P C gQg-' for some g E G. (ii) Since V is a component of U G , VH is a component of ( U G ) H and hence W is a component of (UG)H. Therefore, by Lemma 1.2(i), W is zPz-' n H-projective for some g E G. Hence, by Theorem 1.4(iv), Q C h ( z P z - l n H)h-' for some h E H . Thus Q hzPz-lh-l, as required.

We now apply the foregoing to prove the following useful results. 1.6. Corollary. Let V be an indecomposable RG-module, let H be a subgroup o f G such that V is H-projective and choose an indecomposabte component W of VH such that V is a component of W G (such W always exists, b y Lemma 1.1 (i)). Then V and W have a vertex and

source in common.

304

Green theory

Proof. Let P and Q be vertices of V and W , respectively. By Corollary 1.5, P and Q are G-conjugate, hence Q is a vertex of V . Let U be any source of W corresponding to Q. Then U is an indecomposable RQ-module such that W is a component of U H . Hence W Gis a component of UG. Because V is a component of W G ,it follows that V is a component of UG. Thus U is a source of V , as required.

1.7. Corollary. Let V be an indecomposable RG-module with vertex P , let an RP-module U be a source ofV and let H be a subgroup of G containing P . Choose W to be an indecomposable component of VH such that U is a component of Wp ( b y Theorem l.4(i) and Lemma l . l ( i i ) such that W always exists). Then any vertex Q of W is H conjugate to P . Proof. By Theorem 1.4(i), P is the vertex of U. Hence, by CorolhQh-l for some g E G, h E H , as g P g - l and P lary 1.5(ii), Q required. 1.8. Corollary. Let H be a subgroup ofG and let W be an indecomposable RH-module. Then W G has an indecomposable component V such that V and W have a vertex and source in common.

Proof. We first observe that W is a component of ( W G ) ~ Hence, . by Lemma 1.1(ii), there exists an indecomposable component V of WG such that W is a component of VH. The desired conclusion is therefore a consequence of Corollary 1.6. W

Let U be an indecomposable RG-module with vertex P , let H be a subgroup of G containing P and let V be an indecomposable R H module. Consider the following three conditions: (a) U is a component of V G . (b) V is a component of UH. (c) P is a vertex V . 1.9. Proposition. A n y two o f t h e three conditions above are satisfied by a suitable indecomposable RH-module V .

1. Vertices and sources

305

Proof. Since U is H-projective, conditions (a) and (b) are certainly satisfied by some V (see Lemma l.l(i)). That conditions (b) and (c) are satisfied by some V follows from Corollary 1.7. Let an RPmodule M be a source of U . Since U is a component of M G % (MH)', it follows from Lemma l.l(i) that there exists an indecomposable R H module V such that U is a component of V G and V is a component of M H . Since P is the vertex of M (Theorem 1.4(i)), it follows from Corollary 1.5, that P is a vertex of V . Thus V satisfies (a) and (c), as required. By applying the Green correspondence, it will be demonstrated in the next section (see Corollary 2.8) that all three conditions (a), (b) and (c) are satisfied by some W . For the rest of this section, we assume that R satisfies the following two conditions: (i) R is a complete local ring with charF = p > 0 , where F =

R/ JW (ii) R is a principal ideal domain. Thus R is either a field of characteristic p > 0 or a complete discrete valuation ring with c h a r R / J ( R ) = p > 0. We denote by 1 R G the trivia2 RG-module i.e. 1RG = R and gr = r for all g E G , r E R. For any given RG-module V , we put V = V / J ( R ) V .Since R is a principal ideal domain, every finitely generated indecomposable R-module is cyclic, i.e. is generated by one element. Bearing in mind that F = R / J ( R ) is a field, we see that the following conditions are equivalent: (i) V is indecomposable as an R-module. (ii) V is a cyclic R-module. (iii) dirnFV = 1. Applying the information above, we now record 1.10. Lemma. Let V be an RG-module such that dimFV = 1. Then V is absolutely indecomposable (in particular, 1 R G is absolutely indecomposa ble).

Proof. Since dimFV = 1, v = Rv for some w E v. Let SIR be any finite extension. Then Vs = S @R V is generated by 1 @ w, hence it is indecomposable as an S-module. Thus Vs is an indecomposable

306

Green theory

SG-module, as required. R 1.11. Lemma. Let H be a subgroup of a p-group G and let V be an absolutely indecomposable RH-module. Then V G is absohtely indecomposable. Proof. This is a direct consequence of the transitivity of the induction and Theorem 3.7.15. W 1.12. Corollary. Let H be a subgroup of a p-group G and let V be an RH-module such that d i m F V = 1. Then V G is absolutely indecomposable. In particular, ( ~ R H ) " is absolutely indecomposable. Proof. Apply Lemmas 1.10 and 1.11. W 1.13. Lemma. Let H be a subgroup of G and let V be an R H module. Then ( V G )E (V)" as FG-modules Proof. It suffices to show that the above modules are RG-isomorphic. By definition = V / J ( R ) V so (V)" V G / ( J ( R ) V )by G Theorem 2.2.9(iii). But

v

(J(R)V)"= RG @'RH J(R)V = J( R )(RG@ R H V ) = J ( R ) V G , hence the result. The following important theorem will enable us to take full advantage of the results so far obtained. 1.14. Theorem. Let V be an indecomposable RG-module with vertex Q C_ P , where P is a Sylow p-subgroup of G. If d i m F V = n (e.g. V is R-free of rank n), then ( P : Q ) divides n. I n particular, if p does not divide n, then P is a vertex of V . Proof. For the sake of clarity, we divide the proof into three steps. Step I . Reduction to the case where R = F . Assume that the result is true for R = F . To show that it is true in the general case, let

1. Vertices and sources

307

W be an indecomposable component of V . It suffices to prove that ( P : Q) divides d i m F W . Let an RQ-module U be a source of V . Then UG 2 V @ X for some RG-module X , hence by Lemma 1.13,

Thus W is Q-projective and hence has a vertex Q1 Q. Consequently, ( P : Q1) divides d i m F W and therefore so does (P : Q). S t e p 2. Here we treat the case where R = F is algebraically closed. Let W be an indecomposable FP-module with vertex X and source S. By Lemma 1.11, S p is indecomposable and hence W 2 S p . Thus dimFW = ( P :X)dimFS

(3)

Now write vp

= W, @ . * . @ W,

where the W; are indecomposable FP-modules, and denote by Q; a vertex of W;. Then, by (3), ( P : Qd) divides d i m F W i and, by Corollary 1.5(ii), ( P : Q) divides ( P : Qi), 1 5 i 5 n. Thus, for all i, ( P : Q) divides dim.FWj and therefore (P : Q) divides dirn~V. Step 3. Completion of t h e pro05 Let E be an algebraic closure of F and let an FQ-module U be a vertex of V . Write

where the V , are indecomposable EG-modules. Since, for all i, V , is a component of VE and VE is a component of ( U G ) E 2 (UE)G,we see that V , is a component of ( U E ) G . This shows that each V , is Q-projective and so has vertex Q; C Q. By Step 2, ( P : Q;) divides d i m F V , , 1 5 i 5 m. It follows from (4) that ( P : &) divides dirnEVE = d i m F V , thus completing the proof.

1.15. Corollary. Let V be a n RG-module such that d i m F I / = 1 (hence V is indecomposable, by L e m m a 1.10). T h e n a Sylow p-subgroup o j G is a vertex o j V .

Proof. This is a special case of Theorem 1.14 in which n = 1.

308

Green theory

1.16. Corollary. Let P be a p-subgroup of G and let V be an . some indeRP-module such that d i m F v = 1 (e.9. V = 1 ~ ) Then composable component of V Ghas vertex P and source V . I n particular, every p-subgroup of G is a vertex of some indecomposable RG-module.

Proof. Owing to Corollary 1.8, it suffices to show that P is a vertex of V . Now apply Corollary 1.15 for G = P. H 1.17. Corollary. Let V be an indecomposable RG-module and let P be a p-subgroup of G contained in KerV. Then P is contained in a vertex of V .

Proof. Let W be an indecomposable component of Vp. Since P acts trivially on V , W is an indecomposable R-module, hence d i r n F W = 1. Thus, by Corollary 1.15 (applied to G = P), P is a vertex of W . Now apply Corollary 1.5(ii). We close by proving the following result. 1.18. Proposition. Let S I R be a finite extension and let V be an indecomposable RG-module with vertex P . Then P is a vertex of

every indecomposable component of Vs. Proof. If H is a subgroup of G and W is an SH-module, let W w be the restriction of W to RH. Then we obviously have

Similarly, if U is an RH-module, then

Let W be an indecomposable component of Vs. Since P is a vertex of V , V is a component of ( V P )and ~ so, by (6), Vs is a component of [(Vs)pIG.Thus W is P-projective. Assume by way of contradiction that P is not a vertex of W . Then there exists a subgroup Q of P with (P : Q) = p such that W is Qprojective. Because W is a component of Vs, WRH is a component of

2. The Green correspondence

309

(VS)RH2 ( S : R)V. Thus WRHZ nV for some integer n 2 1. Since by ( 5 ) , [ ( W Q ) G ] ~ ~[(Ws)WIG

it follows that V is a component of [(WQ)RQIG. But this contradicts the assumption that P is a vertex of V , hence the result. W 2.

The Green correspondence

In this section, R always denotes a commutative ring and G a finite group. All modules are assumed to be finitely generated over their ground rings. In addition, we assume that R is a complete noetherian local ring with c h a r R / J ( R ) = p > 0. By Theorem 1.4(v), this assumption ensures that vertices of indecomposable RG-modules are p-subgroups of G. Let P denote a p-subgroup of G and H a subgroup of G containing N G ( P ) . Our aim is to provide a bijective correspondence between the indecomposable RG and RH-modules with vertex P. Such correspondence will be referred to as the Green correspondence. We begin by recording a number of preliminary results.

2.1. Lemma. Let D* D S be a chain of subgroups of G. Then the following conditions are equivalent: (a) D* is G-conjugate to a subgroup ofgDg-lnD for some g E G-S. (ii) D* is S-conjugate to a subgroup ofgDg-' n D for some g E G-S. (iii) D* is S-conjugate to a subgroup of 9Dg-l n S for some g E

G - S. Proof. (i)=+(ii):By assumption, xD*x-l gDg-' n D for some x E G,g E G - S. Thus D* 5 D n s - l D x n x-lgDg-lz. Either x $ S or x - l g @ S and hence D* yDy-' n D for some y E G - S . (ii)+(iii): It is clear, since gDg-' n D C gDg-' n S. (iii)+(i): By assumption, sD*s-l 9Dg-l n S for some s E S,g E G - S . Hence

310

Green theory

for y = s-'g E G - S , as asserted. 2.2. Lemma. Let D S be subgroups o f G . IfV is an indecomposable D-projective RS-module, then V is the only possible indecomposable component of (VG)swhich is not gDg-' n S-projective for all

gEG-S. Proof. By Corollary 2.1.2, there exists an RS-module V' such that (VG)s @ (7) Since V is D-projective, there exists an RD-module W such that W s S V @ L for some RS-module L. Similarly to (1) we can write

= v v'

for some RS-module L'. Since

it follows from the Mackey decomposition that

where X is a full set of double coset representatives for (S,D ) in G with 1 E X . The term x = 1 in (2) is W s E V @ L; all other terms on the right of (2) have x E G - S and are xDx-l n S-projective. Since the sum of these terms is isomorphic to V' @ L', it follows that V' @ L' (and hence V ' ) can be written in the form U1 @ -..@ U,, where each Ui is gjDgr' n S-projective for some gi E G - S. Applying (l), the assertion follows. w 2.3. Lemma. Let Q C D S be a chain of subgroups of G and let U be an indecomposable RG-module with vertex Q . (a) I f & is G-conjugate to a subgroup of XDX-' n D for some x E G - S , then for any indecomposable component X of U s there exists g E G - S such that X is gDg-' n S-projective. (ii) If for all x E G - S, Q is not G-conjugate to a subgroup of XDX-' n D , then there exists a unique indecomposable component V of

2. The Green correspondence

311

Us which is not gDg-' n S-projective for all g E G - S. Moreover, V is the only indecomposable component of Us with vertex Q and U is the component of V G .

Proof. Owing to Proposition 1.9, there is an indecomposable RSmodule V with vertex Q such that U is a component of VG. Because V is D-projective, it follows from Lemma 2.2 that V is the only possible indecomposable component of (V G ) swhich is not gDg-' nS-projective for all g E G - S . (i) If Q satisfies (i), then by Lemma 2.1, Q is S-conjugate to a subgroup of gDg-' n S for some g E G - S. Because in this case V is gDg-' n S-projective, the assertion follows. (ii) If Q satisfies (ii), then by Lemma 2.1, Q is not S-conjugate to a subgroup of gDg-' n S for all g E G - S. Thus V is the only indecomposable component of (V G ) swhich is not gDg-' nS-projective for all g E G - S. By Proposition 1.9, Us has an indecomposable component W with vertex Q. Since Us is a component of (VG)s,W is a component of (VG)s.If W is gDg-lnS-projective for some g E G-S, then Q is S-conjugate to a subgroup of zDz-'n S for some z E G - S. But then, by Lemma 2.1, Q is G-conjugate to a subgroup of gDg-' nD for some g E G - S , a contradiction. Hence W is not gDg-' n Sprojective for all g E G - S. Thus W 2 V and the result is established. H We are now ready to prove the following fundamental result. 2.4. Theorem. (The Green correspondence, Green (1964)). Let P be a p-subgroup of G and let H be a subgroup of G containing N G ( P ) . Then (i) For any indecomposable RG-module U with vertex P , there exists a unique indecomposable component V = f ( U ) of UH with vertex

P. (ii) The map U H f ( U ) induces a bijective correspondence between the isomorphism classes of indecomposable RG and RH-modules with vertex P . (iii) For any indecomposable RH-module V with vertex P , f - ' ( V ) and V satisfy the following properties:

Green theory

312 (a)

f - ' ( V ) is a unique indecomposable component of V Gwith vertex

P.

(b) f-'(V) is a unique indecomposable component of V G which is not XPX-' n P-projective for all x E G - H, while V is a unique indecomposable component of f - ' ( v ) H which is not xPx-' fl H-projective for all x E G - H. (c) f - ' ( V ) is a unique i ~ d e c o ~ ~ o s a component b~e of V Gsuch that V is a component of f-l(V)H. (d) f - ' ( V ) and V have the same source. Proof. (i) Since H 2 N G ( P ) ,P is not G-conjugate to a subgroup of xCPx-' n P for all x E G - H. The assertion (i) now follows by applying Lemma 2.3(ii) for D = Q = P . Note also that, by Lemma 2.3(ii), V is a unique indecomposable component of UH which is not zPx-' n H-projective for all x E G - H. (ii) Let V be an indecomposable RH-module with vertex P . It suffices to show that there exists a unique indecomposable component U of V Gwith vertex P , and also that V is a component of U H . Decompose V Gas a direct sum

V G= u,@ ... E l u,

of indecomposable RG-modules U;, 1 5 i 5 T . Since V is not xPx-' n H-projective for all x E G - H, Lemma 2.2 implies that V is the only indecomposable component of (VG)Hwhich is not 9Pg-l nH-projective for all g E G - H. Hence there exists a unique U E {U,,. . . , U,} such that V is a component of U H . We are thus left to show that U is a unique indecomposable component of V Gwith vertex P . To this end, we first note that U is P-projective because so is V G (see Theorem 2.10.7(v)). Hence we may choose a vertex Q of U with P. If Q is G-conjugate to a subgroup of gPg-' n P for some Q g E G - H, then by Lemma 2.3(i) (with 1> = P,S = H ) for any indecomposable component X of U H , there exists g E G - H such that X is gP9-l n H-projective, a contradiction. Thus, for all g E G - H, Q is not G-conjugate to a subgroup of gPg-' n P . Hence, by Lemma 2.3(ii), P is a vertex of U . Assume that W E {Ul, . . . , U,} has vertex P . Then, by Lemma 2.3(ii), there exists an indecomposable component of WHwhich is not

2. The Green correspondence

313

gPg-' n H-projective for all g E G - H . Hence, by our remark on V , we must have V = W , proving (ii). (iii) Properties (a) and (c) were established in (ii), while the second part of (b) was established in (i). The fact that U = f - l ( V ) is an indecomposable component of V Gthat is not zPz-l n P-projective for all x E G - H was established in (ii). Let W be an indecomposable component of V Gthat is not zPz-' n P-projective for all x E G - H . By the argument of (ii), we may choose a vertex Q of W with Q 2 P . Applying Lemma 2.3(ii) for D = P and S = H , we see that P is a vertex of W . Hence W Z f-l(V), proving (b). To prove (d), let X denote a source of V . Then V is a component of X H and hence V Gis a component of X G . It follows from (a) that f-'(V)is a component of X G with vertex P. Hence X is a source of f - ' ( V ) , as required.

The bijection defined above is called the Green correspondence with respect to (G, P, H ) . In case G, P and H are specified by the context it is simply called the Green correspondence. The remainder of this section is devoted to investigating further properties of this mapping. As a preliminary to our next result, we now record the following useful property. 2.5. Lemma. Let S be a set of subgroups of G and let V be an ind eco mposa ble RG-module . If

lv E

cIm(Tr$)

HES

then V is H-projective for some H E S .

Proof. By hypothesis, for each H E S , there exists V H E EndRH( VH) such that 1V = TrE(VH)

c

HES

Since R is a complete noetherian local ring and V is indecomposable, we know that EndRG(V) is a local ring (see Lemma 3.1.7 and Corollary 3.1.6). Hence, for some H E S, Tr$(cpH) is a unit of EndRG(V). Since,

Green theory

314

by Lemma 2.9.7(ii), IrnTrg is an ideal of EndRG(V),we conclude that T r z is surjective. Thus, by Theorem 2.10.3, V is H-projective. W The following result is a strengthening of the uniqueness statement in the Green correspondence. 2.6. Theorem. ( B u r y and Carlson (1982)). Let P be a p subgroup of G, let H be a subgroup of G containing N G ( P ) and let U and V be indecomposable RG and RH-modules. (i) I f V is a component ofUH with vertex P, then P is a vertex of U . In particular, b y Theorem !?.4(i), V is the Green correspondent of U. (ii) P is a vertex of U if and only if UH has an indecomposable component with vertex P.

Proof. (i) By Corollary 1.5(ii), it suffices to show that U is Pprojective. By Theorem 2.10.3(v), the latter is equivalent to the requirement that the map

Trg : EndRp(Up) -+ EndRG(U) is surjective. By hypothesis, UH = V @ V'. Let cp : UH -+ V and $ : V -, UH be the projection on V and the inclusion map, respectively, and let X = 1c( 0 cp (hence X E EndRH(UH)). Since V is P-projective, 'p = TrF(cp') for some cp' E H o r n ~ p ( UV, ) . Hence X = T r g ( a ) , where u = $ 0 9'. Put p = Trg(u)and S = { z P z - l n H l z E G - H ) . Then, by Lemma 2.9.6(i), p = X 9 where 8 E CQEsIm(Tr;). We claim that X # C Q E Is r n ( T ~ g )Indeed, . if this were not-so, then

+

1v = CP 0

0

1 ~ ,E

C ~rQH(cpE ~ ~ R Q ( u Q ) $1 c C T ~ Q H ( E ~ ~ R Q ( v Q ) ) 0

0

QES

Q€S

and hence, by Lemma 2.5, V is zPz-' n H-projective for some II: E G - H , a contradiction. . I is an ideal of Now let I = E ~ ~ R G (f l UC )B E s I r n ( T r t ) Then EndRc(V) and there is a natural injective homomorphism. ?r

: EndRG(V)/I -+

EndRH(V)/

Irn(Tr;) Q€S

2. The Green correspondence

+

315

+

By the foregoing, ~ ( pI ) = X I # 0 is an idempotent. Therefore p t I is a nonzero idempotent of the local ring E n d ~ c ( V ) l I .Hence p = T r g ( a )is a unit of E n d ~ ~ ( and v ) thus Irn(TrP) = EndRG(U), as required. (ii) If UH has an indecomposable component with vertex P , then by (i), P is a vertex of U . The converse follows from Theorem 2.4(i).

2.7. Theorem. (Burry (1979)). Let P be a p-subgroup of G, let H be a subgroup of G containing P and let f H and f~ be the Green P,NG(P)), recorrespondences with respect to ( H ,P, N H ( P ) ) and (G, spectively. If U and V are indecomposable RG and RH-modules, respectively, both having vertex P , then (i) v is a component of U H if and only i f fH(v) is a component of f G ( U ) N H ( P )*

(ii) U is a component o f V G ifand only if

fG(U) is

a component of

fH ( V ) N G ( P ) .

Proof. Put N = N G ( P )and I( = NH(P). (i) Assume that V is a component of U H . Then f H ( v ) is a component of V,- and VK is a component of ( U H ) K = UK = ( U N ) K . Hence ~ H ( Vis )a component of ( U N ) K .By Theorem 2.4(b), we also have UN =fG(u)@

(@&ZXi),

where each X ; is an indecomposable RN-module which is z;Pzi' n N projective for some z; E G - N . If f H ( V ) is a component of ( X ; ) K for some i E I , then by Corollary 1.5(ii), P is conjugate to a subgroup of z;Pz;', and hence P = ziPzf', a contradiction. Thus fH(V)is a component of f ~U() K ,as required. . Conversely, assume that f ~ ( v is) a component of f c ( U ) ~ Since fc(U), is a component of ( U N ) ,=~ U,- = ( U H ) K , it follows that ~ H ( V ) is a component of ( U H ) K . Therefore there exists an indecomposable RH-module W such that W is a component of U H and f H ( V ) is a component of Wr;. Invoking Theorem 2.6, we deduce that W = fi'(f~(V)) and V is a component of U H . (ii) Assume that U is a component of V G . Since V G is a component of ~ H ( V ) ' 2 ( f H ( V ) N ) G , it follows that U is a component of ( f H ( V ) N ) G . Hence there exists an indecomposable RN-module M

316

Green theory

such that M is a component of ~ H ( V and) U ~ is a component of M G . Applying Corollary 1.5, it follows that P is a vertex of M . Thus, by Theorem 2.4, M = f c ( U ) . Conversely, assume that f c ( U ) is a component of ~ H ( V )Then ~ . fG(U)' is a component of f H ( V ) G . Since U is a component of ~ G ( u ) ~ it follows that U is a component of fH(V)' E ( ( ~ H ( V ) )On ~ the )~. other hand, by Theorem 2.4,

where each I$ is an indecomposable RH-module which is ziPzT' n Pprojective for some zj E H - K . In particular, if Qi is a vertex of Y,, then I&;[ < IPI. If U is a component of KG for some i E I, then by Corollary 1.5(i), Pi is conjugate to a subgroup of Qj,which is impossible. Thus U is a component of V Gand the result follows. H

Corollary. Let U be an indecomposable RG-module with vertex P and let H be a subgroup of G containing P . Then there exists an indecomposable RH-module V such that (a) U is a component o f P . (b) V is a component o f U H . (c) P is a vertex o f V . 2.8.

Proof. We employ the same notation as in Theorem 2.7. Since ~ G ( Uis) I<-projective, f c ( U ) is a component of ( ~ G ( U ) K Hence, ) ~ . by Lemma l.l(i),f c ( U ) is a component of M N for some indecomposable component M of ~ G ( U ) KFurthermore, . by Corollary 1.5, P is a vertex of M . Setting V = fG1(M), it follows that V is an indecomposable RH-module with vertex P. Since, by Theorem 2.7, V is a component of UH and U is a component of V G ,the result follows. Let U be an RG-module and let U1, . . . , U, be a complete set of nonisomorphic indecomposable components of U . Then

where each

ni

is a positive integer, called the multiplicity of U; in U.

,

2.

The Green correspondence

317

2.9. Theorem. ( B u r y (1984)). Let P be a p-subgroup of G , let H be a subgroup ofG containing P and let f H and fG be the Green cor-

respondences with respect to ( H ,P,N H ( P ) )and (G,P, N G ( P ) ) ,respectively. Denote b y T a complete set of those double coset representatives t for ( N , ( P ) , H ) in G for which P C tHt-'. (i) If U is an RG-module, then f H induces a multiplicity-preserving bijection between the nonisomorphic indecomposable components of UH with vertex P and the nonisomorphic indecomposable components of U N H ( p ) with vertex P . f%) If is an RH-module, then fG induces a multiplicity-preseruing bijection between the nonisomorphic indecomposable components of V G with vertex P and the nonisomorphic indecomposable components of

v

with vertex P . (iii) If V is an indecomposable RH-module with vertex P , then f G induces a multiplicity-preserving bijection between the nonisomorphic indecomposable components of V G with vertex P and the nonisomorphic indecomposable components of ( V,H(p))NG(p) with vertex P .

Proof. (i) Let U l , . . . , U, be all nonisomorphic indecomposable We may assume that components of UH and write UH = &niUi. U 1 , ., . ,Us, s 5 T , are all Ui's with vertex P. Then, for i E { 1,.. . ,s}, ( U i ) ~ ~ has ( p )a unique indecomposable component with vertex P , namely f H ( U i ) . On the other hand, if s < j 5 r , then ( U j ) ~ ~ (has p ) no indecomposable component with vertex P by Theorem 2.6. (ii) Owing to (i), (applied to U = V G , H = G ) fG induces a multiplicity-preserving bijection between the nonisomorphic indecomposable components of V G with vertex P and the nonisomorphic indecomposable components of ( V G ) ~ , ( pwith ) vertex P . , ) Let X be a complete set of double coset representatives for ( N G ( P ) H in G. Then, by Mackey decomposition,

Thus it suffices to show that if M is an indecomposable R N G ( P ) module with vertex P and M is a component of [( zV)zHz-lnNG(p)] NG(P))

318

Green theory

then P C sHs-'. Since the latter follows from the fact that P is NG(P)-conjugate to a subgroup of sHz-lflNG(P) (see Corollary 1.5(i)), the required assertion follows. ]NG(P), t E T , (iii) By (ii), it suffices to show that if [( has an indecomposable component, say M , with vertex P, then t E NG(P)H. To this end, choose an indecomposable component W of ( tV)NtHt-l ( P ) such that M is a component of WNG('). Then, by Corollary 1.5, P is tHt-'-conjugate to tPt-'. Hence t h E NG(P)for some h E H and the result is established. 3.

The endomorphism ring of the Green correspondents

Our aim is to exhibit a connection between the endomorphism ring of an indecomposable module and that of its Green correspondent. In what follows, G denotes a finite group and, unless explicitly stated otherwise, R is an arbitrary commutative ring. As a generalization of the concept of H-projectivity, we first introduce the following notion. Let V be an RG-module and let X be a set of subgroups of G. We say that V is X-projective if there exists a decomposition

such that each is Hi-projective for some Hi E X . Let cp : U + V be an RG-homomorphism. We say that cp factors through W if there is an RG-module W and RG-homomorphisms:

a :u +

w,p : w + v

such that cp = p 0 a. The RG-homomorphism cp is said to be X projective if cp factors through some X-projective module. For any RG-module V , we put

3.1. Proposition. Let cp : U + V be an RG-homomorphism. Then 9 is X-projective if and only ijcp E T r i (HomR(U, V ) ) .

3. The endomorphism ring of the Green correspondents

319

E T r g ( H o m ~ ( UV, ) ) ,say

Proof. Assume that

It will be demonstrated that cp factors through the X-projective module W = $HEx(VH)~. To this end, for each H E X , consider the maps

{ {

and

u 9

(VH)G

u

C t E ~ t @ p ~ ( t - ' u )

H

CtcT

@ Z)t

3

V

+

CtETtVt

where T is a left transversal for H in G. By Lemmas 2.10.1 and 2.10.2, both O H and PH are RG-homomorphisms. Furthermore, for all u E U , =

TrZ((PH)(u)

=P

CtpHt-l(U) tET

=

(PH

O

H ( C t t€T

@ (PH(t-lU))

aH)(u),

proving that T r g ( p ~= ) PH o a ~ . Now define a : U + W by a ( u )= &EX ~ H ( u and ) P : W + V by P ( C H E X W H ) = &EX PH(wH), where W H E V$. Then we have

Pea=

C

PHOaH=

HEX

C ~rZ(cp~)=cp, HEX

proving that p factors through W . Conversely, suppose that cp is X-projective. Then there exists an X-projective RG-module W and RG-homomorphisms a : U + W,/3 : W + V such that p = P 0 a. Since W is X-projective, we may write W = @Z1Wiwhere each Wi is Hi-projective for some Hi E X . Owing to Theorem 2.10.3, for each z, there exists $; E E72dRHi(W;) such that G lw, = TrHi((Pi). Let nj : W -+ W; and pi : W; + W be the projection and the inclusion map, respectively, arising from the decomposition of W . Then n

lw =

C pi i=l

n

0

~i

=

C pi i=l

0

TT~,((P 0 rj ;)

n

Tr;,(p; o $; o nj) (by Lemma 2.9.7)

= i=l

320

Green theory

and therefore

HOVLR(U, ( V)). proving that cp E T T ~ 3.2. Lemma. Let U and V be RG-modules. If U or V is X projective, then every RG-homomorphism cp ; U + V is X-projective.

Proof. This follows from the fact that cp = 9 01~= 1v o c p factors through an X-projective module U or V . 3.3. Lemma. Let R be a complete noetherian local ring and let U be a finitely generated RG-module. (i) If U is X-projective, then every indecomposable component of U is H-projective for some H E X . (ii) If U is X-projective, then every direct summand of U is X -

projective. Proof. (i) Our assumption on R guarantees that U has a unique decomposition property. By hypothesis, U = Ul @ - @ Un, for some n 2 1, and each U; is Hi-projective for some Hi E X . If V is an indecomposable component of U , then V is a component of U; for some i , hence V is Hi-projective . (iii) This is a direct consequence of (i). 4

3.4. Proposition. Let U be an RG-module. If U is X-projective, then Tr:( EndR( U ) ) = EndRG( U ) . The converse is true if U is finitely generated and R a complete noetherian local ring.

Proof. If U is X-projective, then by Lemma 3.2, any given RGhomomorphism cp : U + U is X-projective. Hence, by Proposition 3.1, each such cp is in Tr$(End,(U)). Thus T r g ( E n d R ( U ) )= EndRG(u). Conversely, assume that U is finitely generated, that R is a complete

3. The endomorphism ring of the Green correspondents

321

noetherian local ring and that Trz(EndR(U)) = EndRG(U). Since 1~ E T r g ( E n d ~ ( U ) it) , follows from Proposition 3.1 that 1u factors through an X-projective RG-module W . Thus U is a direct summand of W . Since U is finitely generated, we may clearly assume that so is W . Now apply Lemma 3.3(ii). For the rest of this section, H denotes a subgroup of G and X an arbitrary collection of subgroups of G. For the sake of brevity, we put

X A H = {gSg-' n HIS E X , g E G}

V in Given an RH-module V , we shall identify V with its image 1 18 V G .With this identification, Lemma 2.9.4 tells us that the map

induces an R-isomorphism

Inv(V) + Inv(VG) As a generalization of this fact, we next record the following result due to Dade.

3.5. Proposition. For any RH-module V , the map Trg in (1) induces an R-isomorphism

Proof. First we must show that for any u E TrFAH(V),T r g ( u ) E Tr$(VG). To this end, write u = Trsg-lnH(v) for some S E X , g E G,v E I n v ( V g ~ ~ - l Then, ~ ~ ) . by Lemmas 2.9.2 and 2.9.3, TrE(cl) = Tr:(Trsg-l"H(v)) = TrgSg-InH(v) G E TT,Gsg-l(Inv [(vG)gSg-1] ) = Trg(Inv(VG)s) E Tr;(VG)

It follows, from Lemma 2.9.4, that Trg gives rise to an injective Rhomomorphism TTFAH(V)+ TrZ(VG) (2)

322

Green theory

To show that (2) is surjective, we fix S in X and denote by T a full set of double coset representatives for (S,H ) in G. Then, by Mackey’s decomposition,

while, by Lemma 2.9.4,

Hence

as required.

Let U be an RG-module and let V be an RH-module. Our identification of V with 1 8 V gives us the inclusion

v)

HomR( u,

HOmR( u,vG)

(3)

and also the inclusion

HomR( v,u) HOmR( V G ,u )

(4)

( U )is regarded as an element of H O ~ RV(G U , ) where each f E H o ~ RV,

(here T is a left transversal for H in G with 1 E 2').

3.6. Proposition. Let U and V be RG and RH-modules, respectively. Then, in the R-isomorphisms below, (i) and (ii) are induced by the restriction of T r g : HomRH(U, V G )+ HomRG(U, V") to HOmRH(U,V), while (iii) and (iv) by the restriction of

T r z : HomRH(VG,u) + HomRG(VG,u )

3. The endomorphism ring of the Green correspondents

323

Proof. The inclusion of (2) implies the inclusion of HomRH(U, V ) C H O ~ R H ( UV"). , The restriction of

to HomRH(U, V ) is given by

where f E H o ~ R H ( UV, ) ,u E U , T is a left transversal for H in G. On the other hand, by Proposition 2.5.9(ii) and Lemma 2.7.3, there is an R-isomorphism

given by

Since, by Lemma 2.7.3, the map f H CtET t@f is an R-isomorphism of HomRH(U, V ) onto I ~ v [ H o ~ R ( UV)"], H , it follows that T T induces ~ , By the argument an R-isomorphism H o ~ R H ( UV, ) + H o ~ R G ( UV"). of Proposition 3.5, this isomorphism carries T r f A H ( H o m RUH, ( V ) )onto T r z ( H o r n ~ ( U , V " ) )proving , (i) and (ii). Properties (iii) and (iv) are

Green theory

324

established in a similar manner, this time appealing to Proposition 2.5.9(i). Now assume that V and W are RH-modules. Then we have the following inclusions

obtained by identifying W with 18 W and applying (3) for U = W G . Hence the restriction of the map

Trg : HomRH(VG,W G )+ H o ~ R G ( W VG ~ ), to Hom~H(v, W ) induces an R-homomorphism

such that for

f E HomRH(V,W ) and &T

t 8 vt E V G ,

(T is a left transversal for H in G containing 1). 3.7. Proposition. Let V and W be RH-modules. Then the map Trg in (5) induces an injective R-homomorphism

Homm(VG,WG)/Trg(HomR(VG, WG))

If V = W , then this is a homomorphism of R-algebras. Proof. Owing to Proposition 3.6(iv), T r z induces the following R-isomorphism

3. The endomorphism ring of the Green correspondents

325

To prove the first statement, we are therefore left to verify that ~ o m R H ( vW , )n T ~ ; A H ( H ~ ~ R ( v wG)) , = ~ r g A H ( ~ o m Rw ( v ), ) (8) The inclusion 2 being obvious, we fix an element side of (8). Then

c

1c, =

TT,H(+S)

II, of the left-hand

(fs E H 0 ~ R s ( V , W G ) )

SEXAH

Let

7r

: W G= W @ CtET-iI)t @ W + W be the projection map. Then

1c, = 7r

01c,

=

c

Tr;(no$s)

(7ro1c,s

E HOrnRS(V,W))

SEX AH

which shows that II, belongs to the right-hand side of (8). If V = W , then by (6), we have

T&f)

0

TG(1c,)= T G ( f

O

$9

for all f,1c, E E n d R H ( W ) . Thus in this case the given map is a homomorphism of R-algebras, as required.

Until now, R was an arbitrary commutative ring and X an arbitrary collection of subgroups of G. For the rest of this section, we assume that all modules are finitely generated over their ground rings, and use the following additional assumptions and notations: R is a complete noetherian local ring with c h a r R / J ( R ) = p > 0. P is a p-subgroup of G. H is a subgroup of G containing P . X = {&I& gP9-l n P for some g E G - H } . Y = {&I& 9Pg-l n H for some g E G - H } . 3.8. Proposition. Let V and W be RH-modules. (i) If V is P-projective, then

w)>= T T , H ( H O ~ R ( Vw, ) )

T~;,~(HO~R(V,

(ii) If both V and W are P-projective, then the map T r g in (5) induces an R-isomorphism HomRH(V, W ) / T r ; ( H o m R ( V ,

w))

326

Green theory

HOmRG(vG,WG)/Trz(HomR(VG, WG)) IfV = W , then this is a homomorphism of R-algebras.

To establish the opposite inclusion, let $=

TrF (')

E TrfAH(

HomR(

v,W ) ) 7

where S E X A H and X E HomRs(V,W). Since S E X A H , we may write S = xQz-l n H with x E G and Q C yPy-' n P for some y E G- H . By hypothesis, V is P-projective, hence by Theorem 2.10.3, we have l v = T r F ( p ) for some p E E n d ~ p ( V )Setting . D to be a full set of double coset representatives for ( P , S ) in H , it follows that

$ = $ 0 l v = T r f ( X )0 TrpH(p)=

c TrhH,/&-lnp(hX 0

p),

(9)

hED

where the last equality is an easy consequence of the definition of the trace map. Because S 5 ZQZ-I C xyPy-lz-l n zPz-',we have

hsh-l n P 5 h x y ~ y - ~ z - l h - ln h x ~ x - l h - ln P where hz # H or hzy # H since y 4 H . Thus hSh-' n P E X and, by applying (9) we deduce that $ E T?-;(HOmR(v,W ) ) ,as required. (ii) Assume that both V and W are P-projective. Since V is Pprojective, it follows from (i), (7) and (8) that we need only verify that

HOmRH(V,W G )= Homm(V,W )-k Tr?,H(HOmR(I/, W G ) ) To this end, note that by Corollary 2.1.2(ii), ( W G )=~W @ W' where W' is given by W'= t@W

c

t ET,t # 1

(here T is a left transversal for H in G containing 1). Hence

w)@ HOmRH(V,W ' )

HomRH(V,W G )= HomRH(V,

327

3. The endomorphism ring of the Green correspondents

and therefore it suffices to show that

By hypothesis, W is P-projective and therefore W' is also P-projective. It follows from the proof of Lemma 2.2 that W' is Y-projective. Hence, by Lemma 3.2 and Proposition 3.1, every f E Hom&V,W') can be written as

f=

cTrsH(fs)

(fs E HomRS(V, W ' ) )

S€Y

Let a : W' + ( W G )be~ the inclusion map and let D = D ( S ) be a complete set of double coset representatives for ( P ,S ) in H . Then

f = a f 0

=

0

1" =

cc

C Trf(a

0

f s ) TrpH(p) 0

S€Y

TrFSh-lnP(a

O

(hfs) P)

(11)

O

SEY hED

By the definition of Y , S C 9Pg-l n H , for some g E G - H , so hSh-' n P C h9Pg-lh-l n P . Consequently, hSh-l n P E X (since hg 4 H ) and therefore, by (11),

f E TTfAH(HomR(v>

WG)),

proving (10) and hence the result. We have now come to the demonstration for which this section has been developed.

