:(Q)f
} tll1lite G;n:iloups
Walter Feit
CHARACTERS OF FINITE GROUPS
MATHEMATICS LECTURE NOTES
Paul J. Cohen stanford University
Set Theory and tile Continuum Hypothesis
Walter Feit Yale University
Characters of Finite Groups
Marvin Greenberg Northeastern University
Lectures on Algebraic Topology
Serge Lang Columbia University
Algebraic Functions
Serge Lang Columbia University
Rapport sur la Cohomologie des Groupes
Jean-Pierre Serre College de France
Algebres de Lie semi-simples complexes
Jean-Pierre Serre Coll~ge de France
Lie A 1gebras and Lie Groups
CHARACTERS OF FINITE GROUPS
WALTER FEll' Yale University
W. A. BENJAMIN, INC.
New York
Amsterdam
1967
CHARACTERS OF FINITE GROUPS
Copyright © 1967 by W. A. Benjamin, Inc. All rights rese rved Library of Congress Catalog Card Number 67-20769 Manufactured in the United States of America
The manuscript was put into production on September 20, 1966; this volume was pUblished on October I, 1967
W. A. Benjamin, Inc. New York, New York 10016
PREFACE These notes contain the material covered in a course I gave at Yale University during the academic year 1964-1965. A knowledge of linear algebra, Galois theory, and elementary group theory is the only background required of the reader. The primary aim of this book is to familiarize the reader with some of the methods which have proved fruitful in current research in that aspect of group theory which uses the theory of characters. These notes are not meant to replace any of the textbooks on the subject but rather to supplement them. Some overlap with the many group theory books in circulation is. inevitable but I have tried to keep it to a minimum. In Chapter I representations and characters are defined and their basic properties are developed. Nowadays, this is usually done by way of the theory of algebras. However, these topics are here approached from a more classical point of view. This is done partly to illustrate the elementary nature of the material but mainly to derive the basic properties of characters as rapidly as possible. Chapter II centers about R. Brauer's fundamental theorem concerning the character ring of a finite group and some of its generalizations. Several applications are given including some concerned with splitting fields and the Schur index. Chapter III contains various criteria for a group to be nonsimple. Included are several classical results of Burnside and Frobenius. P. Hall's characterization of solvable groups, and J. G. Thompson's criterion for a group to have a normal p complement for an odd prime p. Many of the results in this chapter and their generalizations are of basic v
vi
PREFACE
importance for any further study of the structure of finite groups. As is well known, most of these results can be proved by using the transfer in place of character theory. The material in Chapter IV is mainly of recent origin. Several disconnected topics are discussed. These are meant to provide a random sample of some of the work that has recently been done in this area. Most of the proofs in this chapter utilize the concept of a trivial intersection set whose importance for character theory was first recognized by R. Brauer and M. Suzuki. Some generalizations of this concept are also treated together with the related concept of coherence. Several people who listened to these lectures made helpful comments which were incorporated in these notes. I especially want to express my thanks to G. Seligman and F. Veldtkamp for their suggestions. I am also greatly indebted to Mr. Leonard L. Scott whose critical scutiny of the material prevented several errors from appearing in print. Walter FeU New Haven, Connecticut March, 1967
CONTENTS Chapter I
91. §2. §3. §4. §5. §6. §7. §8. §9. §10.
Representations Characters Complex Representations Integral Representations The Center of the Group Algebra Some Properties of Characte~s Character Tables Induced Representations Induced Characters M-Groups
1 10 18 23 27 32 41 43 45 58
Chapter II
§11. §12. §13. §14. §I5. §16. 917.
The Schur Index A Combinatorial Result and Some Consequences Rational Valued Characters ;;-Elementary Groups The Character Ring Schur Indices and Splitting Fields Equations in Groups
61 67 69 71
78 85 87
Chapter III
§18. §I9. §20. §21. §22.
Criteria for Solvability Quotient Groups Nonsimplicity Criteria Normal 'IT-Complements Thompson Subgroups vii
93 98 103 113 117
CONTENTS
viii
Chapter IV §23. §24. §25. §26. §27. §28. §29. §30. §31. §32. §33.
T.!. Sets Characters of Relatively Small Degree Frobenius Groups An Excursion into Number Theory CN Groups Nonsimplicityof Certain Groups of Odd Order Properties of Involutions Groups with Quaternion S2-Subgroups Coherence A Class of Doubly Transitive Groups Isometries
123 125 133 139 142 147 152 156 158 167 172
Notation
181
References
183
Index
185
CHARACTERS OF FINITE GROUPS
CHAPTER § 1.