3.9. Theorem. Let R be a complete noetherian local ring such that c h a r R / J ( R ) = p > 0, let P be a p-subgroup of G and let H be a subgroup of G containing N G ( P ) . Let f be the Green correspondence gPg-' n P for some g E with respect to ( G ,P, H ) , let X = {QIQ G - H ) and let U , V be finitely generated indecomposable RG-modules with vertex P . Then (i) There is an R-isomorphism:

H o w w ( f ( U ) , f ( v ) ) l ~ r X H ( H o m R ( f ( fu()V, N

+

328

Green theory

Proof. Since both f(U), f ( V )are P-projective, it follows from Proposition 3.8 that T r g induces an R-isomorphism

a :U

--$

f ( U ) G , p: f(lqG +

v

be the inclusion map and projection, respectively. By Theorem 2.4(iii), both U' and V' are X-projective. Hence, by Proposition 3.1 and Lemma 3.2,

Applying (ii), we are thus left to verify that

v)

H O ~ ~ ~ ( nUT ,& H o ~ ~ ( ~ (f u( )~~ ), ~ = T) T) $ ( H O ~ R ( v)) ~, (13) The inclusion 2 being obvious, fix f from the left-hand side of (13). Then we may write

and therefore

f =p f 0

0

CY

=

c

TTg(p 0 fs 0 a) E Try(HomR(U,V ) ) ,

SEX

proving the reverse inclusion. Thus (13) is established and the result follows. I

4. The Green correspondence and Brauer’s induction theorem

329

Keeping the notations and assumptions of The-

3.10. Corollary. orem 3.9, we have

n )h)( f ( U ) ) / J ( E n d R H ( (u))f (i) E ~ ~ R G ( U ) / J ( E ~ ~ RzGE( U (ii) I f U is irreducible, then EndRG(U)

(u))

EndRH(f(U))/J(EndRH(f

(iii) If f ( U ) is irreducible, then E n d m ( f ( U ) )z E ~ ~ R G ( U ) / J ( E ~ ~ R G ( U ) )

(iv) If both U and f ( U ) are irreducible, then EndRG(U)

E&H(f(U))

(v) If H is a subnormal subgroup of G and U is irreducible, then f ( U ) is irreducible and

EndRG(u)

EndRH(f (u))

Proof. Let U be a finitely generated indecomposable RG-module with vertex P . Then f ( U ) is an indecomposable RH-module with vertex P and hence neither U nor f ( U ) is X-projective. Hence, by Proposition 3.4,

+

~ r $ ( ~ n d R ( u ) ~) n d R G ( u )T,r f ( E n d R ( f ( u ) Z ) E~~RH(~(u)) (14) Note also that both EndRG(U) and E n d w ( f ( U ) ) are local rings, by virtue of Corollary 3.1.6 and Lemma 3.1.7. Furthermore, if H is a subnormal subgroup of G and U is irreducible, then UH is completely reducible, by applying Clifford’s theorem and transitivity of the restriction. Since f ( U ) is an indecomposable component of U H , f ( U ) must be irreducible. Thus all the required assertions follow by virtue of (14), Theorem 3.9(ii) and the Schur’s lemma. 4.

The Green correspondence and Brauer’s induction theorem

Our aim here is to demonstrate that the Green correspondence implies the Brauer’s induction theorem. As is well known, this theorem implies

330

Green theory

a fundamental result known as the Brauer's characterization of characters. Throughout, G denotes a nonidentity finite group and p a prime dividing the order of G. We refer to the normalizers of the nonidentity psubgroups of G as the p-local subgroups of G. A subgroup H of G is called local if H is plocal for some prime p dividing [GI. Let IC be an algebraic number field which is a splitting field for the subgroups of G and let R,, be the completion of the integers R of K with respect to a prime divisor of p in R. In what follows, we assume that all RpG-modules are finitely generated and free as Rp-modules. Following Alperin (1981), we say that an RpG-module is p-local if it is the direct sum of modules induced from p-local subgroups of G. Note that our choice of Rp ensures that Rp is a complete discrete valuation ring with residue class field of characteristic p and quotient field It'. As can be seen from the proof, the following result holds with Rp replaced by any field of characteristic p or by any complete discrete valuation ring with residue class field of characteristic p . 4.1.

Theorem.

(Alperin (1981)).

Let

U be an R,,G-module.

Then

U @ W @ M g W ' $ M' for some p-local RpG-modules W,W' and some projective R,,G-modules

M , MI. Proof. We may harmlessly assume that U is indecomposable. We proceed by induction on the order of a vertex P of U . If P = 1, then M is projective and there is nothing to prove. Thus we may assume that the result holds for indecomposable RpG-modules with vertices of order < (PI.Let f be the Green correspondence with respect to (G,P,NG(P))and let V = f ( U ) . Then, by Theorem 2.4, V G % U @ L and the indecomposable components of L have vertices of order < ]PI.Hence, by induction, there are plocal RpG-modules Wl,W and projective R,,G-modules M', M such that

Setting W' = VG$Wl, it follows that W' is a plocal R,G-module such

4. The Green correspondence and Brauer's induction theorem

331

that

W ' $ MI

U @ L @ Wl @ M'

U @ W @M,

as required. H

The following two preliminary results will enable us to take full advantage of Theorem 4.1. 4.2. Lemma. Every I<-representation of G is equivalent (over I<) to an R,-representation of G, that is, a representation by matrices with entries in R,. In particular, any I<-character of G is aforded by some R,G-module.

Proof. Let V be a KG-module which affords a given K-representation of G. Denote by vl, . . . ,v,, a I<-basis of V and put

W = RpGvl

+ R,GV~+ - + R,Gv, * *

Then W is an RG-submodule of V . Since G is finite, W is a finitely generated R,-module. Moreover, W is R,-torsion-free since V is. Because R, is a principal ideal domain, W is R,-free on a finite basis, say wl,. . . ,w,. Since equivalence (over I<) of K-representations of G is the . . ,w, is a result of change of I<-basis of V , it suffices to show that w1,. I<-basis of V. Since no nontrivial R,-linear combination of w1, . . . ,w, is 0 and I
Proof. See Corollary 2.15.11 in Karpilovsky (1987). Let

XI,.

..,xs be all irreducible I<-characters of G. The elements

332

Green theory

of the set

cf:

ZiXilZi

i=l

E

zz 1

are called generalized I(-characters of G. We are now ready to record the following two consequences of Theorem 4.1. 4.4. Corollary. Let x be a K-character o f G . Then them are generalized I(-characters ap and PP ofG such that x = a, PP,where aP is an integral linear combination of K-characters induced from p local subgroups of G and Pp vanishes on p-singular elements of G.

+

Proof. Owing to Lemma 4.2, x is afforded by some R,G-module U . The desired conclusion is therefore a consequence of Theorem 4.1 and Lemma 4.3. R 4.5. Corollary. Every I(-character of G is an integral linear combination of K-characters induced from local subgroups of G.

Proof. Note that the integral linear combinations of characters induced from a subgroup H of G form an ideal in the ring of generalized K-characters of G. Hence we need only verify the result for the principal character 1~ of G. For each prime p , write 1~ = ap Pp where apand PP are as in Corollary 4.4. Then 1~ = np(a, Pp) and each term in the expansion, except for the product of the PP,is obviously an integral linear combination of characters induced from local subgroups. Since the product of the Pp vanishes on each nonidentity element of G , it is a multiple of the regular character and is thus induced from any subgroup of G. So the corollary is true. R

+

+

We have now come to the demonstration for which this section has been developed.

(Brauer (1947)). Every K-character of G is an integral linear combination of characters induced from linear characters of elementary subgroups of G , where b y an elementary subgroup of G we mean one which is a direct product of a cyclic group and a p-group for some prime p . 4.6. Theorem.

4. The Green correspondence and Brauer’s induction theorem

333

Proof. (Alperin (1981)). It clearly suffices to show that every K-character of G is an integral linear combination of K-characters induced from elementary subgroups of G. We argue by contradiction and choose G to be a counterexample of minimal order. Our first aim is to show that (a) The principal character 1~ is not an integral linear combination of characters induced from elementary subgroups. (b) There is a prime p such that O,(G) # 1, where O,(G) is the maximal normal p-subgroup of G. (c) Every proper homomorphic image of G is elementary. To prove (a), we need only observe that the collection of all linear combinations in (a) form an ideal in the ring of all generalized K-characters of G. To prove (b), we note that if every local subgroup of G is proper, then Corollary 4.5 and the minimality of G would contradict the assumption that G is a counterexample to the theorem. To prove (c), assume by way of contradiction that G is a proper homomorphic image of G and G is not elementary. By the minimality of G, the principal character of G must be a n integral linear combination of characters induced from proper subgroups of G; hence the same is true in G. However, the theorem holds in every proper subgroup of G, a contradiction. Thus (c) is established. By (b), we may choose a prime p with O,(G) # 1. We denote by N a minimal normal subgroup of G with N a pgroup (so N is elementary abelian). By (c), GIN is elementary, say a direct product of a q-group & / N , for a prime q, and a cyclic q’-group C / N . It will next be shown that (d) G is not nilpotent. (e) N is the unique minimal normal subgroup of G. ( f ) N = OJG). Assume by way of contradiction that G is nilpotent. Because G is not elementary, it has a homomorphic image of the form Z x Z x 2Z x Z for distinct primes T and s. Therefore, by (c), G must be isomorphic with this group. Denote by R and S the Sylow r-subgroup and Sylow s-subgroup of G. Note that the character T 1~ is the sum of the characters induced from the principal characters of each of the subgroups of order r in R minus the regular character. Thus r . 1~ is an integral linear combination of characters induced from elementary subgroups of

-

334

Green theory

the form S and & x S, where & is a subgroup of order r in R. Since the character s. 1~has a similar expression, we have a contradiction to (a), thus proving (d). To prove (e), assume that M is another such subgroup, so M n N = 1. Then G is isomorphic to a subgroup of G/M x GIN. But, by (c), each of these factors is nilpotent, hence so is G, contradicting (d). Thus (e) is established. O,(G). Because GIN is nilpoTo prove (f), we first note that N tent it follows that G/O,(G) is a p’-group. Hence, if O,(G) is elementary abelian, then by Maschke’s theorem we must have N = O,(G). Assume by way of contradiction that O,(G) is not elementary abelian. The the Frattini subgroup D(O,(G)) of O,(G) is not 1, hence it must contain N, by virtue of (e). But then O,(G)/D(O,(G)) is a central factor of G, hence so is N . Thus N is a central factor and G I N is nilpotent, contrary to (d). This proves (f). By the foregoing, N is a normal p-subgroup of G such that GIN is a p’-group. Hence, by the Schur-Zassenhaus theorem (see Gorenstein (1968)), G = N K is a semidirect product of N and a subgroup K . Furthermore, the uniqueness of N now shows us that N is not a central factor of G. Thus no nontrivial character of N is stabilized by G. Finally, note that the character ( 1 ~of )G~induced by the principal character of I( is the sum of 1~ and characters of G whose restrictions to N do not involve 1 ~Thus, . by Clifford’s theorem, each of these characters is induced from a proper subgroup of G. But this inplies that 1 G is a linear combination of characters induced from proper subgroups, a contradiction. 5.

The Green correspondence and the Brauer lift

Throughout this section, R denotes a complete discrete valuation ring with residue class field F of prime characteristic p , and G a finite group. All RG and FG-modules are assumed to be finitely generated and all RG-modules are also assumed to be R-free. Our aim is to demonstrate how the Green correspondence readily and directly implies the Brauer lift. Let V be an RG-module. Then V = V / J ( R ) V is an RG-module annihilated by J ( R ) ,and hence can be made into an FG-module, with

5. The Green correspondence and the Brauer lift

335

the action of F G given by

+

+

where fi = v J(R)V and i, = xg J ( R ) . We shall refer to V as the FG-module obtained from V by reduction rnodJ(R). If U is an FG-module, then U is said to list if U is isomorphic with T/ for some RG-module V . The FG-module U is said to lift virtually if there are RG-modules V and W such that U @ V and '&I have the same composition factors (counting multiplicity).

5.1. Lemma. (i) If U and U1 are FG-modules with the same composition factors (counting multiplicities), then U lifts virtually if and only if U1 lifts virtually. (ii) If U is an FG-module and each composition factor of U lifts virtually, then so does U . (iii) If U and U1 are FG-modules such that U1 and U @ UI lift virtually, then so does U .

Proof. (i) This is a direct consequence of the definition and the fa.ct that an irreducible FG-module X is a composition factor of Y @ 2 (Y, 2 are FG-modules) if and only if X is a composition factor of Y or 2. (ii) Let U 1 , . . . ,U, be all composition factors of U . Then U and @?==,U; have the same composition factors (counting multiplicity). Since each Ui lifts virtually, then so does @ZlU;. The desired conclusion now follows by virtue of (i). (iii) Assume that both U1 and U @ U , lift virtually. Then there exist RG-modules X , Y,2 and W such that

where M indicates that both sides have the same composition factors. Hence

U a3 and the result follows. H

(m)W e X M

336

Green theory

5.2. Lemma. Let e be an idempotent of an arbitrary ring S and let I be an ideal of S . Then (i) S e n I = Ie. (ii) Se/Ie 2 S E where e denotes the image of e in S = S / I .

Proof. (i) Assume that x E Se n I . Then x = se for some s E S. Thus

x = se = se2 = xe E Ie proving that S e n I 2 Ie. The opposite inclusion being trivial, (i) follows(ii) Applying (i), we have

+

Se/Ie = S e / ( S e n I ) ( S e ] ) / I = (S/I)(e I ) = SE,

+

as required. H 5.3. Lemma. Eve ry projective FG-module lifis.

Proof. It suffices to show that every projective indecomposable FG-module lifts. To this end, we identify FG with R G / J ( R ) G . Let - : RG + RG/J(R)G be the natural map. If E is a primitive idempotent of RG = RG/J(R)G, then E = 2 for some (primitive) idempotent e of RG (see Theorem 3.1.5). Hence, by Lemma 5.2(ii) applied to S = RG and I = J(R)G,we have

RGe = RGe/J(R)RGe= RGe/J(R)Ge R G E -

-

Since RGZ is a typical projective indecomposable RG-module, the result follows.

5.4. Lemma. Let H be a subgroup o f G and let V and W be FH-modules. If V and W have the same composition factors, then V G and W G have the same composition factors.

Proof. Let V,,. . . , V, be all composition factors of V . It suffices to show that V Gand VF @ . - .V: have the same composition factors.

6. The Green correspondence and normal subgroups

337

If n = 1, then V = Vl and there is nothing to prove. We now argue by induction on n. We may assume that Vl is an irreducible submodule of V , in which case &, .. . ,V, are all composition factors of V/Vl. It follows, by induction, that 2 VG/V:: and hG@ - - .@ V$ have the same composition factors. Hence V Gand V p @ * @ Vf have the same composition factors, as required. We are now ready to prove the following classical result. 5.5. Theorem.

(Brauer lift). Eve y FG-module U lifts virtually.

Proof. (Alperin (1986)). We argue by induction on the order of G. It is a consequence of the Green correspondence (see Theorem 4.1) that there exist FG-modules W,W',M , M' such that

U$W$MZW'$M' (1) where M and M' are projective and W and W' are each the direct sum of modules induced from p-local subgroups of G. Owing to Lemma 5.3, both M and M' lift. Furthermore, we claim that if every p-local subgroup of G is proper, then W and W' lift virtually, (hence, by (1) and Lemma 5.l(iii), U lifts virtually). Indeed, if P is a nonidentity psubgroup of G, L is an FNG(P)-module and L e v and N have the same then by Lemmas composition factors, for RNG(P)-mOdUleS V and I?,5.4 and 1.13, the induced modules LG @ (p) and ( N G )are similarly related. By the foregoing, we may assume that there is a nonidentity psubgroup P of G which is normal in G. Since, by Lemma 5.1(ii), it suffices to prove the theorem in the case of irreducible FG-modules, we let U be such a module. Now P acts trivially on U , so U is an F(G/P)module viewed as an FG-module and G / P is of the order smaller than the order of G. Thus U lifts virtually by induction and the result is established. 6.

The Green correspondence and normal subgroups

In this section, G denotes a finite group and R a commutative ring of one of the following types:

338

Green theory

(a) R is a field of characteristic p > 0. (b) R is a complete discrete valuation ring with residue class field of characteristic p > 0. If H is a subgroup of G, then all RH-modules are assumed to be finitely generated. By Theoreml.4(~),these assumptions ensure that vertices of indecomposable RH-modules are p-subgroups of G. In order to describe our general result, we need to introduce one concept. The following lemma will clear our path.

6.1. Lemma. Let N be a normal subgroup o f G and let U and V be indecomposable RG and RN-modules, respectively, such that V is a component of U,. Then there is a vertex P of U such that P n N contains a vertex of V .

Proof. Let PI and Q be vertices of U and V , respectively. By Corollary 1.5(ii) (with H = N ) , we have Q E gP1g-l for some g E G. Setting P = gP1g-l, it follows that P is a vertex of U such that Q P n N , as required. Let U and V be as in Lemma 6.1. Following Alperin (1986), we say that U covers V if there is a vertex P of U such that P n N is a vertex of V . For example, if a vertex of V is a Sylow p-subgroup of N , then U covers V .

(Alperin (1986)). Let N be a normal subgroup of G, let V be an indecomposable RN-module with vertex Q and let W be the Green correspondent of V with respect to ( N ,Q, " ( Q ) ) . Then there is a bijective correspondence between the isomorphism classes of indecomposable RNG(Q)-modules covering W and the isomorphism classes of indecomposable RG-modules covering V . 6.2. Theorem.

Proof. Put K = " ( Q ) and L = N c ( Q ) . Suppose that X is an indecomposable RL-module covering W . Then there is a vertex P of X such that P f l I< is a vertex of W . But Q is a vertex of W and Q K , hence P n I( = Q. Thus P n N = Q and therefore

6. The Green correspondence and normal subgroups

339

Let an indecomposable RG-module Y be such that X is the Green correspondent of Y with respect to (G, P, L). Then X is a component of YL,so W is a component of YK as W is a component of X K . Write YN = V,@. -@V, where the are indecomposable. Then YK = ( V ~ ) K @ - ,@(V,)Kand W is a component of (E)K for some i E (1,. . , ,s}. Since W has vertex Q, it follows from Theorem 2.6(i) that W is the Green with respect to ( N , Q , K ) . Thus V S so V is a correspondent of component of YN and hence Y covers V . We have therefore defined a one-to-one map, which we only need to show to be surjective. Assume that U is an indecomposable RG-module covering V . If P is a vertex of U with Pn N = Q , then NG(P) L by virtue of (1). Let X be the Green correspondent of U with respect to (G, P, L ) and let Y be the Green correspondent of U with respect to (G, P,L N ) . We claim that X is the Green correspondent of Y with respect to ( L N , P , L ) . Indeed, since Y has vertex P , there is an indecomposable component of Y L with vertex P and this module is also a component of X (since Y is a component of U L N ) ,so it must be X by the uniqueness of the Green correspondence. Bearing in mind that W has vertex Q, X has vertex P and P n I( = Q , we are left to demonstrate that W is a component of xh'. Let us first show that V is a component of Y N . Let T be a left transversal for L N in G so that a

Since U is a component of Y Gand V is a component of U,, it suffices to prove that V is not a component of ( t 8 Y ) N for t 4 L N . Now t@Yis an R(tLNt-*)-module and has vertex tPt-l. Hence, each indecomposable component of ( t 8 Y ) Nhas a vertex contained in

However, V has vertex Q so if V is a component of ( t @ Y ) N , then there is z E N with z-lQx = tQt-', that is zt E NG(Q)= L , so t E N L . Since X and Y are the Green correspondents, Y is a component of X N L and therefore V is a component of ( X L N ) N . But

340

Green theory

by Mackey’s theorem, so V is a component of ( X K ) ~Thus . there is an indecomposable component W1 of X K such that V is a component of ( W l ) N However, . X has vertex P and P n K = Q , so Wl has a vertex contained in a conjugate of Q in L , that is, in Q since L = NG(Q). On the other hand, V has vertex Q and V is a component of ( W l ) Nso , Wl must have vertex &. Consequently, W, is the Green correspondent of V, that is W1 S W . Because W1 is a component of X K , the same is true for W and the result follows. H

341

Chapter 6 Simple induction and restriction pairs In this chapter we investigate in detail circumstances under which restriction and induction of irreducible modules are completely reducible modules. After presenting some general background pertaining to block theory, we establish a fundamental result due to Green (1959) and Knorr (1976) which characterizes defect groups of blocks in terms of vertices of indecomposable modules. We then turn our attention to the study of simple induction and restriction pairs introduced by Knorr (1977). As one of the applications, we show that if H is a subgroup of G and F is an arbitrary field of characteristic p > 0, then the following conditions are equivalent: (i) V G is completely reducible for any irreducible FH-module V and M H is completely reducible for any irreducible FGmodule M ; (ii) There exists a normal subgroup N of G such that N H and N has p’-index in G; (iii) J ( F G ) = FG J ( F H ) .

1. Blocks of algebras In this section, A denotes a finite-dimensional algebra over a field F . We begin by recalling the following information. By an F-representation of A (or simply a representation of A ) we understand any homomorphism A EndF(V) of F-algebras, where V is a (finite-dimensional) vector space over F . If n = d i m ~ Vthen ,

Simple induction and restriction pairs

342

EndF(V) is identifiable with the F-algebra Mn(F). Hence the given representation of A defines a homomorphism A + M n ( F ) ;we shall refer to any such homomorphism as a matrix representation of A. If f : A + EndF(V) is a representation of A, define ICW = f ( z ) v for all x E A,v E V . In this way V becomes an A-module, called the underlying module of f. Conversely, if V is an A-module, define

f : A + EndF(V) by f ( s ) v = xv for all IC E A,v E V . Then f is a representation of A; in case V is the regular A-module, we shall refer to f as the regular representation of A. Two representations f; : A + E n d F ( F ) , i = 1,2, are called equivalent if there exists an F-isomorphism $ : Vl + such that fi(z)= $ ~ I ( I c ) $ - ~ for all x E A It is easily seen that two representations of A are equivalent if and only if the underlying modules are isomorphic. We say that a representation f : A + EndF(V) is irreducible (completely reducible, indecomposable) if the underlying module V is irreducible (completely reducible, indecomposable); the same terminology will be applied to matrix represent at ions.

1.1. Proposition. The following properties hold: (i) There exists a direct decomposition

of A into indecomposable two-sided ideals B;

#

0 with B,Bj = 0 for

i#j (ii) Write 1 = el en with e; E B;. Then the e; are mutually orthogonal centrally primitive idempotents and B; = Ae; = e;A. (iii) Z ( A ) = Z(A)el @ @ Z(A)e, is a direct decomposition of Z(A) into indecomposable ideals and Z ( A ) e ; = B; n Z(A). (iw) J(Z(A)e;) = J ( Z ( A ) ) e ;and Z(A)ei/J(Z(A)e;) is a finite field extension of F . (w) Let F; = Z(A)ei/J(Z(A)e;), let di : Z(A) + 6 be the natural

+ -+

343

1. Blocks of algebras

homomorphism and let f; : F; -+ M n i ( F )be the regular representation of F;. Define 7; : Z ( A ) -+ M n i ( F )b y 7; = f;&. Then

Kery; = KerO; = Z( A ) (1 - e ; )

+ J ( Z(A)e;

(1 5 i 5 n )

and (71,. . . ,T,} is a complete set of nonequivalent irreducible matrix representations of Z ( A ) . Each 7; satisfies 7i(e;) = 1 and 7;(ej)= 0 for j # i. Furthermore, i f F is a splitting field for Z ( A ) , then each n; = 1. Proof. (i) and (ii) . Straightforward. (iii) The first statement and the inclusion Z ( A ) e ; E B; n Z ( A ) follows from (ii). Assume that z E B ; n Z ( A ) . Because e; is the identity element of B;, we have z = ze; E Z ( A ) e ; ,proving that B; n Z ( A ) Z(A)e;. (iv) By Lemma 3.18.1, J ( Z ( A ) e ; )= J ( Z ( A ) ) e ; .Because each e; is primitive, Theorem 2.12.4(ii) implies that the Z(A)-module

is irreducible. The latter implies that the commutative algebra

is simple and hence is a field. That this field is a finite extension of F follows from the assumption that A is finite-dimensional over F . (v) It is clear that Kerr; = Kere;. Suppose that z E KerO;. Then ze; = xe; for some z E J ( Z ( A ) ) ,so

+ xe; E Z ( A ) ( l - e;) + J ( Z ( A ) ) e ; and thus I c e d ; Z ( A ) (1 - e;)+ J ( Z ( A ) ) e ; .The opposite containment being obvious, we deduce that K e d ; = Z ( A ) ( l - e;) + J ( Z ( A ) ) e ; . z = z(1 - e;)

It follows from (iii) and (iv) that

Z ( A ) / J ( Z ( A ) )= and that each F; is a field. Moreover, if F is a splitting field for Z ( A ) then each F, is isomorphic to F and so n; = 1. Because f; is the only irreducible representation of Fi, the result follows. W

344

Simple induction and restriction pairs

We now introduce the important concept of blocks. Let

be the decomposition defined in Proposition 1.1. We shall refer to B; as a block and to e; as a block idempotent of A. We shall also write B = B(e)to indicate that B is a block containing the block idempotent e. The representation : Z ( A ) + M n i ( F )described in Proposition 1.1 is uniquely determined up to equivalence by e;. We call yi the irreducible representation of Z ( A ) associated with e;. By a central character of A, we understand any F-algebra homomorphism Z ( A ) + F . If F is a splitting field for Z ( A ) ,then by Proposition 1.1, ylr. . . ,yn are all central characters of A; we shall refer to y; as the central character of A associated with ei (or Bi).

1.2. Proposition. Let A = B(e1) @ -..@ B ( e n ) be the direct decomposition of A into sum of blocks. (i) For every A-module V , V = $y=leiV is a direct decomposition of V into A-modules. In particular, if V is indecomposabte, then V = e;V f o r a u n i q u e i e {I, ..., n } . (ii) For any given left ideal I of A , we have

nB(e;)) I = @z1(I In particular, each indecomposable left ideal lies in exactly one of the B(e;). (iii) If A1 and A2 are ideals of A and A = A1 @ A2, then A1 and A2 are. direct sums of blocks. In particular, the blocks are the only indecomposable ideals of A. (iv) Let I be an ideal of Z ( A ) and let y; be the irreducible representation of Z ( A ) associated with ei. Then ei E I if and only if y;(I) # 0 . Proof. (i) Since e; is central, eiV is an A-module. Because the ei are orthogonal we also have

1. Blocks of algebras

345

Bearing in mind that

the assertion is proved. (ii) Owing to (i), I = $yZle;A and therefore

e;I 5 I n B(e;)= I f~e ; A

eJ

Thus

I = c&(I n B(ei)), a,s

required. (iii) By (ii),

Thus, for all i,

c

a.nd, since B ( e ; )is indecomposable, we have B(e;) Al or B(e;) Hence Ak = @B(eifsAkB(ei) (k = 1,2)

C Az.

as required.

(iv) If ei E I then y;(e;) = 1 # 0 and so n(I)# 0. Conversely, suppose that yi(I) # 0. Since y;(I) is a nonzero ideal of the field yj(Z(A)), we have yi(1) = yi(Z(A))

Thus y;(e;) E ~ i ( 1 ) It . follows, from Proposition l.l(v), that there are elements z E I,y E J ( Z ( A ) )and z E Z(A) such that 2

= e;

+ z(1 - e;) + y

Multiplying both sides by ei gives e; = xe; - ye;. Hence, by raising both sides to the k-th power, we find

Simple induction and restriction pairs

346

with u E I . Because ye; is nilpotent, the result follows. Let el, e2,. . . ,en be the block idempotents of A and let V be an indecomposable A-module. It follows from Proposition 1.2 that there exists i E { 1,.. . ,n} such that e;V= V

and

e j V = 0 for all j

#i

In this case we say that V lies in the block B(e;). Note that if V lies in the block B(e;), then (a) All A-modules isomorphic to V lie in B(e;). (b) v = 1 v = e;v for all v E V. The above gives a classification of all indecomposable (and in particular, all irreducible) A-modules into blocks. Our next aim is to tie together blocks and principal indecomposable modules. Let e be an idempotent of A and let e be the image of e in A = A/J(A). By Lemma 3.5.2, Ae/J(A)e

AE

Hence, by Theorem 3.12.4(ii), A E is an irreducible A-module. 1.3. Lemma. Let A = $r==lAe; be a decomposition of A into principal indecomposable A-modules, and let E , be the image ofe; in A = A/J(A). Then, for any A-module V,

dimFe;V = dimFHomA(Ae;,

v) = mdimFEndA(A?!;),

where m is the multiplicity ofthe irreducible A-module A E ; as a composition factor of V .

v)

Proof. By Lemma 1.5.8, the map HomA(Aei, + e;V, f H f (e) is an isomorphism of additive groups. Because this map is obviously F-linear, we have HomA(Ae;, V )

e;V

as F-spaces

The equality dimFHomA(Ae;,V) = mdimFEndA(AE;) is trivial if V is irreducible. We may thus assume that V has a maximal nonzero

1. Blocks of algebras

347

submodule X . Since Ae; is projective, the exact sequence 0 V 4 V / X -+ 0 induces an exact sequence

4

X

+

Therefore we have

If ml and m2 are the multiplicities of Ae; as a composition factor of X and V / X , respectively, then by applying induction on d i m ~ Vwe , have

Adding up these equalities gives

as asserted. 4

1.4. Corollary. Let e be a primitive idempotent of A, let be the image of e in A = A / J ( A ) and let V be an A-module. Then the following conditions are equivalent: (2) AE is a composition factor o f v . (ii) HomA(Ae, # 0. (iii) eV # 0.

v)

Proof. This is a direct consequence of Lemma 1.3. 4 Let U and V be principal indecomposable A-modules. We say that U a.nd V are linked, written U N V , if there exists a sequence

u, = u,u2,..., u,= v of principal indecomposable modules such that any two neighbouring left ideals in this sequence have a common composition factor. It is

Simple induction and restriction pairs

348

clear that is an equivalence relation on the set of all principal indecomposable A-modules. In what follows, we denote by X I , . . . ,X m the equivalence classes of -. N

1.5, Proposition. Let A = $!="=,ei be a decomposition of A into principal indecomposable modules, and for each j E (1,. . . ,m } , put

Ae; E Xj

(i) I 1 , . . . ,I m are all blocks of A. (ii) Two principal indecomposable A-modules are linked if and only if they belong to the same block. Proof. (i) It is clear that A = $jm=lI,. By Proposition 1.2(iii), it suffices to show that each Ij is a two-sided ideal contained in a block. Il with j # 1. Then Aei and Suppose that Aei C Ij and Aek Aek have no composition factor in common. In particular, Aek has no composition factor isomorphic to A E ; . Hence, by Corollary 1.4, eiAek = 0. This shows that IjIl = 0 for j # 1. Thus

and so Ij is a two-sided ideal of A. If Aei and Aej have the composition factor A e k in common, then by Corollary 1.4,ekAei # 0 and ekAej # 0. By Proposition 1.2(ii), B , Aej C B' and Aek C there exist blocks B , B', B" such that Aei B". It follows that 0 # ekAei & B"B and hence B = B". Similarly, 0 # ekAej C B"B' and so B' = B". Thus Ae; Aej C B and repeated application of this argument shows that Ij is contained in a certain block of A. This establishes the required assertion. (ii) This is a direct consequence of (i).

c

+

We close by recording some properties of separable algebras. An A-module V is said to be separable if VE = E @F V is a completely reducible AE-module for every field extension E / F (here AE = E@=A). Thus a separable module is completely reducible; the converse need not

1. Blocks of algebras

349

be true as we shall see below. For convenience of reference,we now quote the following two standard facts: 1.6. Proposition.

A field F is perfect if and only if every alge-

braic extension of F is separable.

Proof. See Jacobson (1964). 1.7. Proposition. Let V be a completely reducible A-module. Then V is separable i f and only i f for every irreducible submodule W of V the centre of the division algebra EndA(W) is a finite separable field extension of F .

Proof. See Bourbaki (1958). Recall that the algebra A is said to be separable if for every field extension E of F , AE is a semisimple E-algebra. Expressed otherwise, A is separable if the left regular module * A is separable.

The following conditions are equivalent: (i) A / J ( A ) is separable. [ii) For eve y field extension E I F , J(A,y)= J(A),y. (iii) For every field extension E / F , ( A / J ( A ) ) , yE A E / J ( A E ) . In particular, if A/ J ( A ) is separable, then so is AE/J(A,y)for any field extension E / F . 1.8. Proposition.

Proof. Let E / F be a field extension. Then

and the result follows. Let A be an algebra over a field F and let L be a subfield of F . We say that A is definable over L if there exists an F-basis X of A such that for all z 1 , x 2 E X , 51x2 is an L-linear combination of the elements of X. Expressed otherwise, A is definable over L if A 2 F 6 0 A1 ~ for some L-algebra Al.

Simple induction and restriction pairs

350

1.9. Proposition. Let A be a n algebra over a field F and ass u m e that A is definable over a perfect subfield of F. T h e n A / J ( A ) is separable.

Proof. By hypothesis, A F @ L Al for some L-algebra A1 and some perfect subfield L of F. By Proposition 1.8, we need only show that All J(A1) is a separable L-algebra. The latter being a consequence of Propositions 1.6 and 1.7, the result follows. 1.10. Corollary. Let F be a n arbitrary field. T h e n the F-algebras F G / J ( FG) and Z( FG)/J( Z( FG)) are separable. In particular, f o r every field extension E o f F ,

Proof. Let L be the prime subfield of F. Then L is perfect and

Now apply Propositions 1.9 and 1.8.

2.

Defect groups of blocks

In this section, G denotes a finite group, CI(G)the set of all conjugacy classes of G and F a field of characteristic p > 0. Let A be an F-algebra. Recall that the commutator subspace [A,A] of A is defined to be the F-linear span of all Lie products. [ z , y J = z y - ya: with z , y E A.

2.1. Lemma.

W e have

[FG, FG] = { C z g g E FGI

C zg= 0

for all C E CI(G)}

g€C

Proof. This is a special case of Lemma 3.6.10. W Let H be a subgroup of G. Then the natural projection x:FG-+FH

351

2. Defect groups of blocks

defined by

is obviously an F-linear map. In what follows, for each subset X of G, we define Xt E FG to be the sum of all elements in X . The next result due to Brauer, is of fundamental importance. 2.2. Theorem. Let P be a p-subgroup of G, let S = CG(P)and let H be a subgroup o f G for which S C_ H 5 N G ( P ) . Then the natural

projection T : F G --t FS induces a ring homomorphism of Z(FG) into Z(FH) whose kernel is the F-linear span of all C+ with C E C l ( G ) and C n S = 0.

Proof. Let GI,. . . ,C, be all conjugacy classes of G, let X i = and assume that Xr, # 0 for k E (1,. . . , t } and X k = 0 for k E ( t 1,. . .,r } . We know that the elements C t , . . . ,C: form an Ci

nS

+

F-basis for Z ( F G ) . Note also that

and that X 1 , .. . ,X t are mutually disjoint. Moreover, each X i , 1 5 i 5 t , is a union of conjugacy classes of H . Hence T induces an F-linear map from Z ( F G ) to Z ( F H ) whose kernel is of the required form. We are thus left to show that for all i, j E (1,. . . , r } ,

n(C;+Cjf)= T(C;+)T(Cjf) Given s E S , we put T, = { ( x , y ) l z E Ci,y E Cj and xy = s} and denote by T the union of all 2' with s E 5'. If T = 0,then either Xj = 0 or X j = 0, in which case T(C:C:) = 0 = T(C;')T(C~'). Now assume that T # 0. Then 7r(C;'C,')=

c

x y = c

(",Y)ET

c

zy

8ES ( ~ , Y ) E T ~

Because for all h E P and s E S , hsh-' = s, it follows that P acts as a permutation group on T, via h ( x , y ) = (hxh-',hyh-l). Since

Simple induction and restriction pairs

352

churF = p, the sum of all elements xy where (z,y) ranges all orbits of T, of size # 1, is equal to 0. On the other hand, an orbit T, has size 1 if and only if it is of the form ((5, y)} with 5,y E S. Thus

as required. The homomorphism Z ( F G ) + Z ( F H ) defined in Theorem 2.2 is called the Bruuer homomorphism.

2.3. Theorem. (Osima (1955)). Let e be a nonzero centruk idempotent of FG. Then Supp e consists of p'-elements. Proof. (Passman (1969)). Assume by way of contradiction that z E Suppe with z = zy = yz where x # 1 is a p-element and y is a $-element. Let P be the subgroup of G generated by x. Then, by Theorem 2.2, r(e) is a central idempotent of F S , where S = CG(P). Moreover, since z E S, we also have z E Supp?r(e). Hence we may assume that x E Z(G). Applying Lemma 2.1, we deduce that 5

E Suppv

for all v E [FG,FG]

(1)

Choose an integer n with pn 2 [GI and with p" f l(rnodq), where q is the order of y. Put t = y-*e and write t = Ct,g,t, E F . Then, by Lemma 3.6.9, tP" ZE ti"gP" (rnod[FG,FGI) Because gp" is a p'-element, it follows from (1) that x @ SupptP". On the other hand, since e is a central idempotent and since p" = l(rnodq), we have tP" = y+'e = y-*e = t But, by definition of t, x E Suppt, a contradiction. So the theorem is true. Let C be a conjugacy class of G and let g E C. A Sylow psubgroup of CG(g) is called a defect group of C (with respect to p ) . Thus all defect

353

2. Defect groups of blocks

groups of C are conjugate and so have a common order, say p d . The integer d is called the defect of C. We denote by S(C)any defect group of C. If HI and H2 are subgroups of G, we write HI LG H2 (respectively, HI CG H 2 ) to indicate that Hl is G-conjugate to a subgroup of H2 (respectively, to a proper subgroup of H2), while HI =G Hz will mean that HI is G-conjugate to H2. Let e be a block idempotent of F G and let B = B ( e ) . Because e E Z ( F G ) we have

Supp e = C1U C2U . . . U Ct for some C; E CI(G). The largest of the defect groups of C;, 1 5 i 5 T , denoted by 6 ( e ) (or S(B))is called a defect group of e (or of B ) . It will be demonstrated (Proposition 2.5) that all defect groups of e are conjugate and so have a common order, say p d . The integer d is called the defect of e (or of B ) . 2.4. Lemma. Let D be a p-subgroup of G and let C1, C2,. . . , Ct be all elements of Cl(G) with S(C;) CC D . Then the F-linear span ID(G) of C:, C t , . . . ,C t is an ideal o f Z ( F G ) .

Proof. Fix i E { 1,. , . ,t } and denote by Cj any conjugacy class of G. It clearly suffices to show that C;'CT E ID(G). We may assume that C;'C,' # 0, in which case we may choose g E SuppC:C:. Bearing in mind that

we have g = uv for some u E C; and v E Cj. Let of the conjugacy class containing g and let

P

be a defect group

Then P acts on X by conjugation, so by the argument employed in the proof of Theorem 2.2, we have Ci n &(P) # 8. Thus P GG 6(C;) and so P GG D. Hence C;'C[ E ID(G),as required.

Simple induction and restriction pairs

354

If D is a defect group of C E CI(G), we shall write I [ C ]instead of ID(G). The fact that I[C]is an ideal of Z ( F G ) will be needed in the proof of the following result. 2.5. Proposition. Let e be a block idempotent of F G and let y be the irreducible representation of Z ( F G ) associated with e . (i) Suppe = C1 u Cz u . . . u Ct u Ct+l u . . . u Cr. where Cl, . . . ,C, are p-regular classes of G such that (a) For all i E { 1,. . . ,t } , e E I[C;]and S(C;) =G S(Cj) for all i , j E (1,..., t}. (b) S ( C k ) cc S(Ci) for all k E {t 1,. . . ,r } and i E (1,. . . , t } . (ii) If C E CZ(G) is such that 6(C) C G S(C;)for some i E { 1, ... ,t } , then y(C+) = 0. (iii) There ezists i E { 1,. . . ,t } such that 7(C:) # 0. (iv) i f C E CZ(G) is such that -y(C+)# 0, then 6 ( e ) CC S(C).

+

Proof. (i) By Theorem 2.3, e = XC ; : for some nonzero X i in F and some pregular classes C1, . . .,C, of G. Becuase y( 1) = y( e) # 0, there is some k E (1,. . .,r } with y(C$) # 0. But C$ E I[Ck], so we must have y(I[Ck])# 0. Invoking Proposition 1.2(iv), we deduce that e E I[Ck]. By renumbering C1,. . . C, in such a way that {Cl,. . . ,C t } is the subset of {Cl,.. . ,C,} consisting of all Ci with S(Ci) = S ( C k ) , we deduce that (a) and (b) hold. (ii) By (i), 6(C) cc S ( C k ) . Assume that y ( P ) # 0. Because C+ E I [ C ] ,we have y(I[C]) # 0 and so, by Proposition 1.2(iv), e E I[C]. Thus 6(Ck) CG 6(C),a contradiction. (iii) This was established in (i). (iv) By the argument of (ii), e E I[C]. The desired assertion is therefore a consequence of the definition of I[C]. We next provide a useful characterization of defect groups in terms of relative trace maps. Let R be an arbitrary commutative ring and let H be a subgroup of G. Then the centralizer C m ( H )of H in RG is defined by

CRG(H)= {z E RG(& = hz for all h E H } In particular, CRG(G)= Z(RG). Consider the RG-module V = RG

2. Defect groups of blocks

355

where G acts on V by conjugation . Then

I ~ ~ ( V=HCRG(H) ) Hence, if S C H are subgroups of G, then the corresponding relative trace map

T ~ :FCRG(s)

-+

cRG(H)

is given by

T r f ( z ) = Ctzt-'

(z E

CFG(S))

t€T

where T is a left transversal for H in G.

Let S C H be subgroups of G and let T be a left transversal for S in H . (i) Tr? : CRG(S)-+ CRG(H) is a homomorphism of CRG(H)bimodules. In particular, Trf(CRG(S)) is an ideal Of CRG(H). (ii) If D is a p-subgroup of G, then 2.6. Lemma.

h ( G ) = Trg(CFG(D))

where ID(G)is defined in Lemma 2.4.

Proof. (i) We know that T r f is an R-linear map. Fix y E CRG(H) and z f CRG(S). Then, for all z E T , we have yz = zy and so TrF(yz) =

C zyzz-l = y C zzz-l xET

= y~r:(z)

%EX

A similar argument shows that T r F ( z y ) = T r F ( z ) y ,as asserted. (ii) Let z = C+ E I D ( G ) ,where C is a conjugacy class of G. Fix g E C, put L = & ( g ) and choose a Sylow psubgroup Q of L. Since z E I D ( G ) ,we may assume Q C D. Furthermore, by the definition of L, we must have z = T r f ( g ) . Applying Lemma 3.9.2(i) to compute T r z in two different ways, we find

TrLG (Tr&+) = TrE (Tr&))

(2)

Because Tr$(g)= ( L : Q)g and ( L : Q ) is prime to p , we may define

w = ( L : Q)-'Tr;(g)

Simple induction and restriction pairs

356

Then w E CFG(D)and, by (2)) T r g ( w )= z , proving that

ID(G) E Trg (CFG(D)) Let I( be the D-conjugacy class containing g E G and let C be the Gconjugacy class containing 9. Since the elements K+ form an F-basis for CFG(D),it suffices to show that T r g ( K + )E ID(G).But

Trg(Ic+)= ( c G ( g ) : c D ( g ) ) c + and therefore either Trg(I(+)= 0 or CD(9) is a defect of C. Hence in both cases TrE(K+)E ID(G),as desired. H

Proof. (i) This is a special case of Lemma 3.9.6. (ii) It follows from (i) and Lemmas 2.6(i) and 3.9.2(i) that TrE(a)TrE( b ) = T r g (TrE(a )b)

as asserted. (iii) This is a direct consequence of (i). (iv) Apply (ii). 2.8. Lemma. Let A be a jinite-dimensional algebra over F and let e be a primitive idempotent of A . If I l , . . . ,In are ideals of A such that e E Il - - - I,,, then e E Ii for some i E { 1,. . . ,n }.

+ +

2. Defect groups of blocks

357

Proof. Without loss of generality we may assume that n = 2. Since e is primitive, e is the identity element of the local ring eAe and e E 11 + 1 2 implies e E eI1e + e12e. Thus we may assume that A is local and that e = 1. If 1 = a + b with a E I l , b E 12,then not both a and b lie in J ( A ) , say a # J ( A ) . Since A is local, a has inverse a - l , which gives 1 = aa-' E 11,as required. W It is now an easy matter to provide the following useful characterization of defect groups of blocks. 2.9. Proposition. Let B = B ( e ) be a block of FG. Then the following conditions are equivalent: (i) D is a defect group of B. (ii) D is a minimal element in the set of all subgroups H of G such that e

E T r E (CFG(H))

Proof. Assume that D is a defect group of B. Since e E ID(G), it follows from Lemma 2.6(ii) that e E T T E ( C F G ( D ) ) .Now assume that H c D and that e E T T Z ( C F G ( H ) )Then, . by Lemma 2.6(iv), e E IH(G)contrary to the assumption that D is a defect group of e. Thus (i) implies (ii). Assume that H is a subgroup of G such that e E T T ~ ( C F G ( Hand )) let D be a minimal element in the set of all subgroups H of G. By Lemma 2.7(iv), (with K = G), we have

Therefore, by Lemma 2.8, we have

Then the minimality of D implies that gHg-' 2 D. Now suppose that H (and hence gHg-') is a defect group of B. Then gHg-' 2 D and, by the implication (i)+(ii), we obtain gHg-l = D ,as required. H

Simple induction and restriction pairs

358

We close by recording the following deep result due to Brauer (see Curtis and Reiner (1962)). The proof will be omitted since it relies on modular character theory, a topic we do not touch upon in this book. 2.10. Theorem. Let F be an algebraically closed field of characteristic p > 0, let G be a group of order pnm with (p,m) = 1 and let B be a block of F G with defect d. Then d is the smallest integer such that pn-d divides the dimensions of all irreducible FG-modules in B.

3.

Blocks and vertices

In this section, G denotes a finite group and F an arbitrary field of characteristic p > 0. For any psubgroup D of G, ID(G)denotes the ideal of Z ( F G ) defined in Lemma 2.4. All conventions and notations adopted in the preceding section remain in force. Our aim is to examine the vertices of the irreducible FG-modules which lie in a given block B of FG. It will be proved that the defect group of B can be characterized as the maximum among these vertices. A special case of this result, where F is algebraically closed, was established by Green (1959a), while its extension to an arbitrary F is due to Knorr (1976).

Let D be a p-subgroup of G and let e be an idempotent in ID(G) (e.9. e is a block idempotent of F G with D as a defect PUP). (i) For a n y FG-module V , eV is D-projective. (ii) IfV is an indecomposable FG-module in a block B = B ( e ) with defect group D , then V is D-projective. 3.1. Lemma.

Proof. If V satisfies (ii), then eV = V and hence (ii) follows from (i). To prove (i), we employ the relative trace map. By Lemma 2.6(ii), we may choose w E CFG(D) such that e = Trg(w). Consider the map 8 : eV t eV defined by 8(z) = w z for all 5 E eV. Since w commutes with all elements of g with g E D, we have 8 E EndFD(VD). Let T

3. Blocks and vertices

359

denote a left transversal for D in G and let x E eV. Then

and the result follows by virtue of Theorem 2.10.3. H 3.2. Lemma. Let B = B ( e ) be a block of F G and let I be an ideal of Z ( F G ) . I f e 6 I , then Ie is a nilpotent ideal of Z ( F G ) .

Proof. Let y be the irreducible representation of Z ( F G ) associated with e. Owing to Proposition l . l ( v ) Kery = Z ( F G ) ( l- e) + J(Z(FG))e We claim that I s Kery; if sustained it will follow that

l e C (Kery)e= J(Z(FG))e as required. Assume by way of contradiction that y(I) # 0. Because y ( Z ( F G ) ) is a field and 0 # y(I) is an ideal of y ( Z ( F G ) ) ,it follows that

y(Z(FG)) = Y ( I ) Hence y ( e ) = y(x) for some x E I and therefore e - x E Z(FG)(l - e)

+ J(Z(FG))e

Thus e-x = re+z(l-e) for some r E J ( Z ( F G ) )and some z E Z ( F G ) . It follows that e = xe+re and e--ze = re E J ( Z ( F G ) ) . Now J ( Z ( F G ) ) is a nilpotent ideal of Z ( F G ) and Z ( F G ) is a commutative algebra over the field F of characteristic p > 0. Therefore there exists a positive integer n such that

o = (re)P" = ( e - xe)P" = ep" - (xe)P" = e - (xe)P" Thus e = (xe)P" E I , a contradiction. H

Let e be a block idempotent of F G with deject group D and let j D ( G ) be the F-linear span of all C+,C E Cl(G), with 3.3. Lemma.

Simple induction and restriction pairs

360

6(C) CG D. Then ID(G)is an ideal o f Z ( F G ) and jD(G)e is a nilpotent ideal of Z ( F G ) . Proof. By Lemma 2.4, I D ( G )is an ideal of Z ( F G ) . Since j o ( G ) is the sum of all IH(G) with H c D,we see that ~ D ( Gis)an ideal of Z ( F G ) . By the definition of the defect group of e, e $ i D ( G ) . Thus f D ( G ) e is a nilpotent ideal of Z ( F G ) ,by virtue of Lemma 3.2. Let e be a block idempotent of FG with defect group D , let E be any field extension of F and write 3.4. Lemma.

as a sum of block idempotents of EG. Then, for all i E (1,. . . ,n } , ei has D as a defect group.

Proof. Let K and L be the counterparts of I D ( G ) and io(G), respectively, in EG. Fix i E (1,. . , ,n } and observe that, by the definition of the defect group of e;, it suffices to verify that e; E K - L. Since e E ID(G) C I< and since I< is an ideal of Z ( E G ) , we have e; = e;e E I<. On the other hand,

and so, by Lemma 3.3, Le is a nilpotent ideal of Z(EG). Hence if ei E L , then ei = eie E L e , a contradiction. W Let S be a subring of an arbitrary ring R. An R-module M is said to be S-projective if any exact sequence

of R-modules and R-homomorphisms which splits as an S-sequence, also splits as an R-sequence.

3.5 Lemma. Let S be a subring of a n arbitrary ring R and let M be an R-module. Then the following conditions are equivalent: (i) M is S-projective.

3. Blocks and vertices

361

(ii) The canonical m a p II, : R @IS M --t M given b y $(r @3m ) = r m is a retraction, i. e. there exists an R-homomorphism cp : M + R 8 s M with II, 0 cp = 1 (iii) There exists an S-module V such that M is a direct summand of

R @sV . (iv) Consider a n y diagram of R-modules and R-homomorphisms :

with exact row. If there is an. S-homomorphism cp : M + X making the diagram commute, then there is an R-homomorphism $ : M -, X making the diagram commute.

Proof. See Higman (1955, Theorem 4) and Hochschild (1956, Proposition 2). 3.6. Lemma. Assume that the following is a commutative diagram of R-modules and R-homomorphisms.

P

A ~ B - c

0

If Xcp = ly, th.en the following are equivalent: (i) Im(cpXa)G Ima. (ii) There is a 1c, : Z + C such that $a = Pcp and KII, = lz (in particular K is a retraction). Proof. Suppose that Im(cpAa)G Irncr. Since Imcr = li'erp, we

362

Simple induction and restriction pairs

which shows that Pcpp = 0, by virtue of the fact that p is an epimorphism. Thus P'p(Kera) = &(Imp) = 0 and therefore $ = P'pcr-' is a well defined R-homomorphism from 2 to C with the required properties. Conversely, suppose that (ii) holds. Then

and hence Irn(cpXa) C KerP = Ima. Our final preliminary result below will allow us to take full advantage of the results so far obtained.

3.7. Lemma. Let S be Q subring of an arbitrary ring R, let A C S be a subset such that AR = RA and let U be an S-projective R-module. Then M = U/AU is an S-projective R-module (the fact that AU is a submodule of U is a consequence of the equality R ( A U ) = ARU = AU). Proof. Consider the diagram below where all maps are canonical.

0

Taking into account that

3. Blocks and vertices

363

we have

The desired conclusion now follows by virtue of Lemma 3.6. W

We have now come to the demonstration for which this section has been developed. 3.8. Theorem. (Green (1959), Know (1976)). Let G be a finite group, let F be an arbitraryfield of prime chamcteristicp and let B be a block of F G . If P is a p-subgroup of G , then the following conditions are equivalent: (i) P is a defect group of B. (ii) All indecomposable FG-modules in B are P-projective and there is an irreducible FG-module V in B such that P is u vertex of V .

Proof. For the sake of clarity, we divide the proof into four steps. Step 1. Let D be a defect group of B. We claim that it suffices to verify that there is an irreducible FG-module M in B such that D is a vertex of M . Indeed, assume this to be true. By Lemma 3.1(ii), all indecomposable FG-modules in B are D-projective. Hence (i) implies (ii). Conversely, assume that (ii) holds. Since V is D-projective and P is a vertex of V , P is conjugate to a subgroup of D . On the other hand, A4 is P-projective and has vertex D . Hence D is conjugate to a subgroup of P, proving (i). Step 2. Here we establish the case where F is algebraically closed. Let d be the defect of B and let p" be the order of Sylow psubgroups of G. Owing to Theorem 2.10, there exists an irreducible FG-module M in B such that p"-d is the highest power of p dividing dirnFM. Denote by Q a vertex of M . Because M is indecomposable, M is D-projective and so Q c~ D. On the other hand, by Theorem 5.1.14, (S : Q)divides diinFM where S is a Sylow p-subgroup of G. Because ID1 = p d , this means that Q =G D and thus D is a vertex of M.

364

Simple induction and restriction pairs

S t e p 3. Let E be the algebraic closure of F . Here we demonstrate that every irreducible EG-module V is FG-projective. Indeed, by Corollary 1.10, FG/J(FG) is a separable F-algebra and

Denote by U a projective FG-module such that V EZ U / J ( E G ) U . Then, by (l), J ( E G ) U = J ( F G ) U and so V U/J(FG)U. Applying Lemma 3.7 with A = J(FG), we deduce that V is FG-projective. Step 4. Completion of the proof. Let e be the block idempotent of FG contained in B. Because e is a central idempotent of EG, we may write e = el - .. en

+ +

where the e; are block idempotents of EG. But B' = EGel and observe that, by Lemma 3.4, D is a defect group of B'. Applying Step 2, we deduce that there is an irreducible EG-module V in B' such that D is a vertex of V . Now J(EG)V = 0 and so, by (l), J(FG)V = 0. Thus VFGis completely reducible and therefore

for some irreducible FG-modules M;. Since multiplication by e induces the identity on V , all M; are in B. By Step 3, V is FG-projective and so, by Lemma 3.5, V is a direct summand of

But V is irreducible, so V is a direct summand of EG @FG M for some M E {Ml,. .. , M B ) . Denote by Q a vertex of M . Then M is a direct summand of F G @FQ M and thus V is a direct summand of

Invoking Lemma 3.5, we infer that V is FQ-projective. Hence V is EQ-projective and thus D is conjugate to a subgroup of Q. Now M is an irreducible FG-module in B. Consequently, M is D-projective and so Q is conjugate to a subgroup of D. This shows that D and Q are conjugate, thus completing the proof. W

4. Simple induction and restriction pairs

4.

365

Simple induction and restriction pairs

In this section, H denotes a subgroup of a finite group G, F an arbitrary field of characteristic p > 0 and e and f central idempotents of F G and FH, respectively. Recall that, by definition, the zero module will be regarded as completely reducible.

Let V C W be FG-modules. Then eV FG-modules such that eW/eV E e ( W / V ) . 4.1. Lemma.

eW are

Proof. Since e is a central idempotent of FG, eV 5 eW are certainly FG-modules. Let f : W + W/V be the natural homomorphism. Then f restricts to a surjective homomorphism

f’ : eW with K e r f * = eW

-, e ( W / V )

n V . But eW n V = eV, hence the result.

Following Knorr (1977), we define simple induction and restriction pairs as follows. We say that ( e , f ) is a simple induction pair if eVG is a completely reducible FG-module for all irreducible FH-modules 1’ = f V . The pair ( e , f ) is called a simple restriction pair, if fMH is a completely reducible FH-module for all irreducible FG-modules A d = e M . Let us first show that the above definitions may be rephrased in terms of J ( F G ) and J(FH). 4.2. Proposition. (i) The following conditions are equivalent: (a) ( e , f ) is a simple induction pair. (b) e J ( F G ) f C F G . J(FH).

-

(c) f J ( F G ) e J ( F H ) F G . (ii) ( e , f) is a simple restriction pair if and only if

e J ( F H )f

C J(FG)

(iii) Let E F be the smallest field containing the coeficients of e and f . Then ( e , f) is a simple induction (respectively, restriction) pair for F G and F H if and only if it is for E G and E H .

Simple induction and restriction pairs

366

Proof. (i) Put V = F H f / J ( F H ) f . Since V = fV is completely reducible and contains a copy of every irreducible FH-module S = fS, it follows that (e,f) is a simple induction pair if and only if eVG is completely reducible. Invoking Proposition 2.2.1 l(i), we have

V GE F G f / F G - J ( F H ) f and hence, by Lemma 4.1,

e V G Z e F G f / e F G -J(FH)f Consequently, eVG is completely reducible if and only if e J ( F G ) f C_ eFG - J ( F H ) f . Because the latter is equivalent to e J ( F G ) f C F G J ( F H ) , the equivalence of (a) and (b) is established. Assume that (b) holds. Then a

e J ( F G ) = e J ( F G ) f @ eJ(FG)(1- f ) E FG.J(FH)tFG(l-f) = F G [ J ( F H ) F H ( 1 - f)]

+

By applying Lemma 2.3.3(iii), we have

Hence f J ( F G ) e C J ( F H ) - F G , proving (c). The converse follows by symmetry. (ii) Put M = FGe/J(FG)e. Because A4 = eM is completely reducible and contains a copy of every irreducible FG-module W = eW, we deduce that ( e , f ) is a simple restriction pair if and only if fMH is completely reducible. Bearing in mind that, by Lemma 4.1,

fM f F G e / f J ( F G ) e

as FH-modules

it follows that ~ M isHcompletely reducible if and only if

fJ(FH)FGe

fJ(FG)e

367

4. Simple induction and restriction pairs

Because the latter containment is equivalent to e J ( F H ) f J(FG), (ii) follows. (iii) This is a direct consequence of (i), (ii) and Corollary 1.10. Note that if e f = 0 , then obviously (e, f) is both a simple induction arnd restriction pair. From now on, we investigate the case where ef # 0. The following preliminary observation will clear our path.

4.3. Lemma. The following conditions are equivalent: (i) e f # 0. (ii) There exists a projective indecomposable FH-module V = fV such that eVG # 0 . (iii) There is a n FH-module V = f V such that eVG # 0. (iv) There exists a projective indecomposable FG-module P = e P such that f P # 0 . (v) There exists a n FG-module M = eM such that f M # 0 . Proof. (i) +(ii): By Proposition Z.Z.ll(i), we have e ( F H f ) GS eFGf = F G e f

#0

Therefore there exists at least one indecomposable component V of F H f such that eVG # 0. (ii) +(iii): Trivial. (iii) +(iv): Let M be an irreducible FG-module in Soc(eVG)and let P = P ( M ) be the projective cover of M . Because V = f V , we deduce that from Corollary 2.2.7 that

# HOmFG(P,VG) and hence f P # 0. 0

HOmFH(PH,V)

(iv) =+(v): Trivial. (v) +(i): By assumption, 0 a.s required. W

#fM

HomFH(fPH,V)

= f e M = e f M and so ef

#

0,

If e and f are block idempotents, then S(e) and S(f) will denote the defect groups of the corresponding blocks; the vertex of a module M will be denoted by v z ( M ) . We are now in a position to establish our first main result.

368

Simple induction and restriction pairs

(Knorr (1977)). Let e and f be block idempotents of F G and F H respectively, and let ef # 0. (i) If ( e ,f) is a simple induction pair, then f M H # 0 for all FGmodules 0 # M = e M ; moreover 6(f) >G 6(e). (ii) I f ( e , f ) is a simple restriction pair, then eVG # 0 for all F H modules O # V = f V ; moreover, S(f) CG b ( e ) . 4.4. Theorem.

Proof. (i) Let V = eV be an irreducible FG-module such that f VH = 0. We first show that f PH = 0 for the projective cover P of V. Assume the contrary. Because e P = P , it follows from Corollary 2.2.7 that there exists an irreducible FH-module W = fWsuch that

0 # H o ~ F H ( PW H ,) E HomFc(P, W G )Z H o ~ F G ( Pe W , G) However, eWG is completely reducible by hypothesis, so 0

# HonzFG(V, wc)

Hu~FH(VH w) , = HUrnFH( f V H , W ) = 0

a contradiction. Now suppose that

eM = M # 0 and fMH = 0 for some FGmodule M.Then f VH = 0 for any composition factor V of M . Because e is a block idempotent, all indecomposable projective modules in the corresponding block are linked (Proposition 1.5). Thus, by repeated use of the above argument, f P H = 0 for all projective modules P = e P , contradicting Lemma 4.3(v). This proves the first assertion. Owing to Theorem 3.8, there exists an irreducible FG-module V = eV such that vz(V) =G S ( e )

Then, by the foregoing, f v H # 0 and therefore there exists an irreducible FH-module W in Soc( f VH).Because eV = V , it follows from Proposition 2.2.4(i) that 0

#H

v)

o m ~ ~ (VH) w , 2! H o m ~ c ( W ~ Zi , HomFc(eWG,

v)

However, eWG is completely reducible by hypothesis, and e W G is a direct summand of W G ,hence V is a direct summand of W G . Thus, by Theorem 3.8

4. Simple induction and restriction pairs

369

as required. (ii) The first statement is an easy adaptation of the ideas in the proof of (i). To establish the assertion regarding the defect groups, we choose an irreducible FH-module W = fWsuch that

v x ( W ) = H b(f)

Let V = e V be an irreducible FG-module in Soc(eWG). Then

Because f VH is completely reducible and ~ V isHa direct summand of VH, we deduce that W is a direct summand of VH. Denote by Q a vertex of V . Then V is a direct summand of (VQ)', so VH is a direct summand of [ ( V Q ) ~and ] Hthus W is a direct summand of [(VQ>']H. It follows from the Mackey decomposition that W is a direct summand of [(gv)HngQ,-l]H

for some g E G

and consequently

as asserted. R

Two remarks are now in order. 4.5. Remark.

The following assertions are direct consequences

of the above proof. (i) Assume the conditions of Theorem 4.4(i). Then

SoC(fVH)2 f VH/J ( F H )f VH for all irreducible FG-modules V = eV. (ii) Assume the conditions of Theorem 4.4(ii). Then Soc(eWG)%! eWG/J(FG)eWG for all irreducible FH-modules W = fW.

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Simple induction and restriction pairs

4.6. Remark. Write e = Ce; and f = Cfj as a sum of block idempotents of F G and FH, respectively. Then, by the definition, ( e ,f ) is a simple restriction (induction) pair if and only if ( e i ,f j ) are simple restriction (induction) pairs for all i and j .

The rest of the section will be devoted to recording a number of consequences of the results so far obtained. As a preliminary result, we first establish the following useful property.

4.7. Lemma. Let H S be subgroups of G and let W be an irreducible FS-module. (i) If ( e ,f) is a simple induction pair and f w H # 0, then eWG is completely reducible. (ii) If ( e ,f) is a simple restriction pair and eWG # 0 , then fWH is completely reducible.

Proof. (i) By assumption, f WH # 0 and hence we may choose an irreducible FH-module V in Soc( f W H ) .Then f V = V and

Since W is irreducible, there is an exact sequence

vs-+w+o Thus there exists an exact sequence

Because ( e ,f ) is a simple induction pair, eVG is completely reducible. Hence eWG is also completely reducible. (ii) The proof is analogous and therefore will be omitted. W

Let H 2 S be subgroups o f G and let e and f be central idempotents of F G and FH, respectively. Let { e i } denote all block idempotents of FS such that E ; f # 0. Put E = x c ; and assume that ( E , f) is a simple induction pair. Then the following conditions are equivalent: 4.8. Corollary.

4. Simple induction and restriction pairs

371

(i) ( e , f) is a simple induction pair. (ii) ( e ,E ) is a simple induction pair. Assume that W = EW is an irreducible F S module. By Remark 4.6 and Theorem 4.4(i), fW, # 0 since (e,f) is a simple induction pair. Thus, by Lemma 4.7(i), eWG is completely reducible which means that ( e ,E ) is a simple induction pair. (ii) +(i): Let V = f V be an irreducible FH-module. Then e V S and e(eVS)Gare completely reducible, since (E, f) and ( e ,e ) are simple induction pairs. Applying Lemma 4.3, it follows from the construction of E that V s = eVS. Thus eVG is completely reducible and so ( e , f ) is a simple induction pair. f l

Proof. (i) +(ii):

Let H S be subgroups o f G and let e and f be central idempotents of FG and FH, respectively. Let { ~ i } be all block idempotents of FS with t;;e # 0. Put c = C ei and assume that ( e ,E ) is a simple restriction pair. Then the following conditions are equivalent: (i) ( e ,f ) is a simple restriction pair. (ii) ( e , f) is a simple restriction pair. 4.9. Corollary.

Proof. The proof is analogous and therefore will be omitted. H Let N be a normal subgroup of G and let f be a block idempotent of F N . It is clear that the set is a subgroup of G containing N . We refer to G ( f )as the inertia group of f (or b = b ( f ) ) . Note that if T is a left transversal for G ( f ) in G, then the tft-’,t E T , are all distinct conjugates o f f and

f* = Ctft-l tET

is an idempotent of Z( F G ) n Z( F N ) . We say that a block idempotent e E F G covers f (or B = B(e) covers b = b ( f ) ) if e occurs in the decomposition of j” into the sum of block idempotents of FG.

4.10. Lemma. A block B = B ( e ) o f F G covers the block b = b ( f ) of F N if and only i f e f # 0.

372

Simple induction and restriction pairs

Proof. It is clear that e f # 0 if and only if ef* # 0. Since the latter holds if and only if e covers f, the result follows. W 4.11. Theorem. (Know (1977)). Let N be a normal subgroup of G , let b = b( f) be a block of F N and let B; = B;(e;),1 5 i 5 m, be the block of F G covering b. Let Vl,. , .,& be all nonisomorphic irreducible FN-modules in b and let Hi be the inertia group of V;., 1 5 i 2 t . (2) rn 5 min((H1 : N ) , ( H 2 : N ) , . . . , ( H t : N ) } . (ii) If m = ( G ( f ) ): N ) , where G(f) is the inertia group of f, then

FG J ( F N )

J ( FG)f

*

and

b(f) Proof. (i) Let V be a n irreducible module in b and let H be the inertia group of b. Because N d G, it follows from Clifford’s theorem that (1, 1) is a simple restriction pair. Thus, by Remark 4.6, ( e i ,f) is a simple restriction pair, 1 5 i 5 rn. Note also that, by Lemma 4.10, e;f # 0, 1 5 i 5 m. Applying Theorem 4.4(ii), we deduce that Wi = e;VG # 0 , 1 5 i 5 m. Let ci > 0 be the composition length of W; and Mjj its composition factors. Since +i)

(VC)N

=G

= @gETg €3

v

where T is a transversal for N in G, it follows from Clifford’s theorem that

dimFM;j = r;j(G : H)dimFV for some integers

and

rij.

Accordingly,

373

5. Complete reducibility of induced modules

as asserted. (ii) Because H C G(f), it follows that rn = (G(f) : N ) 2 ( H : N ) 2 rnin((H1 : N ) , . . . ,(Ht : N ) } 2 rn and hence ( H : N ) = rn. It follows from (1) that c; = 1 for all i, so V G= $e;VG is completely reducible. Thus (1,f) is a simple induction pair and hence, by Proposition 4.2, J ( F G )f FG J ( F N ) . The assertion regarding the defect groups follows from Theorem 4.4 and Remark 4.6. W

-

5.

C o m p l e t e reducibility of induced modules

Throughout this section, F denotes an arbitrary field of characteristic p > 0 and G a finite group. Our aim is to prove a number of results pertaining to complete reducibility of induced and restricted modules. By the principal block of FG, we understand the block which contains the trivial FG-module, i.e. the irreducible FG-module on which G acts as the identity operator.

5.1. Lemma. The Sylow p-subgroups of G are defect groups of the principal block of F G .

Proof. Let B = B(e) be the principal block. By Theorem 2.3, we may write e in the form e = AICf

+ - + Arc,? *

(0 # A; E F )

for some p-regular classes C1,. . . ,C, of G. Denote by V the trivial FG-module. Since V belongs to B , we have ev = v for all v E V . Hence = (X1IC1I+...+XrIC,l)v for all v E V

which implies that p does not divide ICI for some pregular class C in Suppe. Hence a Sylow p-subgroup of G is a defect group of C, as required. W An application of the above lemma together with Theorem 3.8 now easily yield the following two results.

Simple induction and restriction pairs

374

(Knorr (1977)). Let H be a subgroup of G. Assume that V G is completely reducible for any irreducible FH-module V . Then H is ofp'-index in G. 5.2. Proposition.

Proof. Let B = B(e) be the principal block of FG. By Lemma 5.1, a Sylow p-subgroup P of G is a defect group of B. By Theorem 3.8, we may choose an irreducible FG-module U in B such that P is a vertex of U. Let V be an irreducible submodule of UH.Since eU = U , we have

0 # HOmFH(V, UH) HOmFc(VG, u)

HOmFG(eVG,u )

But eVG is a submodule of V G and VG is completely reducible by assumption. Thus eVG is completely reducible and therefore U is a direct summand of VG. Hence U is H-projective and so H contains a conjugate of P . This shows that H is of p'-index, as desired. W

Let H be a subgroup ofG. Then the following conditions are equivalent: (i) All irreducible FG-modules are H-projective. (ii) H is ofp'-index in G. 5.3. Proposition.

Proof. (i) +(ii): Let B = B(e) be the principal block of FG. By Theorem 3.8 and Lemma 5.1, we may choose an irreducible FGmodule V in B such that a Sylow p-subgroup P of G is a vertex of V. By assumption, V is H-projective and so H contains a conjugate of P. Thus H is of $-index in G. (ii) +(i): This is a consequence of Corollary 2.10.4. Our next aim is to strengthen Theorem 2.3.9 under the additional assumption that R is a field. We need a number of preliminary results contained in a work of Knorr (1977). 5.4. Proposition. Let H be a subgroup ofG and let e and f be central idempotents of FG and F H , respectively. Then the following

conditions are equivalent: (i) (eVG)H is completely reducible for all irreducible FH-modules V.

375

5. Complete reducibility of induced modules

(ii) e J ( F H ) F G C F G . J ( F H ) . (iii) e J ( F H ) F G = F G . J ( F H ) e . Moreover, if these conditions are satisfied, then

e J ( F H ) " F G = ( e J ( F H )- FG)"

for all n 2 1

Proof. (i) +(ii): Put W = F H / J ( F H ) . Then W is completely reducible, so by assumption (eWG)H is completely reducible. Thus e J ( F H ) ann(WG).However, by Theorem 2.3.4(i),

c

ann( W G )= I d ( F G - J ( F H ) )

-

and so e J ( F H ) F G C F G J ( F H ) . (ii) +(iii): It is obvious from (ii) that

eFGJ( F H ) F G C F G - J ( F H ) and therefore, by Lemma 2.3.3(iii),

eFG J ( F H )

c

eFGJ(FH)FG Id(F G * J (F H ) ) = I d ( J ( F H )* F G )

It follows that eFG - J ( F H ) e J ( F H ) F G . The opposite containment is obvious from (ii). (iii) +(i): It is clear from (iii) that F G J ( F H ) e is an ideal of FG. Let V be an irreducible FH-module and let A = a n n ( V ) . Then J ( F H ) C A and FG J ( F H ) e C F G A. Applying Theorem 2.3.4(i), we conlude that

-

J ( F H ) eC F G . J ( F H ) e2 Id(FG * A ) = ann(VG) Hence J ( F H ) ( e V c ) = 0 and so ( e V G ) H is completely reducible. The final assertion follows by induction on n , using (iii). W 5.5. Proposition. Let H be a subgroup o f G and let e and f be central idempotents of F G and F H , respectively. Then the following conditions are equivalent: (i) (fMH)G is completely reducible for all irreducible FG-modules

Simple induction and restriction pairs

376

M. (ii) V G is completely reducible for all irreducible FH-modules V = f V , and fMH is completely reducible for all irreducible FG-modules M. (iii) J ( FG) f = FG - J ( F H )f . (iv) f J ( F G ) = f J ( F H ) F G . Proof. Condition (ii) may be reformulated as: (1,f) is a simple induction pair and a simple restriction pair. The equivalence of (ii), (iii), and (iv) is therefore a consequence of Proposition 4.2. Since (ii) obviously implies (i), we are left to show that (i) implies (ii). Let M be an irreducible FG-module. Then, by assumption, (fM H ) ~ is completely reducible and hence, by Corollary 2.2.10, fMH is completely reducible. Let V = f V be an irreducible FH-module and let M be an irreducible FG-module in Soc(VG). Then

0#

M,VG)

H O ~ F G (

HomFH(MH,

v)

f M H ,v )

H O ~ F H (

Hence there is an exact sequence

which gives rise to an exact sequence

Since (fM H ) is~completely reducible, so is V Gas required. 5.6. Lemma. Let H be a subgroup o f G and let e be a central idempotent of FG. Then the following conditions are equivalent: (i) For any FG-module M = e M , if MH is completely reducible, then so is M . (ii) F G . J ( F H ) F G 2 J(FG)e.

Proof. (i) *(ii): Put M = F G e / F G . J ( F H ) F G e . Then M = eM and M H is completely reducible. Thus, by hypothesis, M is completely reducible. Hence

J ( F G ) e C F G . J ( F H ) FGe 5 F G - J ( F H ) . F G ,

377

5. Complete reducibility of induced modules

as required.

(ii) +(i): Let M = eM be an FG-module such that MH is completely reducible. Then J(FH)M = 0 and if F G J ( F H ) . FG 2 J ( F G ) e , then b

J(FG)M = J ( F G ) e M F G - J ( F H ) - FGM = FG-J(FH)M = 0,

proving that M is completely reducible. 5.7. Lemma. Let H be a subgroup of G and M be an FG-module such that M / J ( F G ) M is H-projective. Then

FG * J(FH)M 2 J(FG)M

+

Proof. Put X = F G - J(FH)M. Then Y = M / ( X J ( F G ) M )is a direct summand of the completely reducible and H-projective module M / J ( F G ) M , so Y is completely reducible and H-projective. Consider the exact sequence

0

+

(X

+ J(FG)M)/X+M / X +Y

4

0

(1)

Because J ( F H ) M 2 X , it follows that ( M / X ) H is completely reducible. Hence (1) splits as an FH-sequence. But Y is H-projective, so (1) splits as an FG-sequence. Thus

MIX 2 [(X + J ( F G ) M ) / X ] CB Y a.nd multiplying both sides by J ( F G ) gives

( J ( F G ) M t X)/X 2 J ( F G ) [ ( Xt J(FG)M)/X] since

Y is completely reducible. Hence (X t J ( F G ) M ) / X = 0

and so J ( F G ) M

X, as required. H

Simple induction and restriction pairs

378

5.0. Lemma. Let H be a subgroup o f G and let e be a central idempotent of F G . (i) If J ( F G ) e F G . J ( F H ) , then all irreducible FG-modules M =

eM are H-projective. (ii) If all irreducible FG-modules M = eM are H-projective, then V H completely reducible implies V is completely reducible for a n y FGmodule V = eV. Proof. (i) Owing to Proposition 4.2(i), (e, 1) is a simple induction pair. Write e = C e; as a sum of block idempotents of FG. Then, by Remark 4.6,each ( e i , l ) is a simple induction pair. Therefore, by Theorem 4.4(i), b(e;) S G H. Let M = eM be an irreducible FGmodule. Then M = e;M for some i and so, by Theorem 3.8, M is b(ei)-projective. Hence M is H-projective, by virtue of Theorem 2.10.7( ii). (ii) Suppose that all irreducible FG-modules M = eM are H projective. By Lemma 5.6, it suffices to show that F G J ( F H ) - F G 2 J ( F G ) e . By assumption and by Theorem 2.10.7(i), F G e / J ( F G ) e is H-projective. Hence, by Lemma 5.7,

-

F G - J ( F H ) FG 2 F G - J ( F H ) F G e 2 J ( F G ) e as asserted. 1 5.9. Lemma. Let H be a subgroup o f G and let e be a central idempotent of F G . If (eVG)H is Completely reducible for all irreducible

FH-modules V , then J ( F H ) e C J ( F G ) . Proof. Suppose that for each irreducible FH-module V , ( e V G ) H is completely reducible. Then, by Proposition 5.4, e J ( F H ) F G is a nilpotent ideal of FG. Accordingly,

e J ( F H )C e J ( F H ) F G C J ( F G ) as required. W

We are now ready to prove our first major result.

5. Complete reducibility of induced modules

379

5.10. Theorem. (Knorr (1977)). Let H be a subgroup o f G and let e be a central idempotent of FG. Then the following conditions are

equivalent: (i) J ( F G ) e = F G - J ( F H ) e . (ii) J ( FG)e = J ( FH)FGe. (iii) eVG is completely reducible for every irreducible FH-module V and M H is completely reducible for any irreducible FG-module M = eM. (iv) eVG and ( e V G ) H are completely reducible for all irreducible FH-modules V . (v) (e V G ) H is completely reducible for every irreducible FH-module V and all irreducible FG-modules M = eM are H-projective. (vi) (eVG)H is completely reducible for any irreducible FH-module V and MH completely reducible implies M completely reducible for any FG-module M = e M . Proof. Condition (iii) may be reformulated as follows: (e, 1) is a simple induction and a simple restriction pair. Thus, by Proposition 4.2, (iii) is equivalent to either of the following conditions:

J ( F G ) e C F G . J ( F H ) and J ( F H ) e C_ J ( F G )

(2)

J ( F H ) - F G and J ( F H ) e C_ J ( F G )

(3)

J ( FG)e

But (2) is equivalent to (i) and (3) is equivalent to (ii), hence (i), (ii) and (iii) are equivalent. The implications (iv)+(v)=+(vi) follow by Lemma 5.8 and Proposition 4.2, while the implication (iv)+(iii) is a consequence of Lemma 5.9 and Proposition 4.2. We are thus left to show that (vi) implies (iv) a.nd (ii) implies (iv). (vi) +(iv): Owing to Proposition 5.4, F G . J ( F H ) e = J ( F H ) F G e and, by Lemma 5.6

F G . J ( F H )- FG 2 J(FG)e Consequently,

J ( F G ) e C F G . J ( F H ) .FGe = F G - J ( F H ) e & F G - J ( F H )

380

Simple induction and restriction pairs

and therefore, by Proposition 4.2, eVG is completely reducible for any irreducible FH-module V . (ii) +(iv): This is a direct consequence of the implications (ii) +(iii), (ii) +(i) and Proposition 5.4. H As an application of the above theorem, we now establish our second major result. 5.11. Theorem. (Knorr (1977)). Let H be a subgroup of G and let F be a field of characteristic p > 0. Then the following conditions

are equivalent: (i) There exists a normal subgroup N of G such that N C_ H and N has p'-index in G. (ii) V G is completely reducible for any irreducible FH-module V and MH is completely reducible for a n y irreduicble FG-module M . (iii) V G and ( V G )are ~ completely reducible for all irreducible F H modules V . (iv) ( V G )is~completely reducible for all irreducible FH-modules V and all irreducible FG-modules are H-projective. (v) ( V G ) H is completely reducible for any irreducible FH-module V and MH completely irreducible implies M completely reducible for any FG-module M . (vi) ( V G )is~ completely reducible for all irreducible FH-modules V and H is of p'-index. (vii) ( M H ) Gis completely reducible for any irreducible FG-module M. (viii) J ( F G ) = F G . J ( F H ) . (ix) J ( F G ) = J ( F H ) F G .

Proof. Applying Proposition 5.5 for f = 1 and Theorem 5.10 for e = 1, it follows that the conditions (viii), (ix), (ii), (iv), (v), and

(vii) are equivalent. Moreover, by Proposition 5.3, (iv) and (vi) are equivalent. It therefore suffices to show that (i) is equivalent to one of the conditions (viii), (ii), (vi). For the sake of clarity, we divide the rest of the proof into two steps.

Step 1. Suppose that for any irreducible FH-module V , (V"), is

5. Complete reducibility of induced modules

381

completely reducible. Our aim is to show that for any given g E G,

J ( F H ) = F H . J ( F L ) where L = H n gHg-' Put V = F H / J ( F H ) . Then V is a completely reducible FH-module, hence so is ( V G ) by ~ ,hypothesis. Setting W to be the restriction of gV to F L , it follows from Mackey's decomposition that W His completely reducible. Put A = ann(W) and S = F L / A . Owing to Theorem 2.3.4(i), J ( F H ) C F H A and hence, by Proposition 2.2.11(i), SH is completely reducible. Hence, by Corollary 2.2.10, S is completely reducible and therefore J ( F L ) C_ A. But

A = F L n g(annV)g-l = F H n F(9Hg-l) n gJ(FH)g-' = F H n gJ(FH)g-l C J(FL) since F H n gJ(FH)g-' is a nilpotent ideal of FL. Thus

A = F H n gJ(FH)g-l = J ( F L ) and hence

J(FH)C FH * J(FL)

Replacing g by g - l , it follows by the same argument that

F H n g-'J(FH)g = J ( F ( H n g-'Hg)) Because

9FHg-l = F(9Hg-l) and gJ(FH)g-l = J(F(gHg-')) the latter implies that

Hence J ( F H ) 2 J ( F L ) and thus J ( F H ) = F H - J ( F L ) , as required. Step 2. Completion of the proof. Suppose that (vi) holds. To prove

382

Simple induction and restriction pairs

(i), we argue by induction on [HI. If H d G then we are done with N = H . Assume that L = H n g H 9 - l is a proper subgroup of H for some g E G. Because (viii) and (vi) are equivalent, it follows from Step 1 that

J ( F G ) = FG * J ( F H ) = FG * F H - J ( F L ) = FG - J ( F L ) Hence, by induction, there is a normal subgroup N of G such that N E L and N has $-index in G. Since L E H , (i) is established. Finally, suppose that (i) holds. Let V be an irreducible FH-module. Because ( V G )is~a direct sum of FN-modules of the form 9V,g E G, ( V G )is~completely reducible. Since N has p'-index in both H and G, it follows from the equivalence of (viii) and (vi) that

J ( F G ) = F G . J(FN) and J ( F H ) = F H - J ( F N ) Hence

FG. J ( F H )= F G . J(FN) = J(FG), proving (viii) and the result follows. 1

383

Chapter 7 Permutation modules This chapter provides a detailed account of a distinguished class of induced modules, namely permutation modules. These modules hold much information about the p-modular representations of G, as well as information about the fusion of p-subgroups of G. The main results presented are due to Alperin (1988), Broue' and Robinson (1986), Robinson (1988), Dress (1975), Saksonov (1971) and Broud (1985). We give a detailed account of the Brauer morphism which enables us to provide a comprehensive coverage of an important type of ppermutation modules, the so called Scott modules. These modules were first discovered by L.L. Scott, and independently by J.L. Alperin. Our method is based on a work of Broud (1985) who employed a systematic use of the Brauer morphism. The chapter ends with the study of ppermutation modules via the Brauer morphisrn. 1.

Preliminary results

In this section, R denotes an arbitrary commutative ring and G a finite group. All RG-modules are assumed to be finitely generated. Given a subgroup H of G, we write 1~ for the trivial RH-module, i.e. 1~ = R as R-module and

hr = r

for all r E R , h € H

Let V be an RG-module. We say that V is a permutation module (respectively, transitive permutation module) if V is R-free with an R-

Permutation modules

384

basis on which G acts as a permutation group (respectively, G acts as a transitive permutation group). Given an RG-module V , we write as usual Inv(V) for the R-module of G-invariant elements of R. 1.1. Lemma. (i) An RG-module V is a transitive permutation module if and only if V Z ( l H ) G for some subgroup H of G. (ii) An RG-module V is a permutation module if and only if it is a direct sum of transitive permutation modules. (iii) Let V be a permutation module and let B be a basis of V on which G acts as a permutation group. If B1,. . . ,B, are G-orbits of B , then Inv(V) is a free R-module with basis Bf,. . . ,B:, where B: = CIEBi 2, 1 I iI r.

Proof. (i) Assume that an RG-module V is a transitive permutation module. Then there exists an R-basis, say v1,. . . ,vn, of V such that G acts as a transitive permutation group on {vl,. . . ,vn}. Set= Rvi it follows that V is an imprimitive RG-module with ting {Kll I: i 5 n } as a system of imprimitivity and with G acting transitively on {Kll 5 i 5 n}. If H is the stabilizer of V,, then V s V y by Proposition 2.1.4(ii). Since & E l ~we, conclude that V E ( 1 ~ ) ~ . Conversely, if V E ( l ~ )then ~ ,by Proposition 2.1.4(i) and Corollary 2.1.2(iii), V is a transitive permutation module. (ii) This is a direct consequence of (i) and the definition of a permutation module. (iii) Straightforward. H For any subset S of G, we define S+ E RG by

s+=cs sES

1.2. Corollary. Let H be a subgroup of G and let Cl, . . . , C, be all H-conjugacy classes of G. Then CRG(H)is a free R-module with basis

c:,

*

. . ,c;.

Proof. We put V = RG and note that V is a permutation RHmodule, where H acts by conjugation on the R-basis G of V . Since the H-orbits of V are precisely the H-conjugacy classes of G and since

1. Preliminary results

385

I n v ( V ) = C R G ( H ) ,the result follows by virtue of Lemma l.l(iii).

Let H be a subgroup of G. (i) 1~ ES R H + . (ii) ( 1 ~E)RG ~ - H+. (iii) If n = [HI is a unit ofR, then e = n-lH+ is an idempotent of RH such that ( 1 ~E)RGe ~ 1.3. Lemma.

(iv) RG+ C RG H+ (hence

1G

is isomorphic to a submodule of

(lH)G).

(v) Let (G : H ) be a unit of R. Then cp E EndRG(RG H + ) given b y v ( H + ) = (G : H)-lG+ is a nonzero idempotent. In particular, (a) RG - H+ = RG+ @ RG(H+- (G : H)-'G+). (b) lc is isomorphic to a direct summand of (1~)'. (vi) If H # G and ( 1 ~ is) indecomposable, ~ then (G : H ) is a nonunit of R . Proof. (i) This is obvious. (ii) Apply (i) and Proposition 2.2.11(i). (iii) This is a direct consequence of (ii). (iv) Let T be a left transversal for H in G. Then

as required.

(v) Consider the RG-homomorphism cp : RG. H+ + RG- H+ given by

cp(gH+) = (G : H)-'G+

By (iv), we have

and hence cp is a nonzero idempotent of EndRG(fiG- H + ) , as required. (vi) Direct consequence of (v). W

386

Permutation modules

We now introduce the following important concept. A G-algebra over R is an R-algebra A on which G acts as a group of R-algebra automorphisms, i.e. g E G acts on a E A to give ga E A , and this G-action transforms A into a left RG-module, and

for all a,b E A and g E G. Let A be a G-algebra. Then A is an RG-module and, for any subgroup H of G, we write AH for the R-subalgebra of H-invariant elements, i.e.

AH = { a E A1 ha = a for all h E H } For example, if A = RG where G acts by conjugation, then A is a G-algebra and

AH = CRG(H) for any subgroup H of G Another typical example of G-algebras is the R-algebra

A = EndR(V) for any RG-module V. Here for any g E G, f E En&(V), gf = gfg-' where (gfg-l)(v) = g(f(g-'v)) for all EV Observe also that for any subgroup H of G, we have

AH = EndRH(V)

If H , I< are subgroups of G such that H C I<, we may define on any G-algebra A , the trace map

by

ta for all a E AH

Tr$(a) = %T

387

1. Preliminary results

where T is a left transversal for H in K . Of course, the above map is a special case of the general trace map applied to the RG-module A. In what follows, given subgroups H , K of G with H C K , we put

A; = T r $ ( A H ) For any subset S of A, we also put

By a permutation G-algebra, we understand a G-algebra A over R such that A is a permutation RG-module, i.e. there exists an R-basis X of A on which G acts as a permutation group. For any given x E X and any subgroup H of G, we write H x and H ( z ) for the H-orbit of x and the stabilizer of x in H , respectively. Thus H x = { hxlh E H )

and

H ( z )= {h E Hlhx = x } It is clear that lHxl = ( H : H ( z ) ) . 1.4. Lemma. Let A be a permutation G-algebra, let X be an R-basis of A which is permztted by the action of G and let H,K be subgroups o j G with H I(. (i) If x l , . . . ,x, are all representatives for the K-orbits of X , then

is an R-basis of A K . (ii) For any x E X ,

Tr$(( %)+) = ( K ( z ): H ( z ) ) (Kx)+ Proof. (i) This is a special case of Lemma l.l(iii). (ii) Let S be a left transversal for H ( x ) in H and let T be a left

Permutation modules

388

transversal for H in K . Then T S is a left transversal for H ( x ) in K and so

as required. H In what follows, we regard 0 as a free R-module on an empty basis.

1.5. Proposition. (Green (1968)). Let A be a permutation Galgebra, let X be an R-basis o f A which is permuted by the action of I(. If xl,.. . ,x, are all G and let H , K be subgroups of G with H representatives for the K-orbits o f X and rn; is the greatest common divisor of ( K ( s ; ): (kHk-'n IC(xi))) with k running over K then the nonzero elements o f t h e set

is an R-basis of A;,

where A;

AX is given b y ( I ) .

Proof. By Lemma 1.4, A; is R-generated by the set

Each ( "x)+ is equal to ( Kx;)+ for exactly one i E { 1,. . . ,n ) ; in fact ( Kx)+ = ( K z ; ) +if and only if x E Kx;, i.e. if and only if x = kx; for some k E I(. Thus A; has R-basis {Xi( Kxi)+ll 5 i 5 n } , where X i is the greatest common divisor of the set of integers

{ ( K (Kx;) : H ( K x ; ) ) l k E K } Note that G( gs)= gG(z)g-' for all x E X , g E G. Hence

('I

gz) =

K n G( '3) = g ( g -l K g n G(x)g-'

I . Preliminary results

389

Applying this formula, we find that, for each k E I(,

( K (kz;): H ( k ~ ; ) ) = ( ( k - ' K k n G ( z ; ) ): ( k - l H k = ( K ( z i ): (k-'Hk

n G(z;)))

nK(zi)))

which gives the result. 1.6. Corollary. Further to the assumptions and notation of Proposition 1.5, assume that p is a prime number such that every p'-number is a unit of R. For any positive integer m, let v ( m ) 2 0 be the largest integer k for which p k divides m and let

xi = m i n ~ ~ , ~ ( ~: < ( z ~ n) K(Z;))

(1 5 i 5 n)

Then the nonzero elements of the set

{ p y Kz;)fll 5 i 5 n} is an R-basis of A$. In particular, if charR = p and K is a p-group, then the elements ( K z ; ) +with K ( z ; )5 kHk-' for some k E I(, form an R-basis of AE. Proof. For any positive integer m, we have mR = p"(")R. The desired conclusion is therefore a consequence of Proposition 1.5.

We next demonstrate that permutation modules hold much information about the p-modular representations of G.

1.7. Proposition. For any RG-module V and any subgroup H of G, H o m R G ( ( l H ) ' , V ) ~ o r n R H ( l HV, H ) Inv(VH) as R-modules.

Proof. The first isomorphism is a particular case of Proposition , have f(1) E lnv(V~). 2.2.4(i). Given f E HomRH(lH,V H )weobviously Furthermore, the map

Permutation modules

390

is clearly R-linear and injective. Since, for any v E InzzI(VH), the map f ( r ) = r v is an RH-homomorphism of 1~ into VH, the desired assertion follows. rn 1.8. Corollary. Let R be a field of characteristic p > 0, let H be a p-subgroup of G and let &, . . . ,V , be all nonisomorphic irreducible RG-modules. Denote b y n; the multiplicity of F/: as a composition factor of (WG. (i) Each E is a homomorphic image of ( 1 ~ ) ~ . (ii) Each ni 2 1 and & n;dimV, = (G : H). (iii) If R is a splittingfield for RG, then n; = dimRP(K)/lHI where P ( K ) is the projective cover of q, 1 5 i 5 r .

Proof. It is clear that (ii) is a consequence of (i). To prove (i), let V be an irreducible RG-module. If W is an irreducible submodule of V H , then W Inv(V'), since charR = p and H is a pgroup. Hence I ~ v ( V H#) 0 and so, by Proposition 1.7,

Since V is irreducible, this shows that V is a homomorphic image of ( l ~ ) ' ,as desired. To prove (iii), note that by Lemma 6.1.3,

On the other hand, by Corollary 2.2.7,

and so n; = dimRHomRH(P(F/:),1 ~ )Now . P ( K ) is RH-free of rank

Thus n; = tidimRHomRH(RH, 1 ~=)ti as required.

(1 5 i

5 r)

1. Preliminary results

391

Let R be a field of characteristic p > 0 or a complete discrete valuation ring with c h a r R / J ( R ) = p > 0 . Let H be a subgroup of G, let P be a p-subgroup of G and let T be a full set of double coset representatives for ( P ,H ) in G. Then 1.9. Proposition.

a n d each ( ltHt-lnp)P is an absolutely indecomposable FP-module.

Proof. Setting V = l ~we, have t K H t - l n p = l t ~ t - ~ n p ,Et 2'. Now apply Theorem 2.7.1 and Corollary 5.1.12. Given a finite set X on which G acts as a group of permutations, we denote RX the corresponding permutation RG-module. If G acts transitively on X and H is the stabilizer of x E X , then obviously R X E ( 1 ~as) RG-modules. ~ For any subgroup S of G, we put

Cx(S) = {z E X ( s z = x for all

s E

S}

(i) R C x ( S ) is a permutation RNG(S)-module. (ii) If G acts transitively on X and H is the stabilizer o f x E X , then 1.10. Lemma.

where T is a complete set of those double coset representatives t for ( N G ( S )H, ) in G for which S C t H t - l .

Proof.

(i) Given g E NG(S) and x E C x ( S ) ,it suffices to show

that

s(gz) = gx

for all s E S

Since s ( g z ) = g(g-'sg)z = gz, (i) is established. (ii) Let g l , . . . ,gn be a left transversal for H in G containing 2'. We may assume that X = ( g I H , . . . ,g n H } and G acts on X by left multiplication. Then

Permutation modules

392

and, in particular, {tHlt E T } & C x ( S ) . If g E N G ( S ) and t E T , then gtH = t H if and only if g E tHt-' n N G ( S )= N t ~ + ( S ) . Hence N t ~ t - ~is(the S )stabilizer of t H in N G ( S ) . Thus we are left to verify that {tHlt E T } is a complete set of representatives of NG(S)-orbits of

CX(S).

H NG(S)tH for some Assume that giH E Cx(S).Then N G ( S ) ~ ; = t E T . Hence gi = 9th for some g E N G ( S ) ,h f H , and therefore 9iH = 9 ( t H )

Finally, assume that t l , t 2 E T are such that gtlH = t2H for some g E N G ( ~ ) Then . N G ( S ) t l H = N ~ ( s ) t 2 Hand hence tl = t 2 , as required. 1.11. Corollary. (Scott (1973)). Let R be a complete noetherian local ring with c h a r R / J ( R ) = p > 0 , let P be a p-subgroup of G and let X be a finite set on which G acts as a group of permutations. Then the Green correspondence with respect to ( G ,P, N G ( P ) ) gives a multiplicity-preserving bijection between the nonisomorphic indecomposable components of R X with vertex P and the nonisomorphic indecomposable components of RCx(P ) with vertex P .

Proof. It suffices to prove the result when G is transitive on X , since then the result can be applied to each transitive component. Since R X S' ( l H ) G , where H is the stabilizer of x E X in G, the desired conclusion follows by applying Lemma l.lO(ii) (with S = P)and Theorem 5.2.9(ii) (with V = 1 ~ ) . 2.

Hecke algebras

Throughout this section, R denotes an arbitrary commutative ring and H a subgroup of a finite group G. Invoking Lemma 1.3, we shall identify the transitive permutation module ( 1 ~with ) ~the left ideal RG H+ of RG where H+ = &H h E RG. If T is a left transversal for H in G, then

-

{tH+It E T } is an R-basis of RG H + which is permuted transitively by G via left multiplication.

393

2. Hecke algebras

The R-algebra E n d ~ c (1 (~ ) ' ) is called the Hecke algebra associated with the triple (G, R, H ) . Our subsequent investigations will show that, in the special case where R is a field of prime characteristic p, this algebra holds much information about the fusion of p-subgroups of G, as well as information about the p-modular representations. Our aim here is to examine the structure of E n d R G ( ( l H ) G ) . The results presented will play a fundamental role in our future study of permutation modules. Given g 1 , g 2 E G and f E EndR( ( 1H ) ' ) , we define

by ( g 1 f s 2 > ( 4=

sl(f(s24)

(v E

(WG)

Recall that E n d R ( ( 1 ~ is ) ~a)G-algebra with the G-action given by

More generally, end^( ( l H ) G ) can be regarded as an R(H x H)-module via

( h l ,h 2 ) f = hlfh,'

(hl,h2 E H,fE E n d R ( ( l H ) G )

Let T be a left transversal for H in G. For any given x E RG, define Q H ( ~ E ) E ~ ~ R ( ( Iby H)~)

a H ( x ) ( t H + ) = xH+ if t E H if t 6H aH(x)(tH+) = 0 Since {tH+lt E T } is an R-basis of ( l H ) ' , a ~ ( x )is indeed a uniquely determined element of E T Z ~ R ( ( ~ HIt) ~is) clear . that c u is~ an R-linear map.

2.1. Lemma. The R-linear map a H : RG + EndR((1H)') satisfies the following properties: (i) aH(x)(gH+)= xH+ i f g E H and aH(Z)(gH+)= 0 i f 9 $ H for all x E RG. (ii) a&zdi) = a H ( Z ) , ha&)h1 = c r H ( h Z ) f o r all h , hl E H , x E RG.

Permutation modules

394

(iii) CYH is a homomorphism of R ( H x H)-modules. (iv) I f g E G and h E gHg-l, then a ~ ( g=) aH(hg) and so a H ( 9 ) E EndRpmgHg-l,((lH)G) for all 9 E G

Proof. (i) This is a direct consequence of the definition of CYH. (ii) The first equality is a direct consequence of the definition and the fact that z h H + = z H + for all h E H , z E RG. To prove the second, it suffices to show that for all g E G, 2 E RG, h , hl E H ,

If g $ H , then both sides of (1) are zero. If g E H , then H+ = gH+ = h l ( g H + ) and so

a H ( h x ) ( g H + )= h x H t = h a H ( z ) h l ( g H + ) , as required. (iii) Given

2

E R G , h l , h2 E H , it follows from (ii) that

W ( ( h 1 ,h2)z) = QH(hl&l) = QH(h1Z:) = hlaH(z)h,' = &I, h Z ) W ( Z ) , as required. (iv) Fix g E G and h = ghlg-' for some hl E H . Then, by (ii),

Hence, if h E H n g H g - ' , then by (2) and (ii),

as asserted.

2.2. Lemma.

erties:

The elements a ~ ( g g) ,E G, satisfy the following prop-

2. Hecke algebras

395

(i) .H(g)(H+) = ( H g H ) + . (ii) For all x,y E G, a ~ ( x=) a ~ ( y if) and only if H x H = H y H . (iiz) The set of distinct a H ( g ) 's is an R-basis of E ~ R G ( 1( ~ ) ~ ) .

Proof. (i) Let T be a left transversal for H

H g H = UteJtgH

n gHg-'

in H . Then

(disjoint union)

and therefore

as required.

(ii) Since a ~ ( x E) E ~ R G ( ( I H ) ~ it ) , is uniquely determined by u ~ ( x ) ( H + The ) . desired conclusion now follows by virtue of (i). (iii) Let T be a full set of double coset representatives for ( H ,H ) in G. Then, by (ii), { a ~ ( t ) lEt T } is the set of all distinct a H ( g ) ' s . If

then by (i),

0=

r,(HtH)+

rtaH(t)(Ht)= tCZ'

t€T

which implies that all rt = 0. Thus the set { a ~ ( t ) lEt T } is R-linearly independent. ) ~ claim ) . that Let f be an arbitrary element of E n d ~ c ( ( 1 ~We

which, by (i), will imply f = C t E T r t a H ( t )hence , the result. Let g l , . . . ,gn be a left transversal for H in G. Then f(Ht) can be

396

Permutation modules

uniquely written in the form n

f ( H t ) = C X ; g ; H + for some A; E R i=l

(4)

Hence, to prove (3), it suffices to show that if HgkH = H g j H , then Ak = Aj. so assume that gj = hgkhl for some h , hl E H . Then gjH = hgkH and

By comparing (4)and ( 5 ) , we see that

Ak

= X j , as required. H

Let H be a subgroup of G and let T be a left transversal for H in G. Since ( 1 ~is )a ~ free R-module with basis {tH+It E H } the R-module E n d ~ ( ( 1 has ~ ) ~an) R-basis

given by

f t s ( x H + )= for all

2

if x H = t H if x H # t H

E G.

2.3. Lemma. For any t , s E T and g E G, we have gfts

= fx,.

where X,p E T are defined b y p H = gsH,XH = g t H . G(ft,) = tHt-' n s H s - l . (ii) Proof. (i) This is a direct consequence of the definition of fts and the action of g E G on f t s . (ii) By (i), g f t s = fts if and only if t H = gtH and sH = g s H , i.e. if and only if g E tHt-l n s H s - l , as required. 2.4. Proposition. Let T be a left transversal for H in G , let S T be a complete set of double coset representatives for ( H ,H ) in G andlethETnH.

397

2. Hecke algebras

(i) EndR( ( 1 ~ ) is~a )permutation G-algebra with permutation basis {f&S E T). are all distinct representatives of G-orbits of { f t s l t , s E (ii) { fhsls E T }. (iii) For all s E S , a H ( s ) is the sum of all elements in the G-orbit of f h s (iv) G(fh,) = H n sHs-'.

s}

Proof. Properties (i) and (iv) follow from Lemma 2.3. By Lemma 2.2, (iii) implies (ii). To prove (iii), note that fhs = a ~ ( s )Hence, . by

(4,

G

= T r H n , H ~ - l ( f h s ) = Trg(fhs)(fhs)7 i.e. a H ( s ) is the sum of all elements in the G-orbit of fh,. H aH(S)

2.5. Corollary. Let S be a complete set of double coset representatives for ( H ,H ) in G, let I( be a subgroup of G and, for each s E S , let m, = g.c.d.{(H

n sHs-') : (9ICg-l n H n s H s - ' ) J g E G }

Then the nonzero elements of the set {msaH(s)lS

E

s}

f o i m an R-basis of T r g ( E n d R K ( l H ) G ) . Proof. Apply Propositions 2.4 and 1.5. H 2.6. Corollary. In the notation of Corollary 2.5, the RG-module (iff)G is I(-projective if and only if each m,, s E S , is a unit of R.

Proof.

Apply Theorem 2.10.3 and Corollary 2.5. H

2.7 Corollary. Let a prime p be a nonunit of R and let H be a p-subgroup of G . If I( is any subgroup of G , then ( 1 ~ is )K-projective ~ if and only if H C 9 K g - l for some g E G . Proof. We keep the notation of Corollary 2.5. Since H is a p group, for any s E S , m, is a nonnegative power of p . Hence, by

Permutation modules

398

Corollary 2.6 and the assumption on p , ( 1 ~is )K-projective ~ if and only if each rn, = 1. Since the latter is equivalent to the requirement that H 9Kg-l for some g E G, the result follows. H The ordinary trace map

tr : ~ n d R ( ( 1 H )-, ~ )R restricts to the R-linear map

tr : E

~ ~ R G ( ( I H -+ ) ~R )

which is given by

Indeed, if g E H , then by Lemma 2.2(i), a ~ ( gis) the identity map and therefore t r a H ( g ) = (G : H ) ( = R-rank of ( l H ) G ) . On the other hand, if g 6 H then for all t E G,

.H(9>(tH+) = t ( H 9 H ) + and the coefficient of t H + in aH(g)(tH+)is zero. Thus in this case we indeed have traH(g) = 0. Keeping the above notation, we now record the following result. 2.8. Lemma.

For all z,y E G ,

Proof. Let T be a left transversal for H

n z H z - l in H . Then

H z H = U t E ~ t z H (disjoint union) and therefore

399

2. Hecke algebras

Hence, if X is the coefficient of H + in a ~ ( y ) a ~ ( x ) ( Hthen +),

( H :( H nxHx-')) 0

A={

if if

H x H = Hy-lH HxN#Hy-'H

Since X is also the coefficient of u ~ ( 1in) a s ( y ) a ~ ( xand )

the result follows by virtue of (6). H We denote by I ( H ) the augmentation ideal of R H , i.e. the kernel of the augmentation map aug : RH -+ R given by a u g ( C ~ h h= )

2.9. Lemma.

C X ~(

~ Eh R, h

E H)

The left annihilator of H+ in RG is RG - I ( H ) .

Proof. Let T be a left transversal for H in G. Since

it follows that

RG - I ( H ) = $ t e ~ t- I ( H )

(8) By ( 7 ) , a typical element x of R G can be written uniquely in the form z = C t E T t x t with x t E R H . Since xtH+ = aug(xt)H+,it follows that zl3' = 0 if and only if xt E I ( H ) for all t E T . The desired conclusion is therefore a consequence of (8). From now on, we write

for the natural homomorphism of R-algebras given by ~ H ( Z ) Z= I

m

for all

x E RG,v E ( 1 ~ ) ~

Given x = C xgg E RG, xg E R,g E G, we also put

x* = C x g g - l

(9)

Permutation modules

400

Then x

H

z* is an antiautomorphism of

RG. Finally, we set

NRG(H+)= (x E RGIH+x* E R G - H') It is clear that N R G ( H + ) is a subalgebra of RG. Furthermore, if H then obviously NRG(H+)= RG. Consider the R-linear map U H : NRG(H+)

Q

G,

EndRG((lH)G)

determined by

It is clear that

The basic properties of the map

are recorded in the proposition below, which is extracted from a work of Broue and Robinson (1986). WH

2.10. Proposition. The map W H : NRG(H+)-+ EndRG((1H)') satisfies the following properties: (i) W H is a surjective homomorphism of R-algebras and Ii'erwH =

RG . I(H ) . (ii) R N G ( H )C NRG(H+)and d H ( g ) = aH(g-') f o r allg E N G ( H ) . (iii) CRG(H) NRG(H+)and for g E G, we have WH

( T r g H ( g ) ( g )=) ( ( H n g H g - ' ) : C H ( g ) ) a H ( g - ' )

(I1)

In particular, WH(X)=

T r $ ( a H ( x * ) ) f o r all x E CRG(H)

(iv) W H ( Z ) = ( T H ( z * ) for all z E Z(RG), where (v) If C is a conjugacy class of G, then

UH

(12)

is given b y (9).

Ic n gHlaH(g)

wH((c-')+) = 9

where g runs over a complete set of representatives of ( H ,H)-double cosets of G.

401

2. Hecke algebras

Proof. (i) Given z, y E NRG(H+) and g E G, we have WH(xy)(gH+) = gH+(zy)*= gH+y*z* = wf(+JH(Y)(YH+)

(by (10)) which shows that W H is a homomorphism of R-algebras. For any x E RG', we have H+x* = ( z H + ) *and thus H+x* = 0 if and only if z H + = 0. Hence, by Lemma 2.9, we have KerwH = RG * I ( H ) . Owing to Lemma 2.2(iii), to prove surjectivity it suffices to exhibit for any given g E G an element of NRG(H+)whose image is u ~ ( g ) To . this end, choose a left transversal tl, . . . ,t , for H n gHg-' in H , and a right transversal ul, . . . ,u, for H n gHg-' in H . Setting

X one readily verifies that

= {tig-luill

5i

n}

Permutation modules

402

proving (12) for x = Tr,!&(ST')(g-l). Since, by Corollary 1.2, the set of all such x forms an R-basis of C R G ( H ) ,(12) must hold for all 2 E

CRG(H ) .

(iv) If z E Z ( R G ) and v E ( l ~ ) ' ,it follows from (10) that WH(Z)V

= vz* = z*v

and thus w H ( z ) = o~(z*). (v) To find the coefficient of aH(g) in w ~ ( ( C - l ) + ) note , that

where X(C,g) is the coefficient of g in the element H+C+ (expressed as a linear combination of the natural basis elements of RG). Hence

and the result follows. H

(Broue' and Robinson (1986)). Assume that R is an arbitrary commutative ring of prime characteristic p and H is a p-subgroup of G . If C is an H-conjugacy class of G and WH(C+)# 0, then for all g E C , C H ( g ) = H n gH9-l 2.11. Corollary.

and WH(C+)

Proof. If g

= aH(s-')

c C , then C+ = Tr&(,)(g)

and therefore, by (ll),

Since H is a p-group and charR = p , it follows that HngHg-l = C H ( g ) , as required. H

403

3. Fusion and permutation modules

3.

Fusion and permutation modules

In this section, F is an arbitrary field of prime characteristic p and G is a finite group. Our aim is to demonstrate that permutation modules ( l p ) ' , P a p-subgroup of G, hold much information about the fusion of p-subgroups of G, as well as information on FG-modules. For convenience of reference, we first recall some notation of the previous section. Let H be a subgroup of G. Then

denotes the natural homomorphism of F-algebras given by a H ( x ) = zu

for all

z E RG,u E ( 1 ~ ) '

Recall, from Proposition 2.10, that the map

defined by

is a surjective homomorphism of F-algebras and KerwH = F G - I ( H )

(by Lemma 2.9, Ii'erwH is the left annihilator of H + in FG). 3.1. Lemma. For anyp-subgroup P ofG, (i) K e r a p C_ J ( F G ) . (iz) Ii'erwp n C F G ( P )C J(CFG(P)).

Proof. (i) If z E Ii'erap, then x annihilates ( l p ) ' . Hence, by Corollary 1.8(i), x annihilates every irreducible FG-module. Thus x E J ( F G ) , proving (i). (ii) It suffices to verify that ICerwp n C F G ( P contains ) no nonzero idempotents. Our assumptions on F and P ensure that FP is local, hence any projective FP-module is free. Assume that e is a nonzero

Permutation modules

404

idempotent of CFG(P).Then eFG # 0 is a left projective (hence free) FP-module. Hence P+eFG is not zero : its dimension is the FP-rank of eFG. Thus eP+ = P+e # 0 and so e # KerwH. W 3.2. Corollary. (Broue' and Robinson (1986)). Let P be a normal p-subgroup of G. Then

C F G ( P ) / ( ~ (n~C~F~GP( P ) ) F ( C c ( P ) / Z ( P ) ) as F-algebras. In particular, C F G ( P )is local if and only if C G ( P )is a p -9ro up. Proof. If C is a P-conjugacy class of G and g E C, then by Proposition 2.10( ii), (iii),

Hence WP(CFG(P)) = W P ( F C G ( P ) )Since . the map 49-7

-+

9Z(P),

9 E CG(P),

induces an F-algebra isomorphism

the first assertion is established. The second assertion follows from the first and Lemma 3.1(ii). Following Brou6 and Robinson (1986), we say that the subgroup

H of G controls the G-fusion of its p-subgroups if whenever P is a p-subgroup of H and g is an element of G such that 9-%7

c H,

then 9 E CG(P)H

(Broue' and Robinson (1986). Let H be an arbitrary subgroup of G. Then

3.3. Theorem.

WH(CFG(H))= EndFG((1H)')

(1)

3. Fusion and permutation modules

405

if and only if H controls the G-fusion of its p-subgroups.

Proof. It follows from Proposition 2.lO(iii) and Lemma 2.2(ii), (iii) that the equality (1) holds if and only i f (a) For any g E G, there exists g1 E HgH such that ((Hng1Hg;') : C H ( ~ is~ not ) ) divisible by p. But if gl = h'gh for h, h' E H , then

and thus (a) is equivalent to: (b) For any g E G, there exists h E H such that ( ( H n 9Hg-l) : C ~ ( 9 h - l )is) not divisible by p. We now claim that (b) is equivalent to: ( c ) For any g E G and any Sylow psubgroup P of H ngHg-l, there exists h E H such that P CH(gh-l). It is clear that (c) implies (b). Conversely, assume that (b) holds. Then, given g E G and a Sylow p-subgroup P of H n gHg-', there exists h E H such that CH(gh-') contains hlPh,' where hl E H n 9Hg-l. If hl = gh29-l with h2 E H , then

which proves (c). By the foregoing, we are left to verify that ( c ) holds if and only if H controls the G-fusion of its p-subgroups. First assume that (c) holds. Let P be a psubgroup of H and let g-'Pg C H for some g E G. Then P C H n gHg-'. Denote by Q a Sylow p-subgroup of H n gHg-' containing P. By (c), we may find h E H such that Q 2 C~(9h-l) Setting z = gh-', we then have z E Cc(Q). Hence z E CG(P) and g = z h proving that H controls the G-fusion of its p-subgroups. Conversely, assume that H controls the G-fusion of its p-subgroups. Let g E G and let P be a Sylow p-subgroup of H n gHg-'. Since g-'Pg G H , we then have g = z h for some z E C c ( P ) , hE H . Since z = gh-', we conclude that P C CH(gh-'), proving (c) and hence the result. H

Permutation modules

406

3.4. Corollary. Let P be a p-subgroup of G which controls the

G-fusion of its subgroups. Then ( 1 ~is)indecomposable ~ if and only if CFG(P)is local.

Proof. This is a direct consequence of Theorem 3.3 and Lemma 3.l(ii). We say that a group G is p-nilpotent if G has a normal p'-subgroup N , called the normal p-complement, such that GIN is a p-group. Thus G is p-nilpotent if and only if G = N S where N is a normal p'-subgroup of G and S is a Sylow p-subgroup of G. We close by providing a proof, due to Robinson (1988), of an extension of a classical result, known as Frobenius normal p-complement theorem. This extension provides a characterization of p-nilpotent groups by the surjectivity of the map US

: CFG(S)

EndFG((lS)G).

3.5. Theorem. Let S be a Sylow p-subgroup of G. Then the following conditions are equivalent: (i) Whenever two elements o f S are conjugate in G, they are conjugate in s. fii) Whenever Q is a p-subgroup of G, NG(Q)/CG(Q)is a p-group. (iii) Whenever Q is a subgroup o f G with Q, g-lQg S, then g = cs where c E C c ( Q ) and s E S. (iv) US(CFG(S))= E n d F G ( ( l s ) G ) . (v) G is p-nilpotent.

Proof. (i) + (ii): Assume by way of contradiction that (i) holds, but that (ii) does not. Then we may find a subgroup Q of S of minimal order subject to Nc(Q)/CG(Q)is not a p-group. Let z E N G ( & ) - CG(Q) be a p'-element. Then, by (i), we must have

Because S is nilpotent, [Q,S]2 Q, so

407

3. F'usion and permutation modules

Due to our choice of Q , NG([Q,z ] ) / c ~ ( [ 21) & , is a pgroup and thus [ Q , s , x ]= 1. Hence, by Theorem 5.3.6 in Gorenstein (1968), we must have [ Q , x ]= 1, a contradiction. (ii) +(iii): We argue by induction on ( S : Q). If Q = S , then N c ( Q ) = QCG(Q)and (iii) holds. Now assume that Q c S and that (iii) holds for subgroups R of S such that ( S : R ) < (S : Q). Choose g E G with gQg-' c S and put U = N s ( Q ) . Then U 3 Q. Let T be a Sylow p-subgroup of NG(Q) with U C T and let V be a Sylow p-subgroup of G with T V . Since z - l V z = S for some x E G, we have U,z-'Uz S,and by the induction hypothesis, x = cs for some c E CG(U),SE S. Then z-'Qz = s-'Qs and Ns(z-lQz) 2 z-lTz, so Ns(s-lQs) is a Sylow p-subgroup of NG(s-'&s) and N s ( Q ) is a Sylow p-subgroup of NG(Q).Similarly, Ns(gQg-l) is a Sylow psubgroup of NG(gQ9-l). Then gNS(g-'Qg)g-l is a Sylow p-subgroup of NG(Q),so

c

S , so by the for some h E NG(Q). Thus Ns(Q),g-lh-'Ns(Q)hg induction hypothesis, hg = clsl for some c1 E C G ( N S ( Q ) ) , SE~ S. Then

gs-' E NG(Q)= CG(Q)NS(Q) since N s ( Q ) is a Sylow p-subgroup of N G ( Q ) and NG(Q)/CG(Q)is a p-group. Thus g E C G ( Q ) S ,proving (iii). (iii) +(iv): Let T be a complete set of double coset representatives for (S,S) in G. By Lemma 2.2, { a s ( t ) l t E 2') is an F-basis of E n d F G ( ( l s ) G ) . Let us first show that for any z E G, there exists c E SzS such that n CSC-* = c,(c) (2)

s

Let Q = S n x5'z-l. Then Q, s-lQx some c E C c ( Q ) , u E S. Then

Sncsc-l

S , so we may write x = cu for

=

sn~ss-l= Q

c

cs(c)

S~CSC-~,

proving (1). Let C be the S-conjugacy class of c, and let X be a left transversal

Permutation modules

408

for S n cSc-l in S. Then

c+s+= c scs+ = (SCS)+ SEX

and therefore a&')

= ws(C+),

proving (iv), by applying (1). (iv) j (v): We first note, that by Lemma 1.3(v), 1~ is a direct summand of ( 1 ~ which ) ~ corresponds to the idempotent, say e, of E n d ~ G ( ( l s )with ~ ) e(S+) = ( G : S)-'G+. By hypothesis, there is an idempotent f of C F G ( S )with w s ( f ) = e. Then

and therefore

S+f = fS+ = ( G : S)-'G+

Now f E CFG(S)and f F G is a direct summand of FG as an FSmodule. Hence f F G is a projective left FS-module, so is free (since S is ap-group). Because S + f F G = FG+, it follows that d i r n ~ ( f F G= ) IS[. Similarly dirnF(FGf) = ISI. On the other hand,

as a left FS-module. Thus f F G f is a projective left FS-module, and d i r n ~ ( f F G f2) ISl. Hence f F G f = f F G = FGf and

proving that f E Z(FG). Since dirnF(FGf) = IS1 and any direct summand of FG f is projective, FGf is an indecomposable left FG-module. Hence f is a primitive idempotent of FG. Let x be a p'-element of G of order n, let X =< 2 > and let e, = n-'X+. Then e: = ex and e,fG+ = exG+ = G+, so e,f # 0. Because f is primitive in FG with e,f = fe,, we have e,f = f . But f E Z ( F G ) and xe, = e, so that xu = u for all u E FGf. Because FGf is a free left FS-module, no nontrivial p-element of G acts as the identity on FGf and thus G has a normal pcomplement,

4. Complete reducibility of ( 1 ~ ) ~

409

as required.

(v) + (i): Let N be a normal p-complement of G. Assume that x,g-lxg E S for some g E G. Since G = S N , we have g = s u for some s E S,u E N . Then z - l g -1 xg = 5 - ~ ( s - ~ 5 s ) ( s - ~ 5 s ) - ~ u - ~ ( s - 1 5 s ) u and ( ~ - ~ x s ) - ~ u - ~ ( s - ~E xSs n ) uN = 1 Thus g - l s g = S-'ZS

and the result follows.

Corollary. Let S be a Sylow p-subgroup of a p-nilpotent group G. Then 3.6.

CFG(S)/(CFG(S) n F G - I ( S ) )2 E ~ ~ F G ( ( ~ s ) ~ ) Proof. Direct consequence of Theorem 3.5 and Proposition 2.10. Complete reducibility of ( 1 ~ ) ~

4.

Let F be a field of characteristic p > 0 and let P be a psubgroup of a finite group G. By Corollary 1.8, each irreducible FG-module is a homomorphic image of (lp)'. It is therefore appropriate to investigate is completely reducible. circumstances under which ( 4.1. Proposition. The following conditions are equivalent: (i) ( 1 ~ is) completely ~ reducible. (ii) J ( F G ) FG - I ( P ) , where I ( P ) is the augmentation ideal of

FP.

Proof. If V = F P / I ( P ) , then V '2.2.11 (i),

!Z l p

and, by Proposition

VG FGfFG * I ( P ) Hence

V G is completely reducible if and only if J ( F G ) 5 F G . I ( P ) ,as

required. Our next observation shows that complete reducibility of (lp)' depends only on the characteristic p of the field F .

Permutation modules

410

Let E be a field extension of F. Then ( 1 ~ p ) ~ is completely reducible i f and only if ( 1 . g ~ is ) ~completely reducible. 4.2. Proposition.

Proof. By Proposition 4.1, it suffices to show that J(FG) E F G I ( P ) if and only if J(EG) EG I ( P ) . We know, from Corollary 6.1.10, that J(EG) = E @F J(FG) Hence, if J(FG) C F G - I ( P ) , then J(EG) C EG . I ( P ) . Conversely, assume that J(EG) E EG - I ( P ) . Then, by Lemma 2.9,

J ( F G ) 5 EG * I ( P )n F G = FG * I ( P ) as required.

Let U and V be finitely generated FG-modules. Recall that the intertwinning number for U and V , written i ( U , V ) is defined to be

i(U,V ) = dirnFHomFc(U,V ) It is clear that

Let { K, . . . ,V,} be a full set of nonisomorphic irreducible FG-modules. Given an FG-module V , we write

k=l

if

vk

appears ak times as a composition factor of V .

4.3. Lemma. Let V x c;=,akVk and W M C;=,b k v k be two FG-modules, where F is assumed to be a splittingfield for FG. Then (i) i(V,W ) 5 EL=,akbk and, in particular, z(V,V ) 5 a:. (ii) V is completely reducible i f and only i f i(V,V ) = a:. Proof. (i) We argue by induction on the value of C;=,ak. If C;=,ak = 1, then V is irreducible and so V % V, for some j E

4. Complete reducibility of ( 1 ~ ) ~

411

{ 1,. . . ,r } . Therefore

and so, by (l),

Now assume that Vo is a submodule of V . By looking at the exact

sequence

we deduce that

If Vo is a nontrivial submodule of V with

then by the induction hypothesis

and

c T

i(V0,W)I

(4)

k=l

as required. (ii) If V is completely reducible, then by (1) i(V,V ) =

C;=,

Permutation modules

412

since z(vk,vk) = 1 for each k E { 1,. . . ,r } . Conversely, assume that a: and write i ( V ,V ) = f

T

VIsoCv M k=l

ikv,,s o c v M

alvk

k=l

Then, by (i), we have

k=l

and

i ( s o c v , v)5

c T

a$ak

k=1

Hence, applying (2), we derive

and i(S0C v,v )=

c T

aiak

k=l

Bearing in mind that r

i ( S o c V , V )= i ( s o c v , s o c v ) = c ( a ; ) z k= 1 it follows from (6) that a” - ak if a! # 0. kll Finally, assume that ak = 0 and ak # 0. Then ak = a; # 0 and so, by ( 5 ) , i ( V / S o c V , V ) # 0. Hence there exists a submodule L with SOCV G L G V such that V / L is isomorphic to a submodule of V . The latter implies the existence of j E (1,. . . , r } with us # 0 and a; # 0.

413

4. Complete reducibility of ( 1 ~ ) ~

But if a; # 0, then aj = a; and so a; = 0, a contradiction. Hence a k = a; for all k E { 1,. . . , r } and thus V = Soc V as required. W

(Sabonov (1971)). Let F be a splitting field for FG, let { &, . . . ,V,.} be a full set of nonisomorphic irreducible FGmodules and, for each k E (1,. . . ,r } ) let U k = dimFP(Vk). If d p is the number of double ( P ,P)-cosets of G, then 4.4. Corollary.

T

c ( u k / I p 1 ) 22

dP

k=l

with equality if and only if (lp)' is completely reducible. Proof. By Corollary l.g(iii), we have (IP)'

k(uk/lPl)vk

k=

1

On the other hand, by Lemma 2.2, i ( ( l p ) G ,( 1 ~ )=~d p) . The desired conclusion is therefore a consequence of Lemma 4.3. W 4.5. Lemma. Let A be a finite-dimensional algebra over a field

and let V be a finitely generated A-module. Then

c

I = (9 E EndA(V)lY(V) J ( A ) V } is a nilpotent ideal of EndA(V).

Proof. It is clear that I is a subspace of the finite-dimensional

7

Permutation modules

414

and so cp" = 0. Thus I is nilpotent, by virtue of Theorem 1.5.12. H 4.6. Proposition.

For any p-subgroup P of G, the following con-

ditions are equivalent: (i) ( is completely reducible. (ii) EndFG( ( 1 ~ )is~semisimple. ) (iii) N&P+)/FG I ( P ) is semisimple. Proof. The equivalence of (ii) and (iii) is a consequence of Proposition 2.10. Since (i) obviously implies (ii), we are left to verify that (ii) implies (i). Put V = ( 1 ~and ) ~assume that J ( F G ) V # 0. Then J ( F G ) V has an irreducible submodule W . By Corollary 1.8(i), there is a surjective FG-homomorphism cp : V + W

J(FG)V

Thus cp is a nonzero element of EndFG(V) such that cp(V) C J ( F G ) V . Hence, by Lemma 4.5, Endf;.G(V)is not semisimple, as required. H 4.7. Theorem. (Saksonov (1971)). Let P be a p-subgroup of G, let r be the number of p-regular classes of G and let d p be the number of double ( P ,P)-cosets in G. Then d p 2 r and, under the assumption that F is a splitting field of FG, the following conditions are equivalent:

(i) d p = r. (ii) ( 1 ~ ) V, ~@ V, @ - - @ V,, where {h, . . . ,V , } is a full set of nonisomorphic irreducible FG-modules. (iii) P is a Sylow p-subgroup of G, ( 1 ~ is) completely ~ reducible and dimFP(K) = IPl for all i E {l,.. . , r ] . Proof. Let F be an algebraically closed field of characteristic p . Then, by Theorem 3.6.13, there exist exactly r , say V,,. . . , V , of nonisomorphic irreducible FG-modules. Put V = ( 1 ~and ) ~write V / J ( F G ) V E $;=,akVk, Soc V Then we have

E $;=lbkVk

(7)

7-

i ( V / J ( F G ) VSOC , V )=

C akbk

k=l

(8)

4. Complete reducibility of ( 1 ~ ) ~

415

where each ak 2 1, by Corollary 1.8(i). Note also that, by Corollary 2.2.7, bk. = q k . , SOCV) = i(Vk,V ) = i ( ( & ) p , lp) # 0 and hence

Ckzla k b k 2 r .

Since

and, by Lemma 2.2, d i m F E n d F G ( V ) = d p , it follows from (8) that dp

2 T.

In what follows, we assume that F is a splitting field for FG and prove the equivalence of (i), (ii) and (iii). (i) j (ii): Assume that d p = r . Then, by the above,

and ak

for all k E {1,...,r }

= bk = 1

(9)

Hence, by (7), V / J (FG)V 2 Soc V and so E n d ~ cV( ) 2 E n d ~ cSoc ( V). But Soc V is completely reducible, hence E n d ~ c ( Vis>semisimple and therefore, by Proposition 4.6, V is completely reducible. Thus V = SOCV and the required assertion follows by virtue of (7) and (9). (ii) +(iii): Let Q be a Sylow p-subgroup of G containing P. Then, by Corollary 1.8, Ci==ldimV, 5 (G : Q ) . On the other hand, by hypothesis, P

C d i m V , = ( G : P ) 2 ( G :Q ) i=l

Hence Q = P and so P is a Sylow psubgroup of G. The remaining assertion follows from Corollary 1.8(iii). (iii) +(i): If u k = d i m ~ P ( V k )then , (uk/lPI) = 1 for all Ic E { 1, . . . ,r } . Since ( l p ) G is completely reducible, it follows from Corollary 4.4 that T

dP

= c(%/l= p f ,I )2 k=l

as required.

Permutation modules

416

5.

Induction f r o m Sylow p-subgroups

Throughout F denotes a field of characteristic p > 0 and Syl,(G) the set of all Sylow psubgroups of G. Given P E Syl,(G), our aim is to examine circumstances under which ( l p ) G is completely reducible. Following Motose and Ninomiya (1975), we say that G is p-radical if (lp)' is completely reducible. Note that, by Proposition 4.2, the notion of a p-radical group depends only upon the characteristic p of a field F (and not on the field F itself). For convenience, we divide this section into two subsections.

A. General results In what follows, given a subgroup H of G , we write I ( H ) for the augmentation ideal of F H . If X is a subset of G, then X + E F G is defined to be the sum of all elements in X . Our aim is to establish some general results on p-radical groups and provide their various characterizations.

5.1. Lemma. Let H be a subgroup of G, let I be a nilpotent lefl ideal of FH and, for each g E G, let 1 9 = g-'Ig. Then J = ngEGFG-Ig is a nilpotent ideal of FG. Proof.

Given x E G,we have

proving that J is an ideal of FG. It follows that for any integer rn 2 1, Jmt'C - Jm(FG.I)= Jm-I

and hence, by induction,

417

5 . Induction from Sylow psubgroups

for any integer rn 2 0. Because I is nilpotent, we conclude that J is a nilpotent ideal of FG. H

5.2. Corollary.

Let P be a Sylow p-subgroup o f G . Then

is a nilpotent ideal o f F G .

Proof. Note that, for any given g E G , I(P)g = I(P9). Since I ( P ) is a nilpotent ideal of F P , the result follows by virtue of Lemma 5.1. H

We are now ready to provide the following characterizations of pradical groups contained in the work of Motose and Ninomiya (1975). 5.3. Theorem.

The following conditions are equivalent:

(i) G is p-radical. (ii) For any subgroup H ofp'-index in G , J ( F G ) FG - J ( F H ) . (iii) J ( F G ) C FG . I ( P ) for some (and hence all) P E SyZ,(G). (iv) J ( F G ) = nSESyl,(G)FG ' I ( s ) * Proof. The equivalence of (i) and (iii) is a consequence of Proposition 4.1. Since (ii) obviously implies (iii), we are left to verify that (iii) =+(iv) and (iv) =+(ii). (iii) +(iv): If J ( F G ) E FG . I ( P ) for some P E Syl,(G), then

J ( F G ) = J ( F G ) g FG - I ( P ) g= FG - I ( P g ) for all g E G and therefore

This proves (iv), by applying Proposition 5.2. (iv) + (ii): Let H be a subgroup of G of p'-index. Then there exists S E Syl,(G) such that S H . By hypothesis, J ( F G ) F G I ( S ) , so V G is completely reducible for any irreducible FS-module V , by Theorem 2.3.7. Since V c E (VH)",it follows from Corollary 2.2.10

-

Permutation modules

418

that V H is completely reducible, for every irreducible FS-module V . Again, by Theorem 2.3.7, J ( F H ) 2 F H . I ( S ) and therefore, by the implication (iii)j (iv),

Thus

as required. 5.4. Corollary. Let H be a subgroup of G of p'-index.

If G is

p-radical, then so is H . Proof. Let S be a Sylow p-subgroup of G with S H . Then, by the proof of the implication (iv) =+ (ii) of Theorem 5.3,

J ( F G ) C FG * I ( S ) j J ( F H )

c FH

*

I(S)

Hence, by Theorem 5.3, if G is pradical, then so is H . The next result, with the exception of the first property of (i), is due to Khatri (1973). 5.5. Theorem. Let N be a normal subgroup of G. (i) If G is p-radical, then so are N and G I N . (ii) If N is a p-group and GIN is p-radical, then G is p-radical. (iii) If GIN is a p'-group, then G is p-radical if and only if N is

p-radical. Proof. (i) Suppose that G is p-radical and let P E Syl,(G). We may choose a Sylow p-subgroup Q of N with Q C P. By Corollary 2.4.3, J ( F N ) 5 J ( F G ) and so, by Theorem 5.3,

J(FN)

c F N n FG - I ( P )

5 . Induction from Sylow psubgroups

I f x E FN

419

n F G . I ( P ) ,then by Lemma 2.9,

But x Q + and x ( P - Q)+ have disjoint supports, so z&+ = 0 and thus, by Lemma 2.9, z E F N I ( & ) . Hence J ( F N ) F N . I ( & ) and so, by Theorem 5.3, N is pradical. I f H = P N , then H is of $-index in G, so by Theorem 5.3,

-

c

J ( F G ) C FG * J ( F H ) Hence, by Theorem 2.3.7, V Gis completely reducible for all irreducible FH-modules V . In particular, ( 1 ~is )completely ~ reducible. Hence ( ~ H I N is) completely ~ ~ ~ reducible. But HIN is a Sylow p-subgroup of G I N , so GIN is pradical. (ii) Assume that N is a normal psubgroup of G such that GIN is pradical. Since N C P E Syl,(G), we have HIN = P I N . By hypothesis, ( l p / ~ ) is~ completely / ~ reducible. Thus ( 1 ~is )completely ~ reducible and G is p-radical. (iii) Assume that GIN is a #-group. If G is p-radical, then so is N , by virtue of Corollary 5.4. Conversely, suppose that N is p-radical. Since GIN is a p'-group there is a P E Syl,(G) with P 5 N . By Theorem 5.3, J ( F N ) F N - I ( P ) . On the other hand, by Theorem 2.10.14, J ( F G ) = FG J ( F N ) . Hence

J(FG)

FG * F N - I ( P ) = FG - I ( P )

and so G is pradical, by applying Theorem 5.3.

(Khatri (1973)). If G = G1 x Gz, then G is p-radical if and only if so are G1 and G2. 5.6. Theorem.

Proof. Suppose that GI and G2 are pradical. Then, by Theorem

5.3,

J(FGi)

FGi . I(Pj)

where P; E SyZ,(Gi),i = 1,2. As is well known (see Motose (1974a)),

Permutation modules

420

and so

+

J ( F G ) FG - I ( Pi) FG - I ( P2) If we put P = Plx P2,then I(Pi) C P and so by the above J ( F G ) C FG * I ( P ) Consequently, G is p-radical, by Theorem 5.3. The converse being a consequence of Theorem 5.5(i), the result follows.

If X is a subset of G, we write r ( X ) = ~ G ( Xand ) l ( X ) = IG(X) for the right and left annihilators of X in FG, respectively. 5.7. Lemma. Let H be a subgroup of G , let I be a left ideal of FH and let J = flgEGFG- 19. Then (i) J = IG{&GrH(I)’FG}. (ii) ~ G ( J=) CgEG r~(1)gFG. (iii) If I is an ideal of F H , then J = rlgEGIgFG.

Proof. (i) By Theorem 3.10.11, I = ZH(TH(I))which implies that

FG * I = ZG(TH(I))FG since FG is a free FH-module. Hence, if g E G , then

It follows that J

= ngEG1G({rH(I)}’FG)

as required. (ii) Applying (i) and Theorem 3.10.11, we have rG(J)

=

fG{lG(

rH(1)’FG))

’a

= xr~(l)’FG, ’€G

5 . ,Induction from Sylow p-subgroups

421

as asserted.

(iii) By Theorem 3.10.11, the left and right annihilators of an ideal in FG (or F H ) coincide. Applying (i), we therefore derive

n,EGIgFG, as desired. W 5.8. Lemma. Let H be a subgroup of G. Then for all g E (2) F G * I ( H )= lc(H+)= {C zgg E FGI C h E H z g h = 0

GI. (ii) I ( H ) - F G= r c ( H + )= { C g E G zgg E FGI

C h E Hz h g

=0

for all g E

GI. Proof. The first equality of (i) is a special case of Lemma 2.9. The second equality is a consequence of the fact that ( C z g g ) H += 0 if and only if ChEH X g h = 0 for all g E G. The proof of (ii) is similar. 5.9. Lemma. Let P be a Sylow p-subgroup of G. Then

ngE~FG * I(Pg) =

x x { axzl

xEG

=

{C

axy = 0 for all

z

E G , S E Syl,(G)

YES

aXzl

xEG

C ayx= 0

for all z E G , S E Syl,(G)

YES

Proof. Put X = ng,GFG- I ( P g )and Y = ngEGI(Pg)FG.Then, by Lemma 5.7(iii),

x =y =

nS€SyZ,(G)FG

'

I(s)= n s ~ S y t , ( G I) ( S ) * FG

and so the desired conclusion follows by virtue of Lemma 5.8.

Permutation modules

422

We are now ready to provide some further characterizations of pradical groups.

Proof. The equivalence of (i),(ii),(v) and (vi) follows from Lemmas 5.9 and 5.7(iii), and Theorem 5.3. That (ii) is equivalent to (iii) follows by taking annihilators. Similarly, since (ii) is equivalent to J ( F G ) = nSESylp(G)FGI ( S ) ,it follows that (ii) is equivalent to (iv) by taking annihilators. W

B. S o m e classes of p-radical groups.

A transitive permutation group G in which only the identity element fixes more than one letter, but the subgroup fixing a letter is nontrivial, is called a Frobenius group. In what follows we shall use the following standard facts for the proof of which we refer to Gorenstein (1968). (a) (Frobenius). Let G be a Frobenius group and let H be the subgroup fixing a letter. Then the subset of G consisting of the identity element together with those elements which fix no letters, forms a normal subgroup N of G of order (G : H ) . (b) (Thompson). Let a group G admit an automorphism of prime order which fixes only the identity element of G. Then G is nilpotent. We shall refer to the normal subgroup N and the subgroup H of G in (a) as the Frobenius kernel and Frobenius Complement, respectively. 5.11. L e m m a . Let G be a Frobenius group with complement H and kernel N . Then

5 . Induction from Sylow p-subgroups

423

(i) G = N H with N n H = 1. (ii) Every h # 1 in H induces b y conjugation an automorphism of N which fixes only the identity element of N . (iii) I HI divides IN I - 1. (iv) N is nilpotent. (9) If g E G - N , then g lies in a conjugate of H . (vi) For every x # 1 in N , C G ( ~ ) N . Proof. (i) It follows from the definition of N that N n H = 1. Since, by (a), IN1 = (G : H ) , we must have IGI = (NHI. Hence G = N H , as asserted. (ii) Because G is transitive and H is the subgroup fixing a letter, our representation is equivalent to that on the cosets of H . Owing to (i), we can take the elements of N themselves as a transversa1 for H in G. Assume that h-lnh = n for some 1 # h E H and 1 # n E N . Then it is immediate that h fixes the coset H n as well as the coset H . But, by definition of a Frobenius group, only the identity fixes more than one letter. Thus h-lnh # n for any 1 # h f H and 1 # n f N , proving (ii). (iii) For a fixed 1 # n E N , the set rn = {h-lnhlh E H } must consist of m = (HI distinct elements of N . But clearly for x,y in N = or n = 0. Thus with x # l , y # 1, we have either IN - {1}1 is a multiple of m and (iii) follows. (iv) Because H # 1, we may find an element h E H of prime order. By (ii), h induces an automorphism of N which fixes only the identity element of N . Thus, by (b), N must be nilpotent. (v) If g E G - N , then g fixes a letter. But the conjugates of H are simply the subgroups fixing a letter. Thus g lies in a conjugate of H . (vi) Owing to (v), N consists precisely of the elements of G which lie in no conjugate of H - (1). Thus CG(x) N for 1 # x E N , as asserted.

ry rX rY

5.12. Lemma. Let F be a field of characteristic p > 0 , let N be a normal p'-subgroup of G and let e = 1NI-l EXEN x. Then e is a central idempotent of F G such that FGe F ( G / N ) .

Permutation modules

424

Proof. It is clear that e is a central idempotent of FG. Let f : FG

+ F(G/N)

-

be the natural homomorphism. Since K e r f = F G I ( N ) and F G = FGe @ F G ( l - e ) , it suffices to verify that FG I ( N ) = F G ( l - e). Because f ( e ) = 1 , we have 1 - e E K e r f and so F G ( l - e ) F G . I ( N ) . Given n E N , we also have ne = e so that ( n - l ) e = 0 and F G . I ( N ) e = 0. Thus F G . I ( N ) F G ( l - e ) and the result is established.

-

s

5.13. Lemma. Let F be a field of characteristic p > 0, let N be a normal p'-subgroup of G and let e = IN[-' CnEN n. If H is a subgroup of G such that G = N H and N n H = 1 , then (i) J ( FG)e = J ( F H ) e . (ii) dimf;.J(FH)e = dim^ J ( F H ) .

Proof. By Lemma 3.18.1 J ( F G e ) = J ( F G ) e , while by Lemma 5.12, FGe 2 F ( G / N ) 2 F H . The latter implies that

{

F H + FGe x H x e

is an isomorphism of F-algebras. Hence J ( F G e ) = J ( F H ) e and

J(FG)e = J(FGe) = J(FH)e, J ( F H )% J(FH)e proving (i) and (ii).

(Motose (1974b)). Let G be a Frobenius group with complement H and kernel N and let F be a f i e l d of characteristic p > 0 . (i) I f p divides IN1 and P is a Sylow p-subgroup of N (and hence of G , by Lemma 5.11(iii,)), then J ( F G ) = F G - I ( P ) (ii) If p does not divide INJ, then 5.14.

Proposition.

J ( F G ) = J ( F H ) e and d i m F J ( F G ) = dirnFJ(FH) where

5 . Induction from Sylow p-subgroups

425

Proof. (i) Suppose that p divides IN1 and let P be a Sylow psubgroup of N . We know, from Lemma 5 . 1 1 ( i ~ that ) ~ N is nilpotent. Thus P d N, so P d G and therefore J ( F G ) = F G I ( P ) . (ii) Assume that p does not divide IN[. Owing to Corollary 6.1.10, we may assume that F is algebraically closed. Note that F N = F N e @ F N ( 1 - e) and FNe 2 1 ~ Hence . F N ( l - e) is a direct sum of nontrivial irreducible FN-modules. Therefore by Theorem 2.8.8(i) and Lemma 5 . 1 1 ( ~ i )F~G ( l - e) E ( F N ( 1 - e))G is completely reducible. Thus J ( F G ) ( l - e) = 0 and so J ( F G ) = J(FGe). Now apply Lemma 5.13. H Turning to p-radical groups, we now prove our first major result. 5.15. Theorem. (Khatri 19?'3), Motose and Ninomiya (1975)). Let G be a Frobenius group with kernel N and complement H and let F be a field of characteristic p > 0. (i) If p divides the order of N , then G is p-radical. (ii) If p divides the order of H, then G is p-radical if and only if so is H.

Proof. (i) By Lemma 5.11(iv), N is nilpotent and so, by Theorem 5.3(iii), N is p-radical. But, by Lemma 5.ll(iii), G/N S H is a p'-group. Hence, by Theorem 5.5(iii) G is p-radical. (ii) Assume that H is pradical. By Theorem 5.5(i), it suffices to show that G is p-radical. If P is a Sylow p-subgroup of H, then J ( F H ) F H . I ( P ) by Theorem 5.3. By Proposition 5.14(ii), J ( F G ) = J ( FH)Nt and therefore J(FG)

FH * I(P)(N+)C F G . I ( P )

Since P is a Sylow p-subgroup of G, it follows from Theorem 5.3 that G is pradical, as required. H The next result provides another class of p-radical groups. 5.16. Theorem. (Tsushima (1986)). Let G be a p-nilpotent group with abelian p-complement N. Then G is p-radical.

Permutation modules

426

Proof. By Proposition 4.2,we may harmlessly assume that F is algebraically closed. Let P be a Sylow psubgroup of G and let

be a decomposition of 1 as a sum of orthogonal primitive idempotents of F N . Because F is a splitting field for F N and N is abelian, each e;FN is one-dimensional. Hence eiFG (e;FN)G is of a dimension lPl. Now G = N P , so FG = ( F N ) ( F P )and thus

e,FG = ( e ; F N ) ( F P )= e;FP It follows that the map

FP

+

z

H

e;FG e;z

is a surjective homomorphism of right FP-modules of the same Fdimension. Consequently,

FP

E e;FG

as FP-modules

(1 5 i

5 n)

and therefore e ; F G . I ( P )is a unique maximal FP-submodule of eiFG. But e;J(FG)is a proper FP-submodule of e;FG, hence

e;FG I ( P ) 2 eiJ(FG)

(I 5 i 5 n )

and thus FG - I ( P ) 2 J ( F G ) . Hence, G is p-radical, by applying Theorem 5.3. H 5.17. Remark. (i) Any p-radical group is p-solvable (see Okuyama (1986)). (ii) It can be shown (see Tsushima (1986)) that a p-nilpotent group is p-radical if and only if [O,! ( G ) ,D]nCo,,(G)(D)= 1 for any p-subgroup D of G (here O,t(G) is the maximal normal p’-subgroup of G ) . 6.

Loewy series for transitive permutation modules

Throughout this section, F denotes a field of characteristic p > 0, H a subgroup of a finite p-group P and P / H the set of all left cosets of

6. Loewy series for transitive permutation modules

427

H in P . We denote by F ( P / H ) the F-linear span of all left cosets of H in P . Since P acts on the set P / H by left multiplication, it is clear that F ( P / H ) is a permutation FP-module which is isomorphic to ( 1 ~ ) To ~ .state our main objective, we first record the following piece of information. Let A be a finite-dimensional algebra over a field and let V # 0 be an A-module. Recall from Sec.18 of Chapter 3 that the descending

chain

V 2 J ( A ) V 2 J ( A ) 2 V2

* * -

of submodules of V is called the lower Loewy series of V . The upper

Loewy series

0 = So(V) G S'(V)

c S2(V) c ...

= Soc(V/S,-l(V)) for n > 0. The Loewy length of V is defined to be the integer k 2 1 such that J(A)"-'V # 0 and J ( A ) k V = 0. The reader may readily verify that

of V is defined by S,(V)/S,-l(V)

V 3 J ( A ) V 2 . * . 3 J(A)k-'V 3 J(A)kV = 0 and

0 = & ( V ) c S,(V)

c

* *

a

c &(V)

=

v

and

Sn(V)= {v E VIJ(A)"v = 0}

( n > 0) Our aim in this section is twofold: first to provide a basis for each I ( P ) " F ( P / H ) / I ( P ) " + ' F ( P / H ) and second to prove that the upper and lower Loewy series for F ( P / H ) coincide. Here and in what follows I ( P ) denotes the augmentation ideal of F P (which coincides with J ( F P ) since P is a p-group and churF = P ) . All the information recorded below may be found in Huppert and Blackburn (1982) or Karpilovsky (198713). The Bruuer - Jennings - Zussenhuus series for P is defined inductively by M'(P) = P and for n 2 2,

M 7 p ) = [Mn-I(P),P]M(P)p where i is the smallest integer satisfying ip 2 n and M,(P)p is the subgroup generated by all X P with x E M ; ( P ) . The subgroups M,(P)

428

Permutation modules

satisfy the following properties: (i) [Mi(P),Mj(P)IG M+j(P) (4.i 2 1). (ii) Mi(P)/Mj+l(P) is elementary abelian ( i 2 1). For each i 2 1, choose elements x;, of P (with s running over an initial sequence of positive integers, depending on i ) lying in M i ( P ) such that the cosets of M;+1(P)which contain them, form a basis for M;(P)/Mi+l(P).If P has order p n , then there are p" products, lexicographically ordered, of the form

where the product with all exponents zero is taken to be the identity element of P . The weight of such a product is defined to be C ia;s. A classical theorem of Jennings asserts that the products of weight rn lie in I(P)" and form a basis of I(P)" modulo I(P>"+l. We begin by proving the following modification of Jennings theorem due to Alperin (1988). 6.1. Theorem. Suppose that we choose an arbitrary linear order for the xis (instead of the lexicographing ordering). Then the p" products

have the property that the ones of weight m lie in I(P)" and form a basis of I(P)" modulo I(P)"+' (xi8 - 1 still has weight i, ( x i s - l)ais has weight iajs, and the weight of a product, no matter what the order of the factors, is the sum of the weights of the factors).

Proof. Since x;,- 1 E I ( P ) ,it is clear that the products of weight rn lie in I(P)". We now claim that every element of P is expressible (and hence uniquely so) as a product nx;;, with the factors in the given order. Choose j with M j ( P ) # 1 and Mj+l(P>= 1. Since the subgroups M i ( P ) are invariant under epimorphisms, it follows by induction on j and passage to P / M j ( P ) ,that if g E P then g = x;js .z with z E M j ( P ) , 1 5 i 5 j - 1. Bearing in mind that z can be expressed as a product of factors xjs and these are central, the claim is established. By the foregoing, the p" products nx;ia (in the given order) span F P . We next claim that this implies that the pn products n ( x i , - l ) a i a

n

6. Loewy series for transitive permutation modules

429

span F P . Indeed, if u1,. . . ,u, are elements of any ring, then it is easily proved by induction that u1 . . . u, - 1 = C ( U i l - l)(ua2- 1) *

-

*

(Uik -

1)

where the sum is over all nonempty subsets I = {il, . . . ,ik} of { 1,.. . ,T } and il < iz < ... < ik. Thus, since the product n ( x i S - 1)' has been taken as 1, it follows that the product nx;:, can be expressed unitri- l ) a i s , as claimed. angularly in terms of the products

n(q,

We shall conclude the proof by demonstrating that the products n(xc;, - l ) a i s of weight rn are a basis of I(P)" modulo I(P)"+'. We argue by downward induction on rn. If I ( P ) k + l = 0 and I ( P ) k # 0, then I ( P ) k c_ SocFP = {ACT E PIX E F } X

and hence dirnFI(P)k= 1. We also have n(z:,-' - 1) E I ( P ) k since k is the product of p - 1 and the sum of the products of i and the number of subscripts s for that i. This product is not zero, since otherwise the remaining pn - 1 products would span F P , (which is impossible) so the assertion holds for rn = k. Assume that our assertion holds for integers greater than rn. Note that the number of our products of weight rn is the same as the number of products of weight rn in lexicographic ordering. Hence if the products in the new ordering are linearly independent modulo I ( P)"+', then they form a basis for I(P)"/I(P)"+'. If these products of weight rn are linearly dependent, then one of them is a linear combination of the others plus (by the induction hypothesis) the products of weight > rn. But this implies that the pn - 1 products (obtained by deleting the expressed product) span F P , which is impossible. The proof is therefore complete. For each i 2 1, let us choose elements x;, in a particular manner: select sets of elements y;, and z;, whose union will be the x;, but with the y;, chosen so that their images in M ; ( P ) / ( Hn M;(P))M;+l(P) are a basis of that group and so that the z;, lie in ( H n M ; ( P ) ) M ; + l ( P ) n H and their images in ( H n M ; ( P ) ) M c ; + l ( P ) / M ; + lform ( P ) a basis of that group.

Permutation modules

430

6.2. Theorem.

(Alperin (1988)). Let f : F P F ( P / H ) be the FP-homomorphism sending each element of P to the left coset containing it. Then the images f(n(y;,-l)"is) ofthe products, lexicographically ordered with 0 5 a;, < p , are a basis of F ( P / H ) such that the images of the products of weight m lie in I ( P ) m F ( P / H )and form a basis of I ( P)"F( P / H ) modulo I ( P)"+' F ( P / H ) . .--)

Proof. It is a consequence of Theorem 6.1 that the products

where the factors yi, - 1 and the z;, - 1 are in lexicographic order, which are of weight m lie in I(P)" and form a basis modulo I(P)m+l. The homomorphism f sends each ziS- 1 to 0 since z;, E H . Therefore, since any product in (1) which has some b;, # 0, can be expressed in the form I C ( Z ; , - 1) with x E F P we have that

f(z(z;, - 1)) = 2 f(z;, - 1) = 0 Thus F ( P / H ) is spanned by the images of all products in the yi, - 1. We claim that these images form a basis. To substantiate the claim, it suffices to verify that there are pn-k of them, where p k = ]HI. The latter will follow provided we show that the total number of y;, - 1 is k. But

and the subgroups H n M ; ( P )obviously filter H , hence the claim. The image under f of a product in the y;, - 1 of weight m lies in I ( P ) " F ( P / H ) , since the product lies in I(P)". We are therefore left to verify that the images of the products of weight m are linearly independent modulo I(P)"+'F( P / H ) . Assume by way of contradiction that this is not the case. Then the image of one of the products of weight m is a linear combination of images of other products in the y;, - 1 of weight m and images of such products of weights > m , since

V ) " F ( P / H ) = f (W") is spanned by the images of all the products in the y;, - 1 of weight of at least m. In particular, deleting this image, that we have just expressed

431

6. Loewy series for transitive permutation modules

as a linear combination, from the set of such images, we would still have a spanning set for F ( P / H ) . But this contradicts the fact that the number of products in the yis - 1 is pk which is the dimension of F ( P / H ) . So the theorem is true. H

As a preliminary to our final result, we first record the following property. 6.3. Lemma.

Let G be a finite group and let V be an FG-module.

Then

where V* is the contragredient of V . Proof. We claim that for all k 2 0,

sk(v*) = (J(FG)"v)'

(2)

where for any submodule W of V , W' = {f E V*If ( W ) = 0). Since

the result will follow by applying (2) for k = i and k = i - 1. We argue by induction on k 2 0. If k = 0, then both sides of (2) are zero. Assume that (2) is true for k. Then SOC(V*/(J(FG)'V)')= Sk+l(V*)/(J(FG)kV)'

(4)

and Sk+l(V*)= W' for some submodule W of V . Since

+

is completely reducible, it follows from (3) (applied to i = k 1) that (J(FG)"'V)'/( J(FG)kV)' is also completely reducible. Hence, by

(4),(J(FG)"'V)'

C Sk+l(V*).On the other hand, by (4),

WL/ ( J (FG)'V)

'

( J (FG)kV /W ) *

Permutation modules

432

is completely reducible. Hence J ( FG)’V/W is completely reducible and therefore J(FG)k+’V C W . Thus

(J(FG)k+’V)’ 2 W’ = Sk+l(V*) and the result follows.

We are now ready to prove the following result. 6.4. Theorem. (Alperin (1988)). Let F be a field of characteristic p > 0 , let P be a finite p-group and let V be a transitive permutation FP-module. Then the upper and lower Loewy series of V coincide.

Proof. We may of course identify V with F ( P / H ) for some subgroup H of P . For each j 0, put % = I ( P ) j F ( P / H )and keeping the notation of Theorem 6.2, denote by t the weight of n ( y i , - 1)P-I. Then, by Theorem 6.2, V, is of dimension 1 and = 0. Note that the weights of the “complementary” elements n ( y ; , - l)a:s and n ( y i s - l)P-l-a:s add up to t. Hence, for all j E {0,1,. . . ,t }

>

dimFT/,/T/,+, = dimFvt-j/V,-j+l (5) This means that the dimensions of the quotients of the lower Loewy series read in reverse order and are the same as read in the usual order. Owing to Lemma 6.3, the quotients of the upper Loewy series of V* are the contragredients of the quotients of the lower Loewy series of V . But V E ( l H ) G , so V* E V and therefore

dim^ (Si+l( V )/ Si ( V )) = dim^ (V ,/ V,+,) for all i E {0,1,. . . ,t } . Hence, by (5), we also have

(6)

dimF(Si+l(V)/Si(V))= dimF(&-i/&-i+l) (7) for all i E {0,1,. . . ,t } . Moreover, we certainly have S o ( V ) = K+’, S l ( V ) 2 V,,S 2 ( V )2 &-’ and so on, which implies dirnF(Si(V))2 dimF(vt-i+l) Applying (6) and (7), we therefore conclude that dimF(S;(V))= dimF(V,-;+I) which in turn implies that S ; ( V ) = V,-i+l, as required. H

7. Characterizations of p-permutation modules

433

7. Characterizations of p-permutation modules In this section, G denotes a finite group and R either a field of characteristic p > 0 or a complete discrete valuation ring such that charR/J(R) = p > 0. All RG-modules are assumed to be finitely generated. We say that an RG-module V is a p-permutation module if the restriction Vp of V to a Sylow psubgroup P of G is a permutation RP-module. Thus V is a p-permutation module if and only if V is R-free of finite rank with an R-basis on which a Sylow p-subgroup of G acts as a permutation group. The following result is essentially due to Dress (1975).

7.1. T h e o r e m . Let V = V1 @ - - .@ V, be a decomposition of an RG-module V into indecomposable submodules. Then the following conditions are equivalent: (i) V is a p-permutation module. (ii) For any p-subgroup Q of G, VQ is a permutation RQ-module. (iii) V is a direct summand of a permutation module. (iv) There exist subgroups H I , . . . ,H , of G such that, for any i E (1,. . . ,n ) V , is isomorphic to a direct summand of ( 1 ~ ~ ) ~ . (v) Each V; is a p-permutation module. (vi) Each V; has trivial source.

Proof. (i) + (ii): Let P be a Sylow p-subgroup of G such that Vp is a permutation RP-module. Then Q C gPg-' for some g E G. If u l , . . . ,vh is a permutation basis for Vp, then gvl,. . . ,gvk is a permutation basis for 4 p S - 1 and hence for VQ,as required. (ii) + (iii): Let P be a Sylow p-subgroup of G. By hypothesis, V p is a permutation RP-module. Hence (Vp)" is a permutation RGmodule. But V is P-projective, so V is isomorphic to a direct summand of (Vp)", proving (iii). (iii) j (iv): By hypothesis, there is a permutation RG-module M such that each V; is a direct summand of M . Since V; is indecomposable, it follows from the Krull-Schmidt theorem that V; is isomorphic to a direct summand of ( 1 ~for~ some ) ~ subgroup H; of G, 1 5 i 5 n, proving (iv). (iv) + (v): Since ( l H i ) G is a permutation module, it is also a p-

Permutation modules

434

permutation module. Hence we need only verify that a direct summand N of a p-permutation module M is a p-permutation module. If P is a Sylow psubgroup of G such that M p is a permutation RP-module, then N p is a direct summand of Mp. But M p is a direct sum of modules of the form (1~)' where Q is a subgroup of P. By Corollary 5.1.12, (19)' is indecomposable. Hence, by the Krull-Schmidt theorem, Np is a direct sum of modules of the form (1~)'. Thus N is a p-permutation module. (v) (vi): Put M = V;; and let P be a Sylow p-subgroup of G such that M p is a permutation RP-module. If Q & P is a vertex of M , then M is Q-projective and so M is isomorphic to a direct summand of ( M Q ) ~By . hypothesis, MQ is a direct sum of modules of the form (1s)Qwhere S is a subgroup of Q. Thus M is a direct summand of (1s)' for some subgroup S of Q. Since M is S-projective and S C Q, S is a vertex of M and hence M has trivial source. (vi) + (i): It suffices to show that each V;; is a p-permutation module. By the implication (i) + (iii), V;; is a direct summand of a permutation (hence p-permutation) module. Hence, by the proof of the implication (iv) j (v), V;; is a p-permutation module.

+

7.2. Corollary. Let H be a subgroup of G and let V and W be RG-modules. (i) If V = V, @ . . - @ V , for some RG-modules K:, then V is a ppermutation module if and only if each T/: is a p-permutation module. (ii) If V and W are p-permutation modules, then so is V @R W . (iii) If V is a p-permutation module, then VH is a p-permutation R H -module. (iv) If U is a p-permutation RH-module, then U G is a p-permutation RG-module.

Proof. (i) Direct consequence of Theorem 7.1. (ii) If {vi} and {wj}are R-bases of V and W , respectively, which are permuted by a Sylow p-subgroup of G, then the same is true for the R-basis {TJ; 8 wj} of V 8~ W . (iii) If Q is a Sylow p-subgroup of H ,then VQ = (VH)Qis a permutation RQ-module, by virtue of Theorem 7.1(ii). (iv) This is obvious.

7. Characterizations of p-permutation modules

435

To provide some further characterizations of p-permutation modules, we need to introduce some terminology. Let F be a field of characteristic p > 0 and let V be an FG-module. Then V is said to be monomial if there exists a basis q ,. . . ,TI, of V such that G permutes the subspaces F v l , . . . ,Fv, of V . Expressed otherwise, V is monomial if and only if V is a direct sum of modules of the form W Gwhere W is a one-dimensional FH-module for some subgroup H of G. Following Dress (1975), we say that V is virtually monomial if there exist monomial FG-modules V, and V2 such that V @ V’ &. It is clear that the direct sums of monomial (virtually monomial) modules are again monomial (virtually monomial) modules. Our second major result will be proved with the aid of the following theorem.

=

7.3. Theorem.

(Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 and let V be an indecomposable F G module with vertex Q and a trivial source. Then there exist subgroups H I , . . . ,H, of G and one-dimensional FHi-module K, 1 5 i 5 n, such that (a) V @ qG@ - - .@ Vf 2 V g l @ . @ V,G (for some li < n). (b) Each H;,1 2 i 5 n, has a normal SyEow p-subgroup P; such that H;/P; is elementary (i.e. a direct product of a cyclic group and a q-group for some prime q) and such that g;P;g,yl C Q for some 9;E G ( l 5 i 5 n ) . m .

Proof. We argue by induction on IQI. First assume that Q = 1 and denote by {E;liE I } the set of elementary subgroups of G. Since V is projective, it follows from Swan’s extension of Brauer’s induction theorem to modular representations (see Swan (1960)) that for each i E I , there exist FE;-modules M; and N; such that

have the same composition factors (counting multiplicities). Since, for any FG-module W , the FG-module V @ F W is projective, we may tensor both modules in (1) with V to obtain the following isomorphism

Permutation modules

436

of FG-modules:

(see Theorem 2.6.1). Hence, since VEi @ M i and V E 8 ~ N; are projective FE;-modules, we may harmlessly assume that G is an elementary group; If p does not divide the order of G, then there is nothing to prove, since in this case the representation theory of G over F is the same as over the complex numbers (see Curtis and Reiner (1981)) and any elementary group is known to be an M-group (see Curtis and Reiner (1962)). Otherwise, we have G = P x H , where P is a p-group and H is an elementary p’-group. Let &, . . . , be all nonisomorphic irreducible FH-modules. Then V y ,. . .,KG are easily seen to be all nonisomorfor some phic indecomposable projective FG-modules. Hence V i E { 1,. .. ,t } and we are again reduced to the case where p does not divide the order of G. This completes the proof of the case where Q = 1. Now assume that IQI > 1 and let H = NG(Q). By the Green correspondence, there exists an indecomposable FH-module W with the same vertex Q and trivial source such that

KG

wG ~ U , @ U , @ . . * @ U ,

(U; is indecomposable, 0 5 i 5 m )

with Uo E V and \&I < I vertex (U;)l for all i E (1,. . . , m } . Because W is a direct summand of a permutation FH-module, W Gis a direct summand of a permutation FG-module. Hence U,, . . . , Urn are direct summands of permutation FG-modules. Thus, by Theorem 7.1 and the induction hypothesis, the theorem holds for each FG-module U;, 1 5 i 5 m. Therefore it suffices to prove the result for W , i.e. we may assume that Q d G. Since Q d G, Q acts trivially on (lQ)G and thus on its direct summand V . Hence V is inflated from an F(G/Q)-module where is also projective. Thus we may apply the case Q = 1 to deduce that V satisfies the conclusion of the theorem (with G replaced by GI&). This means that there exist subgroups Hl/Q,. .. ,Hn/Q of G/Q and one-dimensional F(Hi/Q)-modules E, 1 5 i 5 n, such that (k < n ) . (c) @ @ . . . @ vp @ . .. @ (d) Each Hi/&, 1 5 i 5 n, is elementary of order not divisible by

v,

v :v

= vcl

v;

v

7. Characterizations of ppermutation modules

437

P. Thus if V , is the FG-module inflated from G, then HI, ...,Hn and 6,. . . ,V, satisfy the conclusion of the theorem. This completes the proof. H 7.4. Corollary. (Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 and let V be a projective FG-module. Then there exist subgroups H I , . . . ,H, of G and one-dimensional FH,-module V,, 1 5 i I. n, such that

and such that each H; is an elementary p'-group. Proof. Apply the special case of Theorem 7.2 where Q = 1 to each indecomposable direct summand of V.

7.5. Corollary. (Dress (1975)). Let F be an algebraically closed held of characteristic p > 0 . Then there exist subgroups H I , . . . ,H,, of G and one-dimensional FH;-module V;, 1 5 i 5 n, such that (a) 1~@ V y @ - @ V j 2 V j l @ - @ V,G (for some k < n). (b) Each Hi, 1 5 i 5 n, has a normal Sylow p-subgroup P;,such that H;/P; is elementary. -

a

Proof. Let Q be a Sylow psubgroup of G. Then Q is a vertex of 1~ and 1, is a source of 1 ~Now . apply Theorem 7.3. H

7.6. Corollary. (Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 . Then, for any FG-module V , there exist subgroups H I , .. . ,H , of G and FHi-modules Wi,1 5 i 5 n, such that (a) V @ WF @ , - WkG-N Wt+l @ - - - @ W: (for some k < n). (b) Each H , has a normal Sylow p-subgroup with elementary factor group. Proof. Tensoring the isomorphism of Corollary 7.5 with V, we obtain v @ @i"=l(v 63 !3 @;&+l(V €4

YG)

YG)

Permutation modules

438

Since, by Theorem 2.6.1,

v @ Fc

E)‘

5 n) the result follows by setting Wi = VHi8 V,, 1 5 i 5 n. (vHi @

(1 5 i

7.7. Theorem. (Dress (1975)). Let F be a field of characteristic p

> 0 and let V be an FG-module. Then the following conditions are

equivalent: (i) V is a p-permutation module. (ii) V is a direct summand of a monomial FG-module. Furthermore, under the assumption that F is algebraically closed, each of the above conditions is equivalent to: (iii) V is virtually monomial. Proof. Since each permutation FG-module is a monomial FGmodule, the implication (i) j (ii) follows from Theorem 7.1. The implication (iii) j (ii) follows from the definition of “virtually monomial”. Note also that if F is algebraically closed, then (i) implies (iii), by Theorems 7.1 and 7.3. By the foregoing, we are left to verify that (ii) implies (i) for an arbitrary F . To do this we may, by Theorem 7.1, assume that V 2 W G where W is a one-dimensional FH-module for some subgroup H of G. But W is a p-permutation module, since a Sylow p-subgroup of H acts trivially on W .Hence V is a p-permutation module, by virtue of Corollary 7.2 (iv). 8.

The Brauer morphism

In this section, p denotes a prime, I< a subgroup of a finite group G and R a commutative local ring with c h a r R / J ( R ) = p . In what follows we put F = R / J ( R ) . Let V be an RG-module. Given a subgroup H of I(, we put

and

V ( K )= Inv(VK)/

+ J(R)lnv(V-)

8. The Brauer morphism

439

where H runs over the set of all proper subgroups of K with the convention that for K = 1, &V," = 0 (and hence V ( K )= V / J ( R ) V ) . We remark that in most papers on the subject Inv(V'&) is denoted by V H ;we shall use this notation only in the case V = A is a G-graded R-algebra so as not to conflict with our notation for induced modules. Following Brou6 (1985), we refer to the natural surjection

as the Brauer morphism. Our aim is to record some basic properties of the Brauer morphism with an eye to future applications in the next section. Most of the results recorded are contained in BrouC (1985) and

Brou6 and Puig (1980). 8.1. Lemma. With the notation above, the following properties hold: (i,~~ n v ( v K )is an R(NG(I<)/I<)-modde, v,K)+J(R)I~~(v,) is an R(NG(I<)/I()-submoduleo f I n v ( v ~ )and , V ( K )is an R ( N G ( I ( ) / K ) module annihilated b y J ( R ) . (ii) The Brauer morphism B r g is a homomorphism of R(NG(K)/I(>modules. (iii) If V = U @ W for some RG-submodules U and W , then

(c,,,

(iv) V ( K )= VNG(K)(I{). (v) If I( acts trivially on V and I< is a p-group, then V ( K ) = v/J ( R )v . Proof. (i) If g E NG(I<)and v E Inv(v~),then for any z E I(,

proving that Inv(vk) is an R(NG(I<))-module. Since I< acts trivially on I n v ( v ~ )it, follows that I n v ( v ~ )is an R(NG(K)/K)-module in a natural way. It remains to show that ( x H , - K V,") J ( R ) l n v ( v ~is) an R(NG(K))-submodule of Inv(VK). If K = 1, then there is nothing to prove. Assume that H is a

+

Permutation modules

440

proper subgroup of K and let T be a left transversal for H in K . Given v E Inv(V H )and g E NG(K ) , we have

as required.

(ii) This is a direct consequence of (i). (iii) For any subgroup H of I<, we have

and

which clearly yields the assertion. (iv) This is a direct consequence of the definition of V(IC). (v) If K acts trivially on V , then since K is a p-group, for any proper subgroup H of I(, V$ C J(R)lnv(v~)and I ~ v ( V K=) V . Hence V ( K )= V/J(R)Vas required. 8.2. Lemma.

For a n y RG-module V , we have

Proof. (i) The first equality follows from the fact that I( acts trivially on V ( K ) .We are therefore left to verify that for all v E In’u(VK),

, ) in G Let T be a full set of double coset representatives for ( N G ( K ) K containing 1. Then, by Lemma 2.9.6(i),

8 . The Brauer morphism

441

Now fix t E T - {l},put L = tKt-' n N G ( K ) and denote by S a full set of double coset representatives for ( K , L ) in N G ( K ) . Then each SLS-' n I< is a proper subgroup of I( and hence, by Lemma 2.9.6(ii),

T r y " ) ( tv) E

cTrzs-l

nh' (1nQLs-lnK

))

SES

E I(er(B7-L) This proves ( l ) ,by applying (2). (ii) This is a direct consequence of (i). Let U, V and W be RG-modules. An R-bilinear map

is said to be G-stable if

f(gu,gv) = g f ( u ,v) for all g E G, o E V 8.3. Lemma. Let U , V and W be RG-modules and let f : U x V + W be a G-stable R-bilinear map. F o r each u E U , v E V , dejne fu

E H o ~ R ( VW , ) and

f,,E H o r n ~ ( UW , )

are RG-homomorphisms.

Proof. It is c1ea.r that both maps are R-linear. Fix g E G and u E U . Then for all v E V , f g u ( 4

= f ( s u 7 4 = f ( s u , s ( s - ' v ) ) = 9 f ( u , 9 - l 4 = ( 9f u) ( v> ,

proving that f s u = g f u . Hence u H f u is an RG-homomorphism. A similar argument shows that fg. = gfv for all g E G,vE V and the result follows.

Permutation modules

442

8.4. Lemma. Let U,V and

W be RG-modules and let f

:U x

V -t W be a G-stable R-bilinear map. Then, for any subgroup K of G, the map f K : U ( K )x V ( K )+ W ( K ) given b y

frc (Br;(u),Br;(v)) = W ( f ( . , V ) )

(u

E Inv(W,v E I~v(vK))

is a well-dejned (Nc(IC)/lC)-stableR-bilinear map. Proof. If u E Inv(U~),vE 1nv(V~), then for all g E IC,

proving that f ( u , v ) E I n v ( w ~ ) To . prove that f K is well-defined, it suffices to show that B r F ( f ( u , v ) ) = 0 if u E Ker(Br;) or w E K e r ( B r L ) . Assume that u E Ker(BrK). then

for some X E J ( R ) ,z E I ~ w ( U K ) , U HE U E , where H runs over the set of proper subgroups of K(if (I = 1, then put all U H = 0). Hence

and therefore it suffices to show that Brg(f ( u ~w)), = 0 for all H c IC. Now U H = & t y for some y E Inv(UH),where T is a left transversal for H in K. Since w E 1nv(V~), y E Inv(UH) and f is G-stable, we have f ( y , v) E Inv(lV~).Hence

A similar argument shows that if v E K e r ( B r g ) ,then BrE(f(u,v)) = 0, and thus f is well-defined.

8. The Brauer morphism

443

It is clear that f is an R-bilinear map. Furthermore, for any g E

NG ( K )7 f K ( g B r $ ( u ) , g B r m ) = fK(Br;(P), BrL(gv)) = B.KW(f(P,gV))

= sBrKW(f(., 4) = SfK(Br;w, B d m )

by applying G-stability of f together with Lemma 8.1(ii). This completes the proof of the lemma.

Let V be an RG-module, let P be a p-subgroup of G such that Vp is a permutation RP-module with an R-basis X permuted b y P , and let V = V /J ( R)V. Denote b y x1,x2,. . . ,x, all distinct elements o j X fixed b y P. Then the following properties hold: (i) The set {BrK(x;)Il 5 i 5 n } is an F-basis for V ( P ) . (ii) The F(NG(P)/P)-modulesV ( P ) and v ( P ) are canonically isomorphic. (iii) I f V * is the contragredient ofV and the basisX* = { x * ( xE X } o f V * is dual to X , then there is an F(NG(P)/P)-isomorphism 8.5. Lemma.

V * ( P )-+ V ( P ) * which sends the basis {Br~'(x*)ll5 i 5 n } o f V * ( P ) onto the dual basis o f { B r ~ ( x ; ) 5 ( li 5 n } . Proof. (i) Let x l , x 2 , . . . ,x,, x,+~,. . . ,xt be all representatives for the P-orbits of X, and let Qi be the stabilizer of xi in P , 1 5 i 5 t. By our choice of the xi,we have Q1 = Q2 = = Qn= P and &%+I,. . . ,Qt are all proper subgroups of P . By Lemma l.l(iii), {Trgi(x;)Il5 i 5 t } is an R-basis for Inv(Vp). It therefore suffices to show that for any proper subgroup Q of P , VJ is contained in M , where M is the sum of J ( R ) l n v ( V p )and the R-linear span of T r c k ( x k ) , n 1 5 k 5 t . To this end, fix x E X and denote by H and L the stabilizers of x in Q and P, respectively. Since the elements w E V of the form w = T r QH ( z ) form an R-basis for 1nw(V~), it suffices to verify that for any such w, TT;(V)E M . Since a .

+

~ r g ( ~ r Q H (= x )T) r g ( x ) = Tr,P(Trf;(x))

Permutation modules

444

and Trfi(x) E J(R)lnv(VL)whenever L # H , the required assertion follows. (ii) We first note that V p is a permutation FP-module with F-basis = {?(a: E X},?= x J ( R ) V . Hence, by (i),

x

+

{ ~ r ; ( ~ i )5l il 5 n }

is an F-basis for V ( P ) . Thus Brg(x;) I+ BY:(?;) is an F-isomorphism of V ( P ) onto V ( P ) . Since this isomorphism obviously preserves the action of N G ( P ) / P , the assertion follows. (iii) Let a , b E X and let g a = b for g E P . Then ga*(b) = a * ( g - l b ) = a*(a) = 1 and, for x # b,x E X,ga*(x) = a*(g-'x) = 0. Thus ga* = b*, and similarly if ga* = b', then g a = b. In particular, this implies that x;,.. , ,xi are all distinct fixed points of X * in P . Hence, by (i), {BrF*(xTIl5 i 5 n } is an F-basis for V * ( P ) . It follows that the map

{ B&$;

:

is at least an F-isomorphism. Let F : V x V* + R be defined by f(v, cp) = cp(v) for all v E V,cp E V*. Then, regarding R as the trivial RG-module, f becomes a G-stable R-bilinear map. Since R ( P ) = R / J ( R ) = F, it follows from Lemma

8.4 that the map f-

: V ( P )x

V * ( P )+ F

defined by f P ( ~ 4 4B.pv'(cp)) ,

= CpW

+J(R)

(v E Inw(Vp),cp E Inv(V,*)) is an (NG(P)/P)-stable F-bilinear map. Hence, by Lemma 8.3, the map

V * ( P )+ V ( P ) *

+

defined by B ~ ; * ( ( P H) f,, where f,(Br;(w)) = ~ ( v ) J ( R ) , is an F(NG(P)/P)-homomorphism. Setting cp = x:, we see that this homomorphism coincides with $ and the result follows.

8. The Brauer morphism

445

8.6. Lemma. Let H be a subgroup of G and let V be an H projective RG-module. Then, for any subgroup K of G, V ( K ) = 0 unless K is G-conjugate to a subgroup ofH. In particular, if W is any RH-module, then W G ( K )= 0 unless I( is G-conjugate to a subgroup ofH.

Proof. that

Since V is H-projective, it follows from Theorem 2.10.3

T r g (EndRH( V ) )= EndRG( V ) Hence, by Lemma 2.9.6(ii) l v E TrE(EndRH(V))

TrgngHg-I (EndR(KngHg-l)(v)) g€G

For each g E G, we may therefore choose and thus K that l v = C s E G TrKngHg-l($g)

Inv(Vh-)

c

$g

E EndqKngHg-i)(V)such

Tr$ngHg-l($g)(lnv(vK)) g€G

But for any

z1

E I n z 1 ( V ~ we ) , have $ ~ ~ ( vE)Inv(VKngHg-l)and K

TrKngHg-l ($g)(')

' h

= T r K gHg-' ~

($g

(' )

which implies that

Inv(VK)

v/ngHg-l

i

9EG

thus completing the proof. 8.7. Lemma. Let P be a p-subgroup of G , let H be a subgroup of G such that P is G-conjugate to a subgroup o f H and let T be a set of all representatives t for the double cosets NG(P)tH for which P tHt-'. Then

1

(lRH)G(P) @tET [ l F ( N t H t - l ( P ) N G ( p ) as F(NG(P)/P)-modules Proof. Let X 2 T be a full set of representatives for ( N G ( P )H), double cosets in G. Then, by Mackey decomposition,

Permutation modules

446

where v, = [1R(N2Hz-1(P))

On the other hand, by Lemma 8.6,

V,(P) = o if P

xHx-'

(4)

Furthermore, if P C_ x H x - l , then P acts trivially on V, and thus, by Lemma 8.l(v),

The desired conclusion is therefore a consequence of (3), (4), ( 5 ) and Lemma 8.l(iii), (iv). Given a G-algebra A over R and subgroups H C K of G, we put

A; = T r $ ( A H ) where AH is the subalgebra of H-invariant elements of A and

"

TrH ; AH + A" is the trace map introduced in Sec.1. Note that

Trz(ab)= Trg(a)b for all a E A H ,b E AK

(6)

and

TrE(bu)= bTr$(a) for all a E AH,bE A" (7) It follows from (6) and (7) that A; is an ideal of A". Note also that = A ~ ~ s - for ' all g E G

(8)

Since A is an RG-module, we may define A ( K ) as we did for any RGmodule V , i.e. by putting

A ( K )= AK/ ( C A Z H

+ J(R)AK

8.

The Brauer morphism

447

where H runs over the set of all proper subgroups of K with the convention that for K = 1,& AZ = 0. In particular, the natural surjection

B r i : AK t A ( K ) gives us the Brauer morphism with respect to the RG-module V = A. 8.8. Lemma. Let K be a subgroup ofG and let A be a G-algebra over R. Then (i) Ker(Br;) is an ideal of A K . (ii) AK is an (NG(K)/K)-algebraouer R, and A ( K ) is an ( N G ( K ) / K ) algebra ouer R annihilated b y J ( R ) . (iii) Br; is a homomorphism of (Nc(K)/K)-algebras, i.e. a homomorphismof R-algebras perserving the actions of NG(K ) / K on AK and A(IC).

Proof. (i) This is a consequence of the fact that each A$ and J ( R ) A K are ideals of AK. (ii) The assertion regarding A" follows from (8) applied to H = K . The remaining assertion is a consequence of (9) and the definition of A( K ) . (iii) Direct consequence of (i) and Lemma 8.1(ii). W We close by recording the following observation.

Let A = R X be the group algebra o f a group X over R and let G act on X as a group of automorphisms of X . Then A becomes a G-algebra in the natural way. Let P be a p-subgroup ofG and let H be the subgroup of all elements o f X fixed b y P . Then (i) The map 8.9. Lemma.

is an isomorphism of F-algebras. (ii) If R = F, X = G and G acts on itself b y conjugation, then H = C G ( P )and, upon the identification of FG(P) with FCG(P),the restriction of BrFG to Z(FG) is the classical Brauer homomorphism (see Theorem 6.2.2).

Permutation modules

448

Proof. (i) This is a direct consequence of Lemmas 8.8(iii) and 8.5(i). (ii) Let C1,. . . ,C, be all P-conjugacy classes of G. If IC;l > 1, then BrcG(C:) = 0, while Br;"(h) = h for all h E CG(P).This obviously implies the required assertion. 9.

Scott modules

Let G be a finite group, p a prime and R either a field F of characteristic p or a complete discrets valuation ring with F = R / J ( R ) . All RG-modules are assumed to be finitely generated. Our aim is to examine an important type of indecomposable p permutation RG-modules, the so called Scott modules. Such modules were first discovered by L.L.Scott, and independently by J.L.Alperin; the first published account is in Burry (1982). Our method is based on an,important work of Broud (1985) who utilized a systematic use of the Brauer morphism. We begin by recording the following definition due to Puig (see Brou6 (1985)). Let V be a ppermutation RG-module and let P be a psubgroup of G. Then the Scott coeficient of V associated with P , written s p ( V ) is defined by

Note that, by Lemma 8.2(ii),

s p ( V ) = dim= ( V ( P ) p ( p ) ' p ) 9.1. Lemma. Let P be a p-subgroup of G.

(i) If V and V' are two p-permutation RG-modules, then

(ii) If V is u p-permutation RG-module and V* is the contragredient ofV then s p ( V ) = s p ( V * ) . (iii) For any subgroup H of G, sp [ ( ~ R H ) = ~ ] 1 i f P is G-conjugate to a Sylow p-subgroup of H and sp [(lRH)G] = 0 otherwise.

9. Scott modules

449

Proof. (i) Apply Lemma 8.l(iii) and ( 1 ) . (ii) Owing to Lemma 8.5(iii), s p ( V * ) is equal to the rank of the endomorphism T r NG(p)'p , of V ( P ) * which , is equal to the rank of its transpose, i.e. the endomorphism T r NG(p)'p , of V ( P ) . Thus s p ( V ) = s p ( V * ) , as asserted. (iii) By Lemma 8.6, ( l m ) G ( P )= 0 unless P is G-conjugate to a subgroup of H . By Sylow theorems in H , if g-lPg is a Sylow psubgroup of H for some g in G , then gTIPgl is a Sylow psubgroup of H only for those gl E G for which g1 E NG(P)gH.By Lemma 8.7, it ) ? dimension 1 or 0 according therefore suffices to show that (( ~ F H ) ~ has to whether H is, or is not a p'-group. By Lemma 1.3, we may identify ( ~ F H ) ' with the submodule F G - H + of FG, in which case

as required.

We are now ready to prove the following result. (Scott, Alperin). Let P be a p-subgroup of G. (i) There exists a unique (up to RG-isomorphism) indecomposable p-permutation RG-module Sp(G, R ) such that sp(Sp(G,R ) )# 0 . Furthermore, s p ( S p ( G ,R ) ) = l, and 9.2. Theorem.

Sp(G,R)

[SP(G,R)]*

(ii) If H is a subgroup of G, then Sp(G,R ) is isomorphic to a direct summand of ( ~ R H if) and ~ only if P is G-conjugate to a Sylow p-subgroup of H . In particular, (by taking H = P ) , S p ( G , R ) has vertex P. (iii) If H is a subgroup of G such that P is G-conjugate to a Sylow p-subgroup of H , then (a) H O ~ R G ( ~ SRPG( G , , R ) )% H o ~ R G ( S P ( G , R~)R, G % ) R. (b) S p ( G , R ) is the unique indecomposable direct summand V of ( l R H ) G such that H o ~ R G ( ~ R G#, V 0.)

Permutation modules

450

(c) Sp(G,R) is the unique indecomposable direct summand V of ( 1 ~ ) such ' that HomRc(V,1 R G ) # 0. (d) Sp(G, R)/J(R)Sp(G, R) g Sp(G, F ) . (iv) For any given p-subgroup Q of G, SQ(G,R ) E Sp(G, R ) if and only if Q is G-conjugate to P .

Proof. (i) By Theorem 7.1 (vi), any indecomposable ppermutation RG-module V is a direct summand of ( ~ R Q ) ' for some p-subgroup Q of G. If sp(V) # 0, then by Lemma 9.l(i), (iii), it follows that P is G-conjugate to Q. Furthermore, since ~ p ( ( l R p ) = ~ ) 1, we see that V is the unique indecomposable direct summand of ( 1 ~ p ) ' such that sp(V) # 0. Hence sp(V) = 1 and, by the unicity of V , it follows from Lemma 9.l(ii) that V E V'. (ii) It follows from (i) and Lemma 9.l(i) that Sp(G, R ) is isomorphic to a direct summand of ( ~ R H if) ~and only if s p ( ( 1 ~ ~ ) '#) 0. Hence, by Lemma 9.l(iii), P is G-conjugate to a Sylow psubgroup of H if and only if Sp(G, R ) is isomorphic to a direct summand of ( ~ R H ) ' . (iii) Assume that P is G-conjugate to a Sylow psubgroup of H . By Proposition 2.2.4(i) and Corollary 2.2.7,

HomRG (IRG,

(lRP)G)

R

HomRG

((lRF')G, 1 R G )

Since Sp(G,R) 2 Sp(G,R)*, to prove (a), (b) and (c), it suffices to show that HomRG (IRG, SP(G, R ) ) # 0 (2) To this end, note that by Lemma 8.2(ii), (Sp(G,R))$ is mapped onto ( W G , R))(P)lNQ(p)'p; this last module is not zero by definition of Sp(G, R) since its dimension is precisely sp(Sp(G,R)). Hence

proving (2). Finally, let V be any p-permutation RG-module and let = V / J ( R ) V . It follows from Lemma 8.5(ii) that sp(V)= s p ( v ) . Hence, by the characterization of Sp(G, R) given above, we deduce that

v

10. The Brauer morphism and ppermutation modules

451

proving (d). (iv) This is a direct consequence of (ii). W We shall refer to the RG-module Sp(G,R) as the Scott module of G associated to P. Two extreme cases for R = F are worth noticing: SI(G, F ) is a projective cover of ~ F G while , if P is a Sylow psubgroup of G, then Sp(G,F) 2 ~ F G . The latter follows from Lemma 1.3(v) and Theorem 9.2(iii)(b), while the former is a consequence of Theorem 9.2(iii ) (c). 10. T h e Brauer morphism and p-permutation modules All the notations and conventions adopted in the previous section remain in force. Our aim is to provide a number of important properties of the Brauer morphism applied to ppermutation modules. We begin by recording two auxiliary results, both extracted from a work of Brou6 (1985). 10.1. Lemma. Let V be a p-permutation RG-module and let P be a p-subgroup of G. Then V ( P ) is a p-permutation F ( N c ( P ) / P ) module.

Proof. Let Q be a Sylow psubgroup of NG(P).Then, by Theorem 7.1, VQis a permutation RQ-module. Choose an R-basis X of V which is permuted by Q. Then the F-basis for V ( P )given by Lemma 8.5(i) is permuted by Q / P , as required. W 10.2. Lemma. Let P be a p-subgroup ofG, let V be a p-permutation RG-module, and let A = EndR(V). Then the map

where A$(a) (Br,V(s))= BrpV(a(z))

( a E AP,x E Inv(Vp),

is an isomorphism of F(NG(P)/P)-algebras.

452

Permutation modules

Proof. The natural bilinear map f : A x V G-stable. Hence, by Lemma 8.4,the map

+V

f p :A ( P )x

V(P)

, ( a , v ) H a ( v ) is

V(P)

given by

f P ( B r m , B r 3 x ) ) = B.Pv(4.N is a well defined (Nc(P)/P)-stable F-bilinear map. Thus, by Lemma 8.3, the given map is a homomorphism of F(NG(P)/P)-algebras. Let X be an R-basis of V which is permuted by P.Then the set

where uz,y E A is defined by az,y(.z)= SY,+xfor x,y , z in X is an R-basis of A which is permuted by P . Let C x ( P )be the set of fixed points of X under P. Then we obviously have

S ~ ~ , ~=( ZS g) , z ~ for all

x,y , z E C x ( P )

which, by Lemma 8.5(i), shows that the given homomorphism is an isomorphism. H We are now ready to prove the following result. 10.3. Theorem. (Broue' (1985)). Let V be an indecomposable p-permutation RG-module. (a) The vertices of V are the maximal p-subgroups P of G such that V ( P )# 0. (ii) A p-subgroup P of G is a vertex ofV if and only if V ( P )# 0 and v ( P ) is a projective F(NG(P)/P)-mod.ttle. (iii) The map V H V ( P )induces a bijective correspondence between the isomorphism classes of indecomposable p-permutation RG-modules with vertex P and the isomorphism classes of indecomposable projective

10. The Brauer morphism and p-permutation modules

453

F ( NG (P)/P)-modules. In particular, (Sp(G , R ) ) (P ) is the projective cover of ~ F ( N G ( P ) / P ) . (iv) Let M be a p-permutation RG-module, let E be an indecomposable projective F(NG(P)/P)-module, and let M ( P ,E ) be the corresponding p-permatation RG-module with vertex P . Then M ( P , E ) is a direct summand of M if and only if E is a direct summand of M ( P ) .

Proof. (i) By Lemma 8.6, it suffices to show that if P is a vertex of V , then V ( P )# 0. Suppose V is a component of ( l ~ p ) ' . Then, by Proposition 1.9, Vp is isomorphic to a direct sum of modules of type

where g runs over a certain subset S of G. But V is a component of (Vp)', and since P is a vertex of V it follows that S n NG(P)# 8. Hence 1 ~ isp a component of Vp and so V ( P )# 0. (ii) Assume that V ( P )# 0 and that V ( P )is a projective F ( N G ( P ) / P ) module. Setting A = E n d ~ ( v )it, follows from Lemma 10.2 and Theorem 2.10.3 that

But, by Lemma 8.2(ii), ( A ( P )N) G , ( p ) ' p is the image of A: under the Brauer morphism B r s . Since l v is the unique nonzero idempotent of AG, we deduce that l v E A:, proving that V is P-projective. Hence, by (i), P is a vertex of V . Conversely, assume that P is a vertex of V . Then 1" E A: and so

Thus, by Lemma 10.2, V ( P )is a projective F(NG(P)/P)-module. (iii) It follows from Lemma 8.7 that

The indecomposable p-permutation RG-modules with vertex P corresponds to the indecomposable components of (lRp)Gwith vertex P. Note also that the indecomposable projective F(NG(P)/P)-modules

Permutation modules

454

correspond to indecomposable components of ( l ~ ) ~ G ( ' ) l ~Put . A = f ? ~ ~ d R ( ( l R p ) ~Then, ). by Lemma 10.2,

A ( P ) 3 EndF ( ( 1 ~ ) ~ ~ ( as ~ F(Nc(P)/P)-algebras ) / ~ ) ) ~ vertex P corNow the indecomposable components of ( l ~ p with respond to the primitive idempotents of AF whose image in A ( P ) is nonzero. Since, by Lemma 8.2(ii), Br; maps A: onto A ( P )N,G ( P ) I P

and since AG = AS, we have

(A(P))NG(p)'p = ( A ( P ) N) ,G ( P ) / P Thus the first assertion follows from the standard facts about lifting idempotents: an indecomposable component V of ( 1 ~ p with ) ~ vertex P corresponds to a primitive idempotent e of A; such that Br$(e) # 0, which corresponds to the primitive idempotent Br$(e) of (A(P))NG(P)IP, which in turn corresponds to the direct summand

BrpA(e)

*

( l F ) N ~ ( p ) ' p=

B ~ ~ V ( V ) = V(P)

of ( ~ F ) ~ G ( ' ) / The ~ . assertion regarding Sp(G,R) follows from the fact that by definition, (Sp(G, B)(P))NG(p)Ip # 0 (alternatively, it follows from the equality S p ( G , R ) ( P )= S I ( N G ( P ) / PF, ) . (iv) Let E be any indecomposable projective F ( NG(P)/P)-module. Then, by definition, the RG-module M ( P ,E ) is the indecomposable ppermutation RG-module determined by the condition ( M ( P ,E ) ) ( P )= E . It is obvious that if M ( P , E ) is a direct summand of M , then E is a direct summand of M ( P ) . Conversely, let M be a ppermutation RG-module such that E is a direct summand of M ( P ) , and let A = E n d ~ ( A 4 )Owing . to Lemma 10.2,

A(P)

E~F(M(P))

and therefore, by hypothesis, ( A ( P ) N) ,G ( p ) / pcontains a primitive idempotent e such that e M ( P ) E E. Using the standard facts about lifting idempotents together with Lemma 8.2, it follows that A; contains a primitive idempotent f such that Br$( f) = e. Hence M is an indecomposable component of M with vertex P , and (f M ) ( P )=

-

f

a

10. The Brauer morphism and p-permutation modules

e - M ( P ) Z E , which implies proof of the theorem. W

f-M

455

M ( P ,E ) . This completes the

10.4. Corollary. (Broue' (1985)). The reduction module J ( R ) defines a bijection between the isomorphism classes of p-permutation RG-modules and the isomorphism classes of p-permutation FG-modules.

Proof. By Lemma 8.5(ii), V ( P ) E V ( P ) where V and P are as in Lemma 8.5. The desired assertion is therefore a consequence of Theorem 10.3(iii). 10.5. Corollary. (Broue' (1985)). There is a bijection between the isomorphism classes of indecomposable p-permutation RG-modules and the G-conjugacy classes of pairs ( P ,E ) , whem P is a p-subgroup of G and E is an indecomposable projective F(NG(P)/P)-module.

Proof. Apply Theorem 10.3(iii). W The next observation illustrates that V ( P ) is a familiar classical object for the case where V is an indecomposable p-permutation FGmodule with vertex P . 10.6. Proposition. (Broue' (1985)). Let V be an indecomposable p-permutation FG-module with vertex P . Then the Green correspondent ofV is the FNc(P)-module V ( P ) .

Proof. Let W be the Green correspondent of V with respect to ( G , P , N G ( P ) ) .Since, by Theorem 7.1, V has trivial source, W must also have trivial source (see Theorem 5.2.4(iii)). Hence W is isomor, P acts trivially on W and phic to a direct summand of ( l p p ) N ~ ( p )so thus W ( P ) = W . But, by Theorem 5.2.4, V N G ( p ) E W @ W ' , where W' is a direct sum of indecomposable F(NG(P))-modules with vertex strictly contained in P. It follows from Lemma 8.6 that W ' ( P ) = 0 and therefore, by Lemma S.l(iv)?

'(') as required.

= 'NG(P)(')

W?

Permutation modules

456

We close by presenting some applications of Theorem 10.3. For the rest of this section F denotes an arbitrary field of characteristic p > 0.

10.7. Proposition. (Cabanes (1988)). Let V be a p-permutation FG-moduEe and let P be a normal ~ - s u b g r o uof~ G. Let U be the direct sum of those indecomposable components of V whose vertex contains P and let W be the sum of the others. Then ( i ) V = U @ W , V ( P ) Z UandW(P)=O. (ii) Inv(Up) = U and Inv(Wp) = Ii'er(BrF). (iii) For A = EndF(V),Ii'er(Bri)= { f E E n d ~ p ( V ) l f ( U ) W } . Proof. (i) and (ii). It is clear that V = U @ W . Furthermore, by Theorem 10.3(i), W ( P )= 0 and, by the definition of U , Inv(Up) = U . Since Inv(Up) = U , we have Ker(BrE) = 0 (hence U ( P )= U), while since W ( P )= 0 we have Ii'er(Brr) = Inv(Wp). Thus

V ( P )c U ( P )@ W ( P )c U ( P )= u and

I<er(BrL) = K e r ( B r g )

Ii'er(BrpW) = ~ n v ( ~ p ) ,

proving (i) and (ii). (iii) Let Q be a proper subgroup of P and let f E E ~ F Q ( V To ). prove the "L"containment, it suffices to show that

Since u E Inv(Vp), we have T r g ( f ) ( u ) = T r % ( f ( u ) )with f ( u ) E ITZ~(VQ But ) . then

as required.

To prove the reverse containment, let e : V + U be the projection map, and let 1 - e = C7.1ei be the decomposition of 1 - e associated with a decomposition of W as a direct sum of indecomposable FGmodules whose vertices do not contain P. Let P; be a vertex of e;(W), 1 5 i 5 n. Then Pi does not contain a G-conjugate of P and e; E AFi,

10. The Brauer morphism and ppermutation modules

1 5 i 5 n. But if i E (1,. . . , n } , then gPig-' n P ei E AE

c c A;Eg-I"P

c P for all g

457

E G, so

c - Ker(Br$)

g€G

and therefore 1- e E Ker(Br$). Hence, iff E E n d ~ p ( Vwith ) f(U) C W , then efe = 0 and

f = (1 - elf

+ ef(1-

e) E

Icer(BrAp),

as asserted. 1

As an application of the above, we now prove the following result due to Puig (see Cabanes (1988)). 10.8. Theorem. Let V be a p-permutation FG-module and let P be a p-subgroup of G . Then

( E n d F ( V ) )( P )2 End=(V(P))

as

Nc(P)-aZgebms

Proof. We first note that Inv(Vp) and Ker(Br;) are stable under . the natural action of E n d ~ p ( Von ) every element of E n d ~ p ( V ) Thus V (P ) induces a homomorphism $h : End=p(V) -+ E n & ( V ( P ) ) of NG(P)-algebras. It will be shown that this homomorphism is surjec-

tive and has the desired kernel. Write VNG(p)= U @ W , where U is the direct sum of those indecomposable components of V N c ( p ) whose vertex contains P and where W is the sum of the others. We know, from Lemma 8.1(iv), that V ( P ) = v ~ ~ ( p ) ( P )Hence, . by Proposition 10.7(i), V ( P ) 2 U . It is then clear that the homomorphism 4 corresponds to the map

{ En&p(V)f

--f

I-+

E&(U) efe

where e : VNG(p)+ U is the projection map, by identifying E n d ~ ( u ) with {f E Endp(V)lf(W) = 0 and f(V) U }

c

Permutation modules

458

Hence

Icer+ = {f E End~p(U)lf(U) GW} proving, by Proposition 10.7(iii), that Ker$ = Ker(Br$) for A = hdF(V). We are left to verify that is surjective. By Proposition 10.7(ii), we have U = Inv(Up) and so E n d ~ ( u= ) E n d ~ p ( U ) .Now fix X E E n d ~ ( uand ) denote by A’ an element of E n d ~ p ( Vby ) extending X to V with 0 on W . Then it is clear that 1c, maps A’ to X and the result follows.

+

459

Chapter 8 Permutation lattices This chapter is based exclusively on an important work of Weiss(1988). We are concerned with the detailed study of permutation Z,P-lattices where P is a finite p-group and Z pis the ring of p-adic integers. One of the main results asserts that V is a permutation lattice for P provided there is a normal subgroup N of P such that VNis free and l n v ( V ~ ) is a permutation lattice for PIN. The last section is devoted t o a number of applications of the results obtained. Among them, we prove that if G is a finite nilpotent group, then Z G determines the isomorphism class of G. We also show that if A is an abelian normal subgroup of a finite group G such that G / A is a p-group, then Z G determines the isomorphism class of G. 1.

Generalized permutation lattices

Throughout this section, we fix a prime p and a finite p-group P. We denote by Z p the ring of p-adic integers and put R = Z,[E] where E is a primitive p-th root of 1. We also put

so that F is the field of p elements. Each RP-module below will be assumed to be R-free of finite rank. If V is an RP-module, then we

Permutation lattices

460

Put

v = V/(1 - &)V

Thus V is an FP-module. As usual, we write l p for the trivial R P , Z p P or FP-module (the coefficient ring will always be understood from the context). The following terminology, extracted from Weiss (1988) is nonstandard, but will be very useful in avoiding excessive verbosity. By a permutation lattice for P, we understand any permutation %$-module. Thus, by Lemma 7.1.1, V is a permutation lattice for P if and only if V is a finite direct sum of Z,P-modules of the form (1s)' for some subgroup Q of P . A generalized permutation lattice for P is defined to be a finite direct sum of RG-modules of the form xG for some homomorphism x : Q +< c > of a subgroup Q of P . Here we write xG for R P @RQ R with Q acting on R via x. By a permutation module for P (or a permutation P-module), we understand any permutation FP-module. Our aim is to prove that if V is an RP-module such that V is a permutation module, then V is a generalized permutation lattice. Let M be a generalized permutation lattice for P . Following Weiss (1988), by an eigenfactorof M , we understand a split RP-monomorphism

f:E+M where E is a generalized permutation lattice for P such that the indecomposable summands of E have vertex P and indecomposable summands of Cokerf have vertex not equal to P. Thus, if

M for some homomorphisms

$rx,

P

xr : P, +< c >, then

An eigenfactor of a permutation module for P is defined in a similar fashion. Note that if f : E --+ M is an eigenfactor of the generalized permutation lattice M for P , then the induced map f : E -+ A? is an = (1~)' eigenfactor of the permutation module A?. This is so since

1. Generalized permutation lattices

461

x

for any homomorphism : Q +< c > of a subgroup Q of P. Let V be an FP-module. Given a subgroup Q of P , we let

be the relative trace map and put

v;

= Tr;(Inv(vQ))

and

where Q runs over the set of all proper subgroups of

P

1.1. Lemma. Let Q , S be subgroups o f P and let V be an F P module isomorphic to (1s)'. Then

v;

=

{

I n v ( V ) if gSg-' Q for some g E P 0 otherwise

Proof. We may regard V as an F-space with basis gS,g E P and with P acting by g(g'S) = gg'S, g,g' E P . Let T be a full set of double coset representatives for (Q, S ) in P . Then, given t E T , there exists a set X ( t ) 2 Q such that

QtS = U,,x(t)xtS

(disjoint union)

and hence Q can be written as a disjoint union

t , constitute an F Observe that the elements b, = C z E X ( t ) ~ t SE, T basis for Inv( VQ).Furthermore, TrQp(b,)

=

TrgPnts,-l (ts) tst-1

= TrsL-IT'rQ",St-I ( t S ) = TrlpSt-l ((tSt-l : Q n tSt-*)tS) =

(t5't-l

:Q

ntSt-')Trc(S)

Permutation lattices

462

with T r c ( S )# 0 in I n v ( V ) . Since (tSt-l : Q fl tSt-') = 0 in F unless tSt-l & Q and dirnpInv(V) = 1, the result follows.

1.2. Lemma. Let V be a permutation module for P and let Q be a subgroup of P. (i) VJ # 0 if and only if V has a direct summand isomorphic to (1s)' for some subgroup S of Q . (ii) Suppose that Q is minimal with respect to V[ # 0 , and choose w E Jnv(VQ)be such that Tr;(w) # 0 . Then the map

{

(lQ>' g @ l H gv

is a split monomorphism of FG-modules. Proof. (i) By assumption, there exist subgroups PI, P2,. . . ,P, of P such that v 2 @=l(lP,)' Owing to Lemma 1.1, V$' has a one-dimensional contribution precisely from those Pi which are P-conjugate to a subgroup of Q. Because vertices are well-defined up to G-conjugacy, the required assertion follows. (ii) Write w = o1 - .. o, with wj E ( l ~ ~ 1 5) i ~5 ,r. Then

+ +

T

i=l

implies that TrG(wj) # 0 for some j E (1,. . . ,r } . Applying (i) and minimality of Q, we conclude that Pj is P-conjugate to Q. We may therefore assume that Pj = Q, in which case there is a projection $ : V (19)' with $(v) = oj. Now f is obviously an FP-homomorphism and therefore $ O f : (lQ)' (1Q)' is also an FP-homomorphism. We claim that $ 0 f is injective; if sustained, it will follow that $ o f is an automorphism, say A, of ( 1 ~ ) ' and A - l 0 $ is a desired splitting of f . Assume by way of contradiction that K e r ( $ o f ) # 0. Since P is a p-group, it follows that K e r ( $ o f) contains a nonzero fixed point of P . --f

463

1. Generalized permutation lattices

But Inv( (1~)') is one-dimensional with basis CteTt 8 1, where T is a left transversal for Q in P . Hence

a contradiction. 1.3. Lemma.

Let 8 : E + V be an eigenfactor of the permutation P-module and let Br; : I n v ( V ) + V ( P ) be the Brauer morphism. Then the map BrgoO:E+V(P) is an FP-isomorphism. Proof. Since O(E) I n v ( V ) ,the composite map Brg 0 0 is welldefined. By Lemma 1.1, if Q is a submodule of P and W = ( l ~ ) ' ,then W ( P )= 0 unless Q = P , in which case B r r : I n v ( W ) + W ( P )is an FG-isomorphism. Now there is a splitting exact sequence

O+E&V+U+O

(1)

of FG-modules. By the first paragraph and the definition of eigenfactors, we have U ( P )= 0. Hence the sequence (1) yields an isomorphism

e :E(P)

--f

V(P)

By the same argument,

BrpE : E = ~ n v (+~E)( P ) is also an isomorphism. Taking into account that Br;

the result follows. I

o

8=

e

o

Br;,

1.4. Corollary. Let V be a permutation P-module and let Q be a subgroup of P which is minimal with respect to V l # 0 . If 8 : E + VQ is an eigenfactor for the permutation Q-module VQ,then

Permutation lattices

464

Proof. By Lemma 1.3, we have

On the other hand, the minimality of Q implies that Vg = 0 for all proper subgroups S of &. Hence

T r G ( K e r ( B r F ) )=

C v,P = o

(3)

SCQ

The desired assertion is therefore a consequence of (2) and (3). W

Let G be an arbitrary group and let A be a left G-module, that is, A is an additive group on which the elements of G act from the left as additive homomorphisms such that 1 . a = a and

z(ya) = ( 2 y ) a

for all a E A , z , y E G

A map f : G + A is called a derivation (or crossed homomorphism) if

If f and g are derivations, then their sum f ig defined by

is again a derivation. It follows easily that the set Z'(G,A) of all derivations of G into A is an abelian group. The derivations fa,a E A, given by fa(x) = 2a - a (x E G) are called principal derivations and form a subgroup B'( G, A ) of 2' (G, A ) . The factor group

H'(G, A ) = Z1(G, A)/B'(G,A ) is called the first cohomology group of G with coefficients in A. In what follows, we use some standard properties of cohomology groups (see Atiyah and Wall (1967)).

1. Generalized permutation lattices

465

1.5. Lemma. (i) If V and W are generalized permutation lattices for P , then so is HomR(V,W ) .

(ii) If V and W are permutation lattices for P , then so is H o m z (V,W ) . P

fiii) I f V is a generalized pernutation lattice for P , then (1 - € ) H ' ( P ,V ) = 0

(iv) If V is a permutation lattice for P , then

H'(P,V) = 0 Proof. (i) Applying additivity of both variables, we may harmlessly assume that V = p', W = t+bp where p : Q +< E > and 1c, : S +< E > are homomorphisms for some subgroups Q and S of P. Let V* = H o m ~V, ( R ) be the contragredient of V . Then

by Frobenius reciprocity and Mackey decomposition. Thus HomR(V,W ) is a generalized permutation lattice for P . (ii) Repeat the proof of (i) with p, +, R replaced by l ~I s , Z p re, spect ively. (iii) and (iv). Applying additivity, we may assume that V = pp where cp : Q +< E > is a homomorphism and Q a subgroup of P . In voking Shapiro's lemma, we obtain

This allows us to assume that V has R-rank 1 and that P acts by p. Put I( = K e r p and note that H'(Ii', V ) = H o m ( K , V ) = 0 since V has no torsion. Thus, by the inflation-restriction sequence, we have

H' ( P ,V ) z H'( P / I - ,V )

Permutation lattices

466

with P / K acting faithfully by 'p. In particular, if 'p = 1 then P / K = 1 and H1(P,V) = 0, proving (iv). If p # 1, then P / K is cyclic of order p , say generated by 9. Thus

{w E V l ( l + g

H l ( P / K ,V)

+ +9 * * *

Choosing g such that 'p(g) = E , we have p(1- 9) = 1 - E . But then

P ( P , V)

'p( 1

q v = 0}/(1 - 9)V

+ g + . . . + gp-')

= 0 and

= H l ( P / K ,V) z V/(1 - €)V,

proving (iii). 4 1.6. Lemma. Let G be a finite group, let S be a commutative

ring and let

o-tw+xAv-to

(4)

be an exact sequence of SG-modules. Assume that (4) is S-split and let h : V -, X be an S-homomorphism such that f o h = 111. (i) The map G + Homs(V, W),g H xg where

is a derivation. (ii) The sequence derivation.

(4) is SG-split if and

only if g H

xg is a principal

Proof. (i) Given g E G and w E V, we have f ( h ( v ) g h ( g - ' v ) )= - 9 g - l ~= 0

which shows that xg(v) E W . Given z , y E G, we also have xZy(v) = h(u) - ryh(y-lz-lw) = [h(v)- zh(z-'v)]

=

+ [zh(z-'w) - zyh(y-'z-' 41

x&) + ( z c X y ) ( 4 ,

proving that g H xg is indeed a derivation. (ii) Suppose that the map g H xg is a principal derivation. This

467

1. Generalized permutation lattices

means that there exists cp E Horns(V, W ) such that all g E G. Hence

h ( v ) = gh(g-'w) = gcp(g-'v) - cp(w) and therefore, for all

for all

II

xg = gcp - cp for E V,g E

G

E V,g E G,

+

Setting X(v) = h(v) cp(w), for all w E V , it follows that X is an SGhomomorphism such that f 0 X = l v . Conversely, if X : V + X is an SG-homomorphism such that f o x = l v , then for all g E G , x g = gcp - 'p where cp E Horns(V, W ) is given by cp(v) = X(w) - h ( v ) . The proof is therefore complete. W 1.7. Lemma. RP-modules

Suppose that we are given an exact sequence of

o+w+xtv+o

(5)

such that: (i) V and W are generalized permutation lattices. (ii) The reduction rnod(1 - E )

splits. Then (5) also splits.

Proof. Put A = H o r n ~ ( VW , ) .Since all RP-modules are assumed to be R-free, the sequence (5) is R-split. Hence, by Lemma 1.6, the sequence (5) is RP-split if and only if a suitable 1-cohomology class Ax E H 1 ( P , A ) vanishes. Because A = H o r n ~ ( V , the reduction mod( 1 - e ) of (5) is similarly determined by xx = Ax E H'(P, A ) , where = 0 by virtue of (ii). Now the exact sequence 0 + A % A -+ A + 0, where a ( a ) = (1 - & ) a ,a E A , induces an exact sequence

w),

x,

H 1 ( P ,A ) t H 1 ( P ,A ) t H 1 ( P ,A )

Permutation lattices

468

It follows from (i) and Lemma 1.5 that the homomorphism H1(P,A ) --+ H'(P, A ) is injective. But AX E H1(P,A ) is mapped to xx = 0, so Ax = 0 as asserted. As a final preparatory result, we now prove the following lemma. 1.8. Lemma. Let V be a n R P - m o d u l e such that P acts faithfully o n V and trivially o n = V/(1 - E)V. T h e n P is elementary abelian.

v

Proof. We argue by contradiction. Choose P to be a minimal counterexample. Since the hypotheses of the lemma carry over to the subgroups of P,it follows that every proper subgroup of P is elementary abelian but P itself is not. For the sake of clarity, we divide the rest of the proof into four steps. Step l.Our aim here is to demonstrate that P is one of the following groups. (a) P is a cyclic of order p 2 . (b) P =< X , T J , Z ~ X = ~ y P = ZP = ~ , z z z - ' = z = YZY-',TJXY-' = 2 5 > , p > 2. Suppose that P is not of type (a). Then P contains no elements of order p 2 , so P is nonabelian of exponent p . Hence the centre 2 of P is a proper subgroup, so the pgroup P/Z has a central element zZ of order p . Since z 6 2, yxy-' # x for some y E P . But ZZ is central in P/Z,so yzy-' = ZIC for some 1 # z E 2. Because < z,y,z > is nonabelian of exponent p and order p 3 , the minimality of P forces that P =< z, y, z >. Finally, the case (b) occurs only for p > 2. Indeed, if p = 2 then z = ( y ~ =) 1, ~ which is impossible. Step 2.Here we assume that P is of type (a) or (b) satisfying the hypothesis of the lemma, where V is chosen to have minimal R-rank. Put M = I( @ R V where I( is the quotient field of R. Our aim is to show that the KP-module A4 is irreducible. Assume by way of contradiction that N is a nontrivial KP-submodule of M . Then V' = N n V , V'' = V/V' are RP-modules of smaller R-rank. The RP-exact sequence

1. Generalized permutation lattices

469

is R-split, hence its reduction mod(1 - e) remains exact. Because P

acts trivially on V , it must also act trivially on V ' a n d V . Applying minimality of V , we deduce that P does not act faithfully on V',V''. However, in both cases (a) and (b), P has a unique minimal normal subgroup Po # 1 and so Po acts trivially on both V' and V". Taking the Po-fixed points in (6)) we obtain an exact sequence 0+

v'+ I1ZD(Vpo)+ v"+ H'(P0, V')

Because H1(Po,V') = Horn(P0, V ' ) = 0, it follows from (6) that

1n0(Vpo)=

v

Thus Po acts trivially on V , a contradiction.

Step 3.0ur aim here is to eliminate the case where P =< g > is cyclic of order p 2 . The field 1((6), where 6 P = e, is totally ramified of degree p over K , and R[6]is integrally closed, hence a valuation ring. Observe also that (6- l ) P / ( 1 -E) is a unit of R[6].Now every irreducible KP-module on which P acts faithfully is of the form M j = [ K ( S ) ] ,as a I<-space, with P acting by

Hence, if M is as in Step 2, then M 2 Mj for somej, and RP acts on V via the ring homomorphism RP + R[6],sending g to 6j. Because R[6] is a valuation ring, every finitely generated R[G]-submoduleof K ( 6 ) is isomorphic to R[6].Thus V 2 [R[S]] and therefore (g - 1)V/(1-&)V

r

(Sj

- l)R[S]

This shows that P does not act trivially on V = V/(1- E ) V ,a contradiction.

Step 4.Here we complete the proof by eliminating the case where P is of type (b). Assume that P is of type (b). Since P' =< z >, there

Permutation lattices

470

are p - 1 absolutely irreducible nonisomorphic KP-modules on which P acts faithfully. Given j E (1,. . . , p - l), denote by M j the I(-space with basis { V k I k m O d p ) and P-action: ZVk

= & j V k , XVk = & - j k v k ,

yvk

= vk+1

By computing the characters afforded by these modules, one immediately verifies that M I ,M2, . . . ,Mp-l are all nonisomorphic absolutely irreducible KP-modules on which P acts faithfully. Fix j E { 1,. . . ,p - 1) and let NO C Mi be the R-linear span of { v k l k m o d p } . Then NO is an RP-module and, since z , x act trivially on No/( 1- & ) N o ,it follows that

No/(l - &)No% F P

as FP-modules

where p = P/ < z,x > and P acts on F P via P + p . The latter, together with the fact that p =< 9 > is cyclic of order p , impiies that the P-submodules of

FP

= F[y]/(y -

1)P

are all in the chain 0 = (y

- 1)P

c (y - 1 ) P - 1 c - - c (y - 1) c F P = (y - 1 ) O *

Setting

+

N; = (y - 1)'No (1 - &)No (0 5 i L p ) we thus obtain all the P-modules between No and Np = (1 - &)No. If r is any integer, say r = i p s with 0 5 i < p , then put

+

N, = (1 - &)W; Then we obtain a sequence of P-modules on M j such that for all integers r, (i) N , 3 %+I. (ii) (1 - E)N,= N,+P. (iii) ( y - l)NT= N,+,. Now take M as in Step 2. Then M E M j for some j E { 1,.. . , p - 1). Choose r to be minimal with respect to N, C V. Because P acts trivially on V , we have ( y - l)V

c (1 - &)V

471

1. Generalized permutation lattices

whence

But then iV,.-(p-l~ V , contrary to the minimality of r. This demonstrates that V cannot exist, thus completing the proof. We have now accumulated all the information necessary to prove the following result. 1.9. Theorem. (Weiss (1988)). Let P be a finite p-group, let Z p be the ring ofp-adic integers and let R = iZp[€] where E is a primitive p-th root of 1. If V is an RP-module such that V = V/(1 - E)V is a permutation module for P , then V is a generalized permutation lattice for P .

Proof. We argue by induction on [PI and n = ranlcR(V). We thus assume that the result is true for all p-groups of order < [PIand all RP-modules of R-rank < n. By assumption, there exist subgroups PI, Pz,. . . ,P, of P such that

v E @f,(lp,)P

as FP-modules

where F = Z p [ e ] / ( l - E ) Z ~ [Observe E ] . also that, since every unit of R of ppower order is in < E >, every RP-module of R-rank 1 is given by a homomorphism cp : P -+< E >, i.e. P acts on R via cp. We divide the rest of the proof into two steps.

Step 1.Here we treat the more difficult case where Pi # P for some i E { 1,. . . ,r } . Let Q be a subgroup of P minimal with respect to V$ # 0. Owing to Lemma 1.2(i) and the assumption that Pi # P , it follows that & is a proper subgroup of P . Now (VQ) = ( V ) , is a permutation module, hence by induction VQ is a generalized permutation lattice for &. Let 0 : E 4 VQ be an eigenfactor of VQ. Then, as has been VQ observed at the beginning of the section, the induced map 8 : E is an eigenfactor for the permutation module VQ. Applying Corollary 1.4, we therefore infer that

v;

= Tr;(

e(E ) )

(7)

Permutation lattices

472

We may choose an R-basis of E consisting of eigenvectors e for Q in the sense that for all q E Q,qe = cp(q)e for a suitable homomorphism ‘p :-Q +< E >. Reducing this basis modulo (1 - e ) gives an F-basis of E . Hence, applying Tr; 0 8 to this basis, we obtain a spanning set of # 0, by virtue of (7). Therefore E contains an eigenvector e for Q such that Trc(8(E))# 0 in I n v ( v ) . In what follows, we write cp : Q +< E > for the homomorphism corresponding to e . Consider the RP-homomorphism f : cpp --+ V given by

v[

If

f : ( 1 ~ ) ‘-+ 3 is the reduction of f

modulo 1 - E , then

Because Trg(g(E))# 0, it follows from Lemma 1.2(ii) that f is a split monomorphism of FG-modules. In particular, setting R = 1 - e, we have a commutative diagram

v0-

7t

-

-0

with exact rows and with K e r f = 0. Here T and - denote, respectively, multiplication by R and reduction modulo R. It follows from the snake lemma t.hat R : K e r f + K e r f is surjective and R : Cokerf 3 Coker f is injective. Because R is a valuation ring, we deduce that f is injective and Cokerf is an RP-module, say L , which is R-free of finite rank. Consequently, we obtain an exact sequence

2. Permutation lattices and normal subgroups

473

of RP-modules whose reduction modulo 7r splits. It follows that 1 is a direct summand of the permutation module and so, by the KrullSchmidt theorem, is a permutation module. But L has smaller Rrank than V , so L is a generalized permutation lattice, by the induction hypothesis. Applying Lemma 1.7, we deduce that (8) splits. Thus V L @ (pp is a generalized permutation lattice for P , as required.

v

Step 2.Here we treat the case where all Pi = P (hence P acts trivially on V ) . By induction, we may assume that P acts faithfully on V . Hence, by Lemma 1.8, P is an elementary abelian pgroup. Let K be the quotient field of R and put M = I( @ R V . Then all irreducible KG-modules are of the form K @ R (pp for a suitable homomorphism cp : P -+< E >. Hence M has such a constituent, say N . It follows that V n N is an RP-module isomorphic to 'pp and that V / ( Vn N ) is an RP-module which is R-free of finite rank. This, as in (8), yields an exact sequence 0 + 'pp -+ v -+ L + 0

whose reduction modulo 7r splits, since V is trivial. The desired conclusion now follows by the argument following (8). 2.

Permutation lattices and normal subgroups

All conventions, notation and terminology introduced in the previous section remain in force. In particular, p is a prime, P a finite p-group and Z,the ring of p-adic integers. Let N be a normal subgroup of P and let V be a finitely generated Z,P-module. Our aim is to exhibit circumstances under which V is a permutation lattice for P. Namely, we prove that V is a permutation lattice for P , provided the following two conditions hold: (i) The Z,N-module VN is free. (ii) I ~ w ( V is N a) permutation lattice for P I N . 2.1. Lemma. Let A , B and C be algebras over the same commutative ring S and let cy : A + C and p : B -+ C be algebra homomorphisms. Then there exists an 5'-algebra D , unique up to isomorphism, and there are algebra homomorphisms y : D + A , 6 : D + B such that

Permutation lattices

474

(i) the diagram D

r

-A

a

6

is commutative; and (ii) if

(2)

is a commutative diagram, then them is a unique homomorphism cp : D'+ D s u c h t h a t y o c p = y ' a n d 6 o c p = 6 ' . Proof. Given a,@,define D by

D = { ( a , b ) l aE A , ~ B E , a ( a ) = @(b)} and let y : ( a , b) H a

(3)

and S : ( a ,b) w b

be the corresponding projections. Then (1) obviously becomes commutative.

2. Permutation lattices and normal subgroups

475

Suppose that (2) is a commutative diagram, and define cp : D' + D by

cp(d') = (~'(d'),S'(d')) for all

d' E

D'

(here (y'(d'), S'(d')) E D since cu 0 7' = ,B 0 6'). Clearly, rcp(d') = ~ ' ( d ' ) and Scp(d') = S'(d') for every d' E D',and therefore 'p is of the stated kind. It follows from (3) that Kery = (0, K e r p ) and

K e r b = (Kercu,0)

(4)

Therefore, if cp' : D' + D also satisfies y 0 cp' = 7' and S o cp' = S', then

Kery n K e r b = 0. This shows that cp - cp' = 0, and thus Irn(cp - cp') and cp is unique. The uniqueness of D can be established by considering a D with the same properties; we denote the corresponding maps in (1)by 7 : D + A and 8 : D + B. Then we have unique homomorphisms cp : D + D,Cp : D + 0 , such that

Hence ycpq = y and Sp(P = 6,that is 'pp : D + D preserves both the A and B-coordinates and so cpp = 1 on D. Hence Cp(D)= Cpcp(q(D))is a direct summand of the additive group of D , D = p ( D )@ X . Since in (ii), a unique map D' + D was required for every D',X must vanish. Thus (/3 is an isomorphism and the result is established. W We next present some applications of the above lemma. It will be convenient to introduce the following standard terminology. A commutative diagram (1) satisfying (ii) is called a pullback square (or pullback diagram) of algebras. The algebra D defined in (3) is called the fibre product of the pair ( a ,p}. Observe that Lemma 2.1 is valid in case A , B and C are modules and a : A + C, ,G' : B + C are module homomorphisms. For this reason, we use the same terminology for modules, namely pullback square of modules and the fibre product of the pair ( a ,p } . The following is a typical example of a pullback square of

4 76

Permutation lattices

algebras. Let I and J be ideals of an algebra A. We construct a pullback square of algebras, where all the arrows in the diagram below represent natural homomorphisms. We leave it to the reader to verify that A / ( I n J ) is isomorphic to the fibre product of the pair { a ,p}.

In what follows, p is a prime, F is a field of p elements and R = Z,[E], where E is a primitive p t h root of 1. 2.2. Lemma. Let C =< g > be a cyclic group of orderp, let @ : Z,C t R be the homomorphism of Z,-algebras induces b y g H E , and let a : R 4 F , ,8 : !Up --+ F be reductions modulo T = 1 - c and p , respectively. Then

is a pullback square of !U,-algebms. Proof. Owing to Lemma 2.1, we need only verify that Z p C is identifiable with the fibre product of { a ,p}. Consider the homomorphism

2. Permutation lattices and normal subgroups

477

A;gi(A; E Z,), then x H (Eyzt A ; E ' , , ~ ~A;).~ Because 1 E - cP-' = 0 and l , ~. .,. , F 2is a 23,-basis of R, it follows that the given map is an injective homomorphism of Z,-algebras. Moreover, we obviously have

If

J:

=

+ +

+

: A;&; E R(A; f Z,) and let A E Z,be such Conversely, let r =::C that a ( r ) = p(X). Then A; = A(modpZ,) and so::C : A; = A+pp for some p E Z p . Setting J:

= (A,

-p)

*

1

+ (A,

-p)g

+ + * * *

(Ap-2

- p)gp-2 - pgp-l,

it follows that $(x) = r and aug(z) = A, as required. 2.3. Lemma. Let C be a cyclic normal subgroup of order p in a finite p-group P , and let V be a Z,P-module such that the Z,C-module Vc is free. Put V, = V/I(C)V,Vz = V/C+V and V3 = V/(C+V I ( C ) V ) , where C+ = CzECx and I ( C ) is the augmentation ideal of Zp C . Then

+

V

is a pullback square of Z,P-modules, where all the maps are natural horno mo rp hisms.

Permutation lattices

478

Proof. Consider the homomorphism

{

v

vz x v1

+

(r(v),W)

If ~ ( v = ) S(v) = 0, then v E C+V f~I(C)V = 0 since C+I(C) = 0. Hence the given homomorphism is injective. Moreover, we obviously have ar(z))= PS(w) for all w E V Conversely, let (v2,vl) E Vz x V, be such that a(v2) = P(v1). Denote by a l , . . . , a , a Z,C-basis of V. Because a ( q )= P(vl), we may write v2

= A1a1

and

+ - + A,a, + c+v *

*

+

+

p l a l + * * . prar I ( C ) V for some Ai,pi E Z p C such that x;=:=,(Xi-pi)ai E C + V + I ( C ) V (hence A; - pi E I ( C ) Z p C + ) .Because Z,C+ and I ( C ) are ideals of Z p C with intersection 0, there is a pullback square ~1 =

+

where all the arrows represent natural homomorphisms. Therefore there exist t l , . . . , t r in Z p C such that t i = Ai(rnodZ,C+), t; pi(rnodI(C)). Setting v = &tia;, it follows that y(v) = v2 and 6(v) = vl, as asserted. W

=

Let S be a local commutative ring and let A be an S-algebra which is finitely generated as S-module. Then, for any ideal I of A , the natural map U ( A ) + U ( A / I ) is surjective. 2.4. Lemma.

2. Permutation lattices and normal subgroups

479

Proof. Owing to Lemma 3.1.3(i), A / J ( A )is artinian. The desired conclusion is therefore a consequence of some standard facts on lifting units (see Corollary 7.1.5 in Karpilovsky (1988)). 2.5. Lemma. Let N be a normal subgroup of a finite group G and let S be a commutative ring. If V is an SG-module such that the SN-module VN is free, then

{

V/I(N)V v+I(N)V

+

I~V(VN) N+v

where N + = C I E N x, is an isomorphism of SG-modules.

Proof. If n E N and v E V , then n ( N + v ) = N+v and so N+v E 1 n v ( V ~ )Since . N + I ( N )= 0, the given map is well-defined. Moreover, it is an SG-homomorphism since N + is central in SG. Let B be an SN-basis of V . If v E I ~ ~ ( V N then ) , v = zlbl . - . x,b, for some bl,. . . ,b, in B and some zl,. . .,x, in I n v ( ( S N > N ) Since .

+ +

I ~ ~ ( ( S N= ) NS )- N+ the given map is surjective. Finally, let v E V be such that N+v = 0. We may write v = xlbl * . - xmbm for some xi E S N , b; E B , in which case N+x; = 0 for all i E { l , . .. , m } . Because N+x; = aug(z;)N+, it follows that xi E I ( N ) and hence v E I ( N ) V , as required.

+ +

We have now come to the demonstration for which this section has been developed. 2.6. Theorem. (Weiss (1988)). Let p be a prime, let Z pbe the ring of p-adic integers and let N be a normal subgroup of a finite p group P . Assume that a finitely generated ZpP-module V satisfies the following two conditions: (if The Z,N-module VN is free. (ii) Inv(VN) is a permutation lattice for P I N . Then V is a permutation lattice for P .

Proof. It will be convenient to divide the proof into two steps.

Permutation lattices

480

Step 1.Reduction to the case where N = C is cyclic of order p and central in P. Suppose that the result is established under the above hypothesis. The case IN1 = 1 being trivial, we argue by induction on IN[. Assume that 1 # N CI P satisfies (i) and (ii) and choose a central subgroup C of P of order p such that C C N . Put W = Inv(Vc),P = P / C and N = N / C . Then W is a Z,P-module and N is a normal subgroup of p . Because V is a free Z,N-module, Inv(Vc) is a free Inv((Z,N)c)-module. But, by Lemma 2.5,

Z,N g Z , N / I ( C ) Z , N g I n u ( ( Z , N ) c ) and therefore Wjq is a free Z,N-module. Because

Inv(WN) = Inv,,C(Inv(VC)) = Inv(V,>, we see that 1 n v ( W ~is>a permutation lattice for P I N = P / N . Bearing in mind that IN1 < I N ] , it follows from the induction hypothesis that W is a permutation lattice for P = P / C . Since restricting a free Z , N module V , to C trivially yields a free Z,C-module? it follows from our assumption that V is a permutation lattice for P , as required.

Step 2.Here we complete the proof by treating the crucial case where N = C is cyclic of order p and central in P . Then the hypotheses of Lemma 2.3 are fulfilled and we use the notation of that lemma. Owing to Lemma 2.5, V, Z Inv(Vc) as Z,P-modules. Thus hypothesis (ii) says that V, is a permutation lattice for P . The images of P ( V / ~ ( C ) V=)(PV W V ) / W ) V

+

and

(C+V

+ I(C)V)/I(C)V

under the isomorphism of Lemma 2.5 coincide. Hence

and therefore & / p x E V3. Thus V3 is permutation module for P (over the field F of p elemets). But now C+ annihilating Vz implies that we may consider V2 as an RP-module, via the isomorphism 1c, : Z,C/Z,Cs -+ R, since C C Z ( P ) ensures that the R and P-actions

2. Permutation lattices and normal subgroups

481

commute. Then V./rV2 E V,,where r = 1 - E , so Theorem 1.9 implies that V2 is a generalized lattice for P . By the following, we may write

V, E @

k ( l ~ ~ (as ) ~

Z,P-modules)

(5)

= @TVT

(as RP-modules) (6) for some subgroups Q k of P and some homomorphisms 9, from a subgroup P, of P into < .E >. Bearing in mind that v 2

v, E V1/pK 2 V,/rv, we deduce that @ k ( l ~ ~ ) ' 3 @,(lp,)' as FP-modules. Since our permutation modules are indecomposable, it follows from the KrullSchmidt theorem that we can re-index the k's as T ' S in such a way that (1Qr)' 2 ( I P ~ ) for all T Moreover, uniqueness of vertices up to P-conjugacy ensures that we can replace Q, by P,,i.e. we can choose our initial isomorphisms ( 5 ) and (6) as V, Z @ , ( I P , ) ~ (as Z,P-modules) (7)

v, ="

@T(PT

(as RP-modules)

(8)

Now C acts on V, via II, : Z,C + R, by the definition of the R-action, so every (P, has the same restriction II, to C. Hence Q , = Kercp, has index p in P, 2 C, Q, n C = 1. Thus P, = C x Q, for all T and therefore by Lemma 2.2, we have the Z,-algebra pullback square

On applying induction from P, to P , this becomes a P-module

Permutation lattices

482

pullback square

Letting W = mT(lQr)', as Z,P-module, the direct sum of the above squares over T can be viewed as a pullback square for Wi

with

WI = W/I(C)W,w 2 = W/C+W

and w 3

+

= W/(C+W I ( C ) W )

and all the arrows being the natural homomorphisms. By construction of W , the isomorphisms (7) and (8) may be viewed as P-module isomorphisms f : W2 4 V, and h : Wl + V, over R and Z,, respectively. We therefore obtain the following diagram with f,h isomorphisms and W,V faces pullback squares, all of P-modules (the maps fi and f are induced by h and f in the obvious way and we seek to construct A). Suppose that fi = f,i.e. there is only one arrow W3 + V,. Then the above diagram commutes and so a diagram chase shows that X exists

2. Permutation lattices and normal subgroups

483

and is uniquely determined by the diagram. Thus X : W -+ V is an isomorphism of Z,P-modules, proving that V is a permutation lattice for P.

I\

'\

I

A

h

is a P-module Because i,fare isomorphisms, the map ,f3 = f=' o automorphism of W3.We claim that there exists a P-module automorphisn p of Wl such that jl = ,B; if sustained, we simply replace h by h o p-' in the above diagram, in which case h

( h 0 $1

= i0 ($1

*

=h0

A

(h)-lo

f=f

But then the result follows by the preceding paragraph, so we are left to prove the claim. Wl/pWl. Put By construction, ,f3 is a P-automorphism of W3 E = E n d Z (Wl). Then I n v ( E ) = Endzp,(W1) and

EndFG(Wl/PWl) = fnv(End~(Wi/pW~)) = Inv(E/pE) Thus we are left to show that the ring homomorphism

is surjective on units. Because Wl is a finitely generated Z,P-module, it follows from Lemma 3.1.7 that Endzp,(Wl) = I n v ( E ) is a 22,algebra which is finitely generated as a Z,-module. Therefore, by

Permutation lattices

484

Lemma 2.4, it suffices to show that the ring homomorphism (9) is surj ect i ve . Now Wl is a permutation lattice for P, hence by Lemma 1.5(ii), (iv), H ' ( P , E ) = 0. But then the exact sequence

0 + E -+ E

E/pZ

i

+0

induces an exact sequence

0

+ I n v ( E ) + In?@) + I n v ( E / p E )+ 0

as desired. 3. Some bimodule isomorphisms

Throughout this section, unless explicitly stated otherwise, G and H denote arbitrary groups and R an arbitrary commutative ring. If V is a finitely generated (left or right) free RH-module, then any two bases of V have the same number of elements (called the rank of V ) . This is so since RG admits a homomorphism into a field. Given a positive integer n, we write M,(RH) for the ring of all n x n-matrices over R H . The unit group of M,(RH) will be denoted by GL,(RH). We say that V is an (RG, RH)-birnodule if V is a left RG-module and right RH-module such that (xv)y = x(vy)

for all x E RG, y E RH, w E V

and rv = vr

for all

T

E R, w E V

Two (RG, RH)-bimodules V and W are called isomorphic if there exists an isomorphism f : V + W of additive groups such that f ( m y ) = xf(v)y

for all

5

f RG, y E

RN,w E V

It is straightforward to verify that V E W if and only if there exists an R-isomorphism f : V + W such that

f(gvh)=gf(v)h

for all g E G, h E

H,vE V

3. Some bimodule isomorphisms

485

Observe also that if V is a left RG-module, then V can be viewed as an (RG, R)-bimodule by setting v r = rv for all r E R,v E V . Furthermore, if W is another left RG-module, then V 2 W as RGmodules if and only if V % W as (RG, R)-bimodules. Let S be a subgroup of G and let V be an (RS,RH)-bimodule. Because RG is an (RG, RS)-bimodule, RG @RS V is an (RG, R H ) bimodule, where the bimodule structure is given by t(x 8 v ) s = tx 8 vs

(t, x E RG, v E V,s E R H )

We refer to R G @ , R ~as V the induced bimodule and denote it by V G .The special case where H = 1 corresponds to the usual notion of induced module. Let V be an (RG, RH)-bimodule and let < V >= { < 2) > 1v E V } be an R-module copy of V . Then < V > can be viewed as an R ( G x H ) module by setting ( g , h )< v

>=< gvh-l >

(g E G , h E

H,v E V )

Let V and W be two (RG, RH)-bimodules. Then V S W as (RG, RH)-bimodules ifand only if< V >2< W > as R(Gx H ) -

3.1. Lemma.

modules.

Proof.

Let $ : V

+W

be an R-isomorphism such that

f ( g v h ) = gf(v)h

for all g E G, h E H , v E V

Then the map $* :< V >+< W an R-isomorphism such that $*((g, h) < v >) =< f(gvh-l)

> given by $*(< v >) =< f ( v ) > is

>=< g f(v)h-l >= (9,h)+*(< ?>) I

for all g E G , h E H,v E V . Because the argument is obviously reversible, the result follows.

A matrix representation of G over RH is a group homomorphism p :G

--f

GL,(RH)

(1)

486

Permutation lattices

for some n 2 1. We refer to n as the degree of p. Two such representations p1 and p2 of degree n are said to be equivalent if there exists A E GL,(RH) such that

Given p as in (1))define p* : RG + M,(RH) by

Then p* is a ring homomorphism. Hence, if V, is the free right RHmodule of all n x 1-matrices over RH, then the formula

defines Vp as an (RG,RH)-bimodule. We refer to V, as the (RG, RH)bimodule corresponding to p. It is a standard fact that one may treat the terms "matrix representation of G over R' and "R-free left RG-modules of finite rank" as interchangeable. However, a modification is required when R is replaced by RH. Namely, the following property holds.

3.2. Lemma. The map p H Vp induces a bijection between the equivalence classes of matrix representations of G over RH and the isomorphism classes of (RG, RH)-bimodules which are free of finite rank as right RH-modules.

Proof. Let Q and p be matrix representations of G over RH of degree n. A map f : V, --t Vp is an isomorphism of right RH-modules if and only if there exists A E GL,(RH) such that f ( v ) = Av for all v E V,. On the other hand, for any such f , f(gv) = gf(v) for all v E V,g E G if and only if P ( g ) = Aa(g)A-' for all g E G. This demonstrates that (Y is equivalent to p if and only if V, E Vp. Let V be an (RG,RH)-bimodule which is free of finite rank as a right RH-module. Because (gv)x = g(vs)for all g E G, v E V, x E RH, the map pg : V + V given by pg(v) = gv is an automorphism of V . Thus, by identifying V with the right RH-module of n x 1-matrices E GL,(RH). It follows over RH, we have pg(v) = rgvfor a unique rg

487

3. Some bimodule isomorphisms

that p : G + G L n ( R H ) ,g that V, = V , as required.

I+

rS is a representation of G over R H such

We are now ready to prove our main result, in which p denotes a prime and Z pthe ring p-adic integers. 3.3. Theorem. (Weiss (1988)). Let G and H be finite p-groups GLn(ZpH) be a matrix representation of G over Z p H . let p : G Define p* : G + G L n ( Z p )b y composing p with the homomorphism GL,(Z,H) + GLn(Zp) induced b y the augmentation map. Let M be the (Z,G,Z,H)-bimodule corresponding to p and let N be the Z,Gmodule corresponding to p*. If

for some subgroups GI,. . . , G, of G, then ME@T r=l

M piG

as (Z,G, Z,H)-bimodules

for suitable homomorphisms pi : G; + H , where Mpi is the ( Z p G i ,Z , H ) bimodule corresponding to p;.

Proof. By definition, M is viewed as n x 1 column vectors with entries in Z p H on which G acts by (left) matrix multiplication via p and Z , H acts by (right) scalar multiplication. Let K =< M > be the Z p ( Gx H)-module defined by (2,y

)

< M >=< p(x)rny-l >

(z E

G, y E H , rn E M )

Then, by Lemma 3.1, the isomorphism classes of the ( Z p G , Z p H ) bimodule M and the Z p ( Gx H)-module < M > determine each other. Put P = G x H and L = 1x H . Because M is a free right Z,H-module, A'L is a free Z,L-module. Therefore, by Lemma 2.5, K / I ( L ) K Inv(l{~)as Z,P-modules. Now I% N as Z,(P/L)-module, via the identification P / L = G induced by the

projection G x H

--+

G. It follows, from our hypothesis on N , that

K/I(L)I< E $i==l(lGi)G

via

P/L =G

(2)

Permutation lattices

488

Invoking Theorem 2.6, we deduce that

K 2 $$=l(lpj)p

(3)

(Pjis a subgroup of P) is a permutation lattice for P.

Because L a P and KL is a free Z,L-module, it follows from (3) and Mackey decomposition that

Pjn L

for all j E { I , .

=1

. . ,k}

(4)

Observe also that

and

Z P ( P / L )@ Z p P ( l P J P = %(P/L) @ Z p P ( Z P P @ Z p P 3%> Z P ( P / L )@ Z P P j ZP

= 2

PIL Iq LIL

(by (4))

Applying (2) and (3) and using identification P / L = G, we conclude that G @:=l(lG,)G

(lpr~(P>))

Invoking the Krull-Schmidt theorem, we have r = k and, after suitable renumbering, (lGi)'

= ( l V G ( p i ) )G

for all i E ( 1 , . . . , r }

Moreover, replacing P;by a P-conjugate, we may assume that prG(Pi)

= G;

for all i E { 1 , ..., r }

(5)

Therefore, by (4)and ( 5 ) , composing the inclusion Pi 4 P with the projection p r G induces an isomorphism P;--+ G;. Thus, for each z E G;, there exists a unique p;(z) E H such that ( z , p ; ( z ) ) E Pi. It follows that p; : G; --+ H is a group homomorphism such that

4. Applications

489

Let Mp, be the (Z,G;, Z,H)-bimodule corresponding to pi. Then the induced (Z,G, Z,H)-bimodule ME corresponds to the Z p ( G x H ) module < M f > satisfying

with P, determined by (6), as is easily verified. Consequently, by (3) and the fact that k = T ,

(e3L1M.)

=

e3;=l(lPi)p

E

< M >

as Z,(G x H)-modules. Invoking Lemma 3.1, we finally deduce that

as ( Z p G ,Z,H)-bimodules, thus completing the proof. W

4.

Applications

In this section, we apply Theorem 3.3 to obtain some results pertaining to the isomorphism problem for group rings described below. Let RG be a group ring of a finite group G over a commutative ring R. There has been a considerable amount of work over the years dedicated to the following problem: To what extent does RG determine the group G? A favourite gambit of group ring theorists has been to impose some conditions on the ring R in the expectation that RG determines G up to isomorphism. There is a striking example of Dade (1971) of two nonisomorphic metabelian groups G and H such that such that for all choices of the field R , RG and RH are R-isomorphic. Therefore, generally speaking, a field is not a suitable candidate for R. The rings R for which the group ring RG yields the most information on the structure of G are integral domains of characteristic 0 in which no rational prime divisor of the order of G is invertible. Typical examples of such rings are as follows: (a) The ring R of algebraic integers in some finite extension of the rationals (in particular, the ring Z of rational integers). (b) The ring Z(n)= { a / b J a ,b E Z , b # 0,n = /GI, (b,n) = l}.

Permutation lattices

490

(c) The rings 22, and Z(,in )case G is a p-group. Of special interest is, of course, the case where R = Z . The isomorphism problem in this case may be stated as follows: The isomorphism problem. Is it true that Z G 2 Z H implies

G Z H? To present some results pertaining to the above problem, we first record a number of preliminary observations. A homomorphism from the group ring RG1 to the group ring RG2 is, by definition, a ring homomorphism which is also an R-homomorphism. The augmentation ideal I ( R G )is the kernel of the homomorphism from RG to R induced by collapsing G to 1. Explicitly, I ( R G ) consists of all 2

=L 9 ,

x9 E R,g E G

for which

aug(x) = E x g = 0 We shall write I ( G ) instead of I ( R G ) when there is no danger of confusion. A unit u in RG is called trivial (respectively, normalized) if u = u g g for some ug E U ( R ) ,g E G (respectively, if aug(u) = 1). A normalized group basis of RG is, by definition, a group basis consisting of normalized units. We shall write RG = R H to indicate that H is a normalized group basis of RG. Note also that if H1 is another group basis of RG, then

RG = RH

where H = {aug(t-l)tlt E H I }

and H E HI. Therefore the isomorphism problem may be stated as follows: Does R G = RH imply G Z H ? Let J be an ideal of RG. Then the multiplicative kernel of the map G RGIJ is G n (1 t J ) = { g E GJg- 1 E J } --f

and G + J will denote the image of G under this map. In other words,

G + J = ( g t J J gE G }

4. Applications

491

With this notation, we obviously have

G / ( G n (1

+J))2 G +J

(1)

Finally, note that if R G = R H then aug = aug' where aug' is a homomorphism RG + R induced by collapsing H t o 1. Hence I ( G ) = I ( H ) and every unit normalized with respect to G is also normalized with respect to H . 4.1. Lemma. Let a be an algebraic number and let n be a natural number such that na is an algebraic integer. If (a1 = a,a2,. . . ,at} is the set of all Q -conjugates o f a , then either a is an algebraic integer or in the ring 5% [al,a2, . . . ,at]at least one rational prime divisor of n

is invertible. Proof. Assume that a is not an algebraic integer. Then there exists an elementary symmetric function f of t variables such that f (a1,. . . at) @ i2 . Since na is an algebraic integer )

f ( a l , a 2 ., . , a t )= a j b

for some a , b E Zt

such that ( a ,b) = 1,b > 1 and all prime divisors of b are divisors of n. If p is one of these divisors, then because of (a,p) = 1, there exist c, d E Z such that ac dp = 1. It is clear that

+

and hence

a l p E 5%

[a19a2,

- ,at] *

But then

as required. H

4.2. Theorem. (Saksonov (1972)). Let R be an integral domain of characteristic 0 in which no rational prime divisor of IGl is a unit. If u = Cu,g(u, E R,g E G ) is a unit of a finite order in R G with

Permutation lattices

492

u1 # 0, then u = u1 . 1. In particular, all central units offinite order in RG are trivial.

Proof. Let z = C z g g ( z gE R,g E G) be a central unit of finite order. Then zt # 0 for some t E G and therefore u = zt-' is a unit of finite order with u1 = zt # 0. Hence it suffices to show that if urn= 1 and u1 # 0 then u = u1 1. To prove the required assertion, let t r ( z ) be the trace of z E RG in the regular representation of RG. Then the matrix of u is conjugate to diag(e1,. . . ,E,), n = IGl, and

-

-

t r ( u ) = u1 n = E~

+ + . - .+ ~2

E~

where E ; , 1 5 i 5 n , is an rn-th root of unity (belonging to a sufficiently large field containing R). By looking at the tr(u') where ( r , m ) = 1 , we deduce that the set {PI = u l ,p2,.. . ,pa}of all (Q -conjugates of u1 is a subset of R and thus

Because lGlu1 is an algebraic integer, Lemma 4.1 may be employed to infer that u1 = (el . e n ) / n is an algebraic integer. Since u1 # 0, the latter implies ~1 = e2 = ... = cn, hence the result. 1

+ +

4.3. Proposition. Let R be an integral domain of characteristic 0 in which no rational prime divisor of [GI is a unit. (i) I f H is a torsion subgroup of normalized units of RG, then H is a linearly independent set (in particular, H is finite) and the order of H divides the order ofG. (ii) Let RG = RH, let N be a normal subgroup of G and let N* = H n ( l + R G - I ( N ) ) .Then R G . I ( N ) = R G . I ( N * ) , I N I = IN*[ and

R(G/N) R(H/N*) (iii) If RG = RH and G is p-nilpotent, p a prime, then H is p nilpotent and RP RQ where P and Q are Sylow p-subgroups of G and H , respectively.

4. Applications

493

Proof. (i) Assume by way of contradiction that C;"=lcvihi= 0 where hi E H , a ; E R,i E (1,...,n } and cuj # 0 for some j E { l , ..., n } . Then aj - 1 = -C'Y;(h;h;') i#j

and if we express the elements hihj', i # j , in terms of the elements of G then at least one of them, say hkhj', k # j , has a nonzero coefficient of 1. Hence, by Theorem 4.2, h&J' = a * 1 for some cu E R. Since a u g ( h k ) = aug(hj) = 1 we have h k = hj, a contradiction. Put e = IHJ-' ChEH h. Then e is obviously an idempotent of FG, where F is the quotient field of R. For any 2 = Cz,g E RG, put t r ( z ) = xl. If h # 1, then since aug(h) = 1, we have t r ( h ) = 0 by Theorem 4.2. Thus ire = /HI-'. On the other hand, it is easy to see that t r ( e ) = IGI-'rn for some rn 2 1. Hence /GI = lHlm as asserted. (ii) Let RG = R H and let T : RG + RG, G = G / N , be the natural homomorphism. Then, by (i), the image H of H in RG is a linearly independent set in RG which obviously spans RG, hence RG = RH. Moreover, since T can be regarded as the extension of the epimorphisn H + H (whose kernel is N * ) by R-linearly, we have

K e r T = RG . l ( N ) = RH I ( N * ) Furthermore, since IG/NI = [GI = = IH/N*I, we have IN1 = IN* which proves the required assertion. (iii) Let N be a normal p'-subgroup of G such that GIN 2 P . Then, by (ii), N* Q H is such that [NI = IN*] and H / N * 2 Q, where Q is a Sylow p-subgroup of H . Since, by (ii), R ( G / N )E R ( H / N * ) ,the result follows. 4

Let R be an integral domain of characteristic 0 in which no rational prime divisor of /GI is a unit. If S is a subgroup of G , then for I ( G ) = I ( R G ) and I ( S ) = I ( R S ) , 4.4. Lemma.

G n (1 + I ( G ) I ( S ) )= S' *

Proof. We first consider the special case where S = G. By taking the case n = 2 in Theorem 2.1 of Sandling (1972) we see that

s n (1 + I ( S ) ~=)S'

Permutation lattices

494

whenever Tp(SmodS') = S for all primes p for which peR = p"+'R for some nonnegative integer e(Tp(SmodS') denotes the subgroup of S generated by all elements of S some p t h power of which is in 9).It is clear that Tp(SmodS') = S'

if p does not divide 15'1. If p is a prime such that p"R = pe+lR for some e, then p"( 1- p z ) = 0 for some z E R and since R has no zero divisors, p is a unit of R. This shows that p does not divide IS1 and completes the proof of the special case. To prove the general case, let T be a left transversal of S in G containing 1 and let g = t s be a typical element of G ( t E T ,s E S). Consider the R-linear map

(p:RG+RS which is the R-linear extension of g z E I ( S ) and the equality

H

s. Then p(z) = z for any

shows that cp(I(G) I ( S ) )= I(S)'. Consequently, a

I ( G ) I ( S ) n I ( S ) = I ( S)' which obviously implies

+

G n (1 I ( G ) - I ( S ) )= S n (1 + I ( S ) ' ) , thus completing the proof by applying the special case established above. H

Let m be the exponent ofGIGI and let R be any commutative ring such that R/mR Z / m Z . Then, for any t E I(RG), there ezists g E G szlch that t = g - l(rnodI(RG)2). 4.5. Lemma.

Proof. The identities

+

+

~b - 1 = ( a - l ) ( b - 1) ( a - 1) ( b - 1) ( a , b E RG) [ a ,b] - 1 = a-lb-' [ ( a- l ) ( b - 1) - ( b - l)(a - l)] ( a , b E U(RG))

4. Applications

495

show that the map p : G -+ I ( R G ) / I ( R G ) 2given by

+

‘ p ( g ) = ( g - 1) I ( R G ) 2

for all g E G

is a homomorphism with G’ Kerp. Since gm E G‘ for any g E G, we have I ( R G )= ~ ‘ p ( g m ) = mcp(g)= m(g - 1) I ( R G ) ~

+

and therefore mr(g - 1) E I(RG)2for all r E R,g E G. Since a typical element of R is r = t - 1 mrl where rl E R and t E {0,1,. . . ,m - l}, we have p ( g t ) = r(g - 1) I ( R G ) ~

+

+

This shows that

‘p

is surjective and completes the proof.

4.6. Lemma. Let A be an abelian normal subgroup of G and let R be an integral domain of characteristic 0 in which no rational prime divisor of [GI is a unit. Assume further that R/mR E Z / m Z where m is the exponent of A . If H is a normalized group basis of RG and the image of each h E H in R ( G / A ) is a trivial unit, then G E H .

Proof.

By Lemma 4.4,we have

+

G n (1 I ( G ) I ( A ) )= 1 and therefore, by (I), G + I ( G ) . I ( A )E G. If A* is the normal subgroup of H corresponding to A (see Proposition 4.3(ii)), then RG . I ( A ) = RH I ( A * ) and hence I ( G ) * I ( A ) = I ( H ) . I ( A * ) . The latter (as the reader may easily verify) implies

H n ( l + I ( G ) . I ( A )= 1 It follows that H

+ I ( G )- I ( A )Z H . H

It will next be shown that

+ I(G) I(A)C G + I(G) *

which will complete the proof. Fix h E H . By hypothesis, h Since

*

I(A),

= g(modRG. I ( A ) )for some g

+

RG I ( A ) = I ( A ) I ( G ) * I ( A ) *

E G.

Permutation lattices

496

= +

-

we have h g t ( m o d l ( G ) I ( A ) )for some t E I ( A ) . Furthermore, by Lemma 4.5, t G a - l(modl(A)’) for some a E A. Hence

h

= g + ( a - 1 ) = (1 - g ) ( a - 1) + ga G g a ( m o d l ( G )- I ( A ) ) ,

thus completing the proof.

4.7. Corollary. (Whitcomb (1968)). Let G be a finite metabelian group. Then ZIJG determines the isomorphism class o f G .

Proof. Apply Lemma 4.6 and Theorem 4.2. From now on, we concentrate on some applications of Theorem 3.3, which will provide us with another class of groups determined by their integral group rings.

Lemma. Let R be a commutative ring and let G and H be finite groups such that G is a subgroup of U ( R H ) . If V is the 4.8.

(RG,RH)-bimodule corresponding to the inclusion G + U ( R H ) , then G is conjugate to a subgroup of H if and only i f there exists a homomorphism u : G + H such that V E V, (equivalently, b y Lemma 3.1, < V >E< V, > as R(G x H)-modules). CY : G t H is a homomorphism such that V,. The inclusion G --t U ( R H ) and CY : G t U ( R H ) are matrix representations of G over R H of degree 1. Hence, by Lemma 3.2, these representations are equivalent. Thus there exists a unit u E R H such that a ( g ) = ugu-* for all g E G. This shows that G is conjugate to the subgroup u(G) of H . Conversely, assume that there exists a subgroup S of H and u E U ( R H ) such that S = uGu-l. Define u : G + H by u ( g ) = ugu-l for all g E G. Then u : G + U ( R H ) and the inclusion G + U ( R H ) are equivalent matrix representations of G over R H . Hence, by Lemma 3.2, V E V, as required. 4

Proof. Suppose that

V

Z

Let G be a finite group and let R be a commutative ring. The set of all normalized units of RG constitutes a subgroup of U ( R G ) which

4. Applications

497

we denote by V ( R G ) . Since U(RG) = U ( R ) x V ( R G ) ,the study of U(RG)is equivalent to that of V ( R G ) .

(Weiss (1988)). Let p be a prime, let Z,be the ring ofp-adic integers and let P be a finite p-group. If Q is a finite subgroup o f V ( Z , P ) , then 4.9. Theorem.

uQu-'

P

for some u E V ( Z p P )

Proof. Owing to Proposition 4.3(i), Q is also a pgroup. We now apply Theorem 3.3 for n = l , Q = G , P = H and p : Q + V ( Z , P ) being the inclusion map. Then N = l ~so, T = 1,G; = G and hence, by Theorem 3.3, M E Adpl. Thus, by Lemma 4.8, zQ2-l P for some 5 E U ( Z , P ) . Setting u = aug(z-l)x, the result follows. 4.10. Corollary. (Roggenkamp and Scott (1987')). Let p be a prime, let G be a finite p-group and let H be a subgroup of V ( Z p G ) with \HI = /GI. Then H = u-lGu for some u E V ( Z , G ) .

Proof.

This is a direct consequence of Theorem 4.9.

4.11. Remark.The above result was established by Roggenkamp and Scott under more general circumstances. Namely, instead of Z p one can take a local or semilocal Dedekind domain of characteristic 0 with a unique maximal ideal containing p . a

4.12. Corollary.

Let G be a finite nilpotent group. Then

ZGZ ZH

implies G % H

Proof. If G is a finite p-group, then the result follows from Corollary 4.10. The general case follows from this special case by applying Proposition 4.3(iii). 4.13. Theorem. Let A be an abelian normal subgroup of a finite group G such that GIA is a p-group for some prime p. If Z G % ZH , then G E H .

498

Permutation lattices

Proof. Let n = (GI and R = { a / b l a , b E Z , b # O,(b,n) = 1). Then RG E RH and we show that G E H . We may harmlessly assume that H is a normalized group basis of RG. It is clear that R is an integral domain of characteristic 0 in which no rational prime divisor of n is a unit. Moreover, for any natural number rn dividing n, R / m R E Z/ m Z . Consider the natural map f : RG + R(G/A). Then, by the proof of Proposition 4.3(iii) RG = RH, where G = G / A and H = f ( H ) . Hence, by Remark 4.11, H = u-lGu for some u E U(RG). Since R is semilocal, there exists a unit v E U(RG)such that f(v) = u. Hence, replacing H by uHu-', if necessary, we may assume that H = G. The desired conclusion is now a consequence of Lemma 4.6. We close by remarking that Roggenkamp and Scott (1987) announced that any finite abelian-by-nilpotent group is determined by its integral group ring. A more general version of Theorem 4.13 can be found in Roggenkamp and Scott (1987, Corollary 5 ) .

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510 Tucker, P. A. (1962 )

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511

Not ation N u m b e r Systems

the natural numbers the rational integers the padic integers the rational numbers the real numbers the complex numbers the integers mod rn Set theory C

E

1x1

x-Y

proper inclusion inclusion the cardinality of the set X the complement of Y in X

Number theory a divides b the greatest common divisor of a and b the p-part of n the p’-part of n

512

Not a tion

Group Theory the multiplicative group of a field F the subgroup generated by X the cyclic group of order n direct product of GIand G2 N is a normal subgroup of G the centralizer of X in G the normalizer of X in G - x-l -1 Y XY the commutator subgroup of G the centre of G = {9"l9 E GI = {g E Glg" = 1) the group of all nonsingular linear transformations of V '

the the the the

image of J: under the action of g E G stabilizer of J: E X automorphism group of X p-component of A

Rings and modules the F-dimension of V projective cover of V the opposite ring the centre of R tensor product the characteristic of R direct sum direct product of rings polynomial ring on X over R the Jacobson radical of a ring R the radical of an R-module V the submodule of invariant elements

Not at ion

VH VG

513 the restriction to H induced module relative trace map dual module contragredient module outer tensor product the algebra of R-endomorphisms of V the principal ideal generated by r n x n-matrices with entries in R

=L@Fv =L@FA the Loewy length of V the annihilator of M G-conjugate of V the intertwinning number inflated module extension of V the matrix unit n-th direct power of V support of z the g-component of A the crossed product of G over R the unit group of A the degree of u graded units of A the inner automorphism induced by u crossed system the class sum the sum of all elements in X the trivial RH-module = a for all h E H} = { u E A1 = Trg(AH) right annihilator of X left annihilator of X

514

Not ation the group algebra of G over R the augmentation ideal of RG the group of normalized units of R G the augmentation map the socle of V = TrE(Inv(VH)) = WVK)/(CHCK VHK J ( R ) W V K ) ) the Brauer morphism the Martindale ring of quotients of S the twisted group algebra of G over F the block containing e the commutator subspace HI is G-conjugate to a subgroup of H2 HI is G-conjugate to Hz the defect group of e

+

Cohomology theory the group of all A-valued 2-cocycles of G a coboundary the subgroup of ZZ(G,A ) consisting of all = Z2(G,A)/B2(G, A ) coboundaries the cohomology class of f = Z'(G, A)/B'(G,A) the corestriction map the restriction map obstruction cocycle inflation map Field theory field extension degree of F over I< Galois group of F over Ii the smallest subfield containing S and Ii the smallest subfield containing Ii and Ql,.

. . ,Q,

Notation

515 the algebraic closure of F the norm of p over F the trace map

516

Index algebra 2 definable over a subfield 349 graded 142 homomorphism of 2 strongly graded 142 Alperin 299,330,333,337,338,383, 428, 430, 432,448, 449 annihilator 22 A sano-0sima-Takahasi theorem 179 Atiyah 464 basis 3 bilinear form 210 associative 21 1 nonsingular 21 1 symmetric 211 bimodule 3 Blackburn 42, 43,210, 427 block 344 block idempotent 344 Bourbaki 5 Brauer 332, 337, 358 Brauer’s induction theorem 332 Brauer-Jennings-Zassenhaus series 42 7 Brauer lift 337 Brauer morphism 439 BrouC 383,400, 402,404,439,448, 451, 452, 455 Burnside 44

Burry 314, 315, 317, 448 Cabanes 456,457 canonical injection 55 Carlson 314 Cauchy sequence 132 centralizer 3 character 40 degree of 40 module of 40 characteristic polynomial 222 Clarke 268 class function 40 Clifford 75, 245, 257 Clifford’s theorem 75 Cline 164 coboundary 167 cocycle 167 Cohen 186 cohomology class 167 cohomologous cocycles 167 crossed homomorphism 464 commutator subspace 174 complete discrete valuation ring 186 composition length 13 conjugate characters 101 conjugate modules 74 Conlon 153, 184 control 404 crossed product 145

INDEX crossed system 146 Curtis 42, 102, 222, 223, 358, 436 Dade 141, 149, 153, 229, 235, 263, 489 derivation 464 Dickson 118 direct product 2 disjoint modules 93 Dornhoff 42, 191 Dress 383, 433, 435, 437, 438 eigenfactor 460 extension of domains 186 extension of modules 154 Feit 191 fibre product 475 finite extension 186 first cohomology group 464 Fong 265 Frobenius 422 Frbenius algebra 212 Frobenius complement 422 Frobenius kernel 422 Frobenius reciprocity 65 Frobenius reciprocity for characters 66 fully invariant submodule 18 G-algebra 386 generalized permutation lattice 460 generalized I<-character 332 G-grading 233 Gorenstein 422 graded homomorphism 148 graded submodule 233 Green 191,299,311,341, 358,363, 388

517 Green correspondence 311, 313 group 3 centre of 3 cyclic 4 Frobenius 422 multiplicative 3 p-nilpotent 406 p-radical 416 p-solvable 262 group algebra 37 Harris 209, 225, 228, 232, 240, 242 head 64 Hecke algebra 393 Higman 110, 361 Hochschild 361 Huppert 42, 43, 210, 253, 427 I-adic topology 132 ideal 1 G-invariant 149 graded 148 nil 1 nilpotent 1 primitive 22 i dempot ent 1 centrally primitive 1 orthogonal 1 primitive 1 induced bimodule 485 induced character 52 induction 45 inflation map 245 intertwinning number 63 invariant element 83 invariant submodule 83 Jacobson 26, 135, 284

518 Jennings 428 Jordan-H6lder theorem 12 Karpilovsky 102,209,271,272,331, 427, 479 Kharchenko 199 Khatri 418, 419, 425 KnGrr 341, 358,363,365,368,372, 374, 379, 380 Krull-Schmidt theorem 122 Kiilshammer 275 Lang 220, 224 left transversal 3 lift 335 linked modules 347 Loewy length 266 Loewy series 266 Lorenz 203, 205, 207 lower Loewy series 427 Mackey 90, 91, 93 Mackey decomposition 90 Mackey intertwinning number theorem 93 Mackey tensor product theorem 91 Martindale 194 Martindale ring of quotients 194 Maschke’s theorem 41 matrix units 29 maximal submodule 3 metric space 132 minimal prime ideal 192 module 2 absolutely indecomposable 186 artinian 8 coinduced 56 completely reducible 15

INDEX contragredient 84 dual 79 faithful 22 finitely cogenerated 8 finitely generated 8 free 3 graded 223 H-injective 108 homogeneous 17 homogeneous component of 17 H-projective 107 imprimitive 47 indecomposable 2 induced 46 irreducible 2 kernel of 47 noetherian 8 permutation 97, 383 p-permutation 433 p-local 330 principal indecomposable 15 projective 7 separable 348 strongly graded 233 strongly indecomposable 120 totally indecomposable 180 transitive permutation 383 trivial 373 Motose 416, 417, 419, 424, 425 multiplicity 63 Nakayama 24, 65 Nakayama’s lemma 24 natural projection 67 nilpotent element 1 Ninomiya 416, 417, 425 norm map 222

INDEX normalized group basis 490 normalized unit 490 normalizer 3 Okuyama 426 orthogonality relations 43 Osima 53, 102, 352 outer tensor product 87 Passman 203, 205, 207, 352 permutation G-algebra 387 permutation lattice 460 p-element 3 p’-element 3 plocal subgroup 330 p-regular conjugacy class 3 p-regular element 3 prime ideal 192 principal block 373 principal derivation 464 projective cover 123 projective matrix representation 243 pullback square 475 projective representation 243 proper submodule 3 psingular conjugacy class 3 Puig 439, 448, 457 radical 19 of module 19 of ring 19 ramification index 186 reduced characteristic polynomial 222 Reiner 42, 102, 222, 223, 358, 436 relative trace map 104 representation 38 completely reducible 39

519 indecomposable 39 irreducible 39 tensor product of 41 restriction of modules 45 Reynolds ideal 277 right transversal 3 Ring 1 artinian 8 associative 1 centre of 1 complete 134 local 116 noetherian 8 opposite 1 semisimple 19 socle of 218 Robinson 273, 283, 284, 287, 288, 289,292,295,296,383,400, 402, 404, 406 Roggenkamp 497, 498 Saksonov 383, 413, 414, 491 Sandling 493 S-cancelable ideal 203 Schur’s lemma 34 Scott 383, 392, 448, 449, 497, 498 Scott-Alperin theorem 449 Scott coefficient 448 Scott module 451 second cohomology group 167 simple induction pair 365 simple restriction pair 365 skew group ring 145 socle 64 source 301 stabilizer 4 superfluous submodule 21

520 support 38 symmetric algebra 212 system of imprimitivity 47 Swan 263, 435 tensor product 4, 87, 246 Thivenaz 294 Thompson 422 trace map 222 transitive action 4 transitivity of induction 54 trivial unit 490 Tsushima 425, 426 twisted group algebra 145 twisted group ring 145 unique decomposition property 122 unit 24 left 24 right 24 unramified extension 186 upper Loewy series 427

INDEX vertex 301 virtual lift 335 Wall 464 Ward 139, 153 Ward-Willems theorem 139 weak division 153 weakly G-invariant module 153 weakly isomorphic modules 153 weight 428 Wedderburn- Artin theorem 35 Wedderburn's theorem 36 Weiss 459, 460, 471, 479, 487, 497 Willems 139, 253, 265 Witt-Berman theorem 43 Whitcomb 496 X-inner automorphism 199 X-projective module 318