REPRESENTATIONS
Let N be a group and let If be a field. An If-representation ~ of (~is a homomorphism of (\) into the group of nonsingular linear transformations of some finite-dimensional vector space 'U over If. 'U is called the underlying vector space of ~. The dimension of 'U is the degree of ~. An Ifrepresentation is faithful if it is an isomorphism. Two Ifrepresentations if l ' 32 of N are similar if they have the same underlying vector space'U and if there exists a nonsingular linear transformation S of'U such that i'ij(G) S-l ij2 (G)S for all G E: 6). ~ representation of ® is an ij -representation of (~for some field ~. If N is a group and If is a field, then the group algebra If(N) of (~) over If is the ring consisting of all elements of the form LNa(G)G, where a(G) E: If for all G E: (~ and where addition and multiplication are defined in a natural way as follows; L;a(G)G + :Eb(G)G "" L; {a(G) + b(G)} G N
(\)
6(.. a(G)GI L;("lb(H)H ., •
'6
(\)
= L;
G,HE:N
I
L; a(GH- 1 )b(H) G
GE:('! HE:N
a(G)b(H)GH
2
CHARACTERS OF FINITE GROUPS
If ~ is an ~-representation of N with underlying vector space '0, then ~ can be extended to a ring homomorphism of ~'N) into the ring of linear transformations of '0 by defining ~{~(,\a(G)G} ~Na(G) ~(N). In this way '0 has a unital ~(N)-module structure imposed on it. Conversely a finitely generated .unital ~«~)-module gives rise to a uniquely determined ~-representation of ~. Thus the study of ~-repre sentations of N is equivalent to the study of finitely generated unital ~(~)-modules. The module point of view has been treated extensively in the literature, see for instance the books by Curtis and Reiner or M. Hall. In these notes we will however adhere to the representation approach. If '0 is an n-dimensional vector space over the field ~, then the group of nonsingular linear transformations of '0 is isomorphic (in many ways) to the group of nonsingular n x n matrices with coefficients in ~. Thus an ~-representation of degree n of (\\ could equally well have been defined as a homomorphism of ~ into the group of nonsingular n x n matrices with coefficients in 5=. It will often be convenient to consider it as such. Let ~ be any field. The following are some examples of ~ -representations of a group N. (i) The unit ~-representation ~ of N of degree 1 is defined by ~(G) = 1 for all G E: ()). (ii) If il is a homomorphism of ~ onto a group of permutations on the n symbols 1, ... , n, then il may be interpreted as an 5=- representation in the following manner. Let '0 be a vector space over 5= with basis {VI' ... , vn}. Define vi il(G) vi ~(G) for G E: N, i 1, ... , n. Such an ~-repres entation is called a permutation ~ - representation of (~. (iii) The (right) regular ~-representation ~ of ~ is the permutation ~-representation of (';\ arising from the (right) regular representation of N. More explicitly let '0 be a vector space over ~ with basis {vHI H E: ('I}. Define vH9{(G} vHG for all G,H E: (~. (iv) If ill' il2 are ~-representations of (~ with underlying vector spaces'O,'W respectively. Then ~'I ® ~2' ill ® il2 are ~-representations of N with underlying vector spaces '0 ® 'W, '0 ® 'W respectively.
REPRESENTATIONS
3
(v) If I" is an 9=-representation of (~ define the contragredient representation 1"* of il by 3* (G) = I1(G-l Y for G E: (\), where the prime denotes transpose. Let '0 be the underlying vector space of an 9=-representation Ii of N. An invariant subspace of '0 is a subspace 'W such that'W I"(G) c 'W for all G E: N. If 'W is an invariant subspace of '0 then I" defines l'F-representations of (~with underlying vector spaces 'Wand 'O/'W. These representations are called constituents of Ii. An l'F - representation I" of N or its underlying vector space '0 is l'F -irreducible if 0, '0 are the only invariant subs paces of '0. I" or 'U is l'F - reducible if it is not I" -irreducible. I" or '0 is completely reducible if '0 = '0 1
(W~G) ~~gD for all
G
E:
N, where I,
~ are constituents
of 1". Furthermore I" is completely reducible·if and only if there exists a nonsingular matrix S with coefficients in l'F such that S-1 ~(G)S =
for all G
E: (\),
(~1 (G)
o .• I" n0(G) )
where each I"i is l'F -irreducible.
(Maschke) Let (\~ be a group and let l'F be a field such that char (F YINI . Then every l'F-representation of N is completely reducible.
(1.1)
Proof. Let I" be an (F - representation of N of degree n. The proof is by induction on n. If n = 1, then I" is (F-irreducible. Assume now that n > 1. If ty is l'F-irreducible there is nothing to prove. Thus it may be assumed that il(G) =
4
CHARACTERS OF FINITE GROUPS
W(G) (s(G)) ~(G) for all G ( o see that (S(GH) =
~(G)
Multiply this by (S(GH)
E:
. _ 6J. Smce (~( GH)
(S(H) + (S(G)
~(H-1)
~(H)
= iH G) ii(H) we
for G,H
E:
N
on the right. Thus
~((GH)-1) ~(G)
= (S(GH)
~(H-1)
1I(G) (S(H) Sum this over all H
~(H-1)
E:
+ (S(G)
N, divide by
INI and define
Thus M~(G)
Let S =
= 1I(G)M
(~ ~)
+
G:(G)
for G
E:
• Then for G E: N
( ~((oG)
1I(G)M + (s(G)) ~(G)
= S
Therefore S-1 duction 11 and
(~((oG)
~(G)S ~
N
=
M~( G)) ~(G)
0)
~(G)
C(~G) ~(OG)) for
G
E:
(\j.
By induc-
are completely reducible and hence so is
~.
(1.2) (Schur's lemma) Let ~1 and ~2 be rf-irreducible rf - representations of (\\ for some field rf. Suppose that S is
REPRESENTATIONS
5
a nonzero matrix with entries in ff such that iY 1(G)S "" S ~z (G) for all G E: N. Then S is nonsingular and til is similar to iYz Proof. Let '0, W be the underlying vector space of iY 1 , iYz respectively. Hence S is a linear transformation sending '0 into W. Let '0 0 be the kernel of S and let Wo = '0 S. Since viY1(G)S =:: vSiYz(G) for v E: '0, G E: N it follows that '0 0 , Wo is an invariant subspace of '0, W respectively. Since S c;t! 0 by assumption we get that '0 0 c;t! '0 and W 0 c;t! O. Hence '0 0 0 and W0 =: W by the irreducibility of '0 and W. ThusSisnonsingular. Consequently S-l~l(G)S::= trz(G) for G E: N. If iY is an ff - rep res entation of N define C ff ( t\') '" 0::
{sis iY(G) =: iY(G)S for G E: N, S has coefficients in ff} . We remark that Cff(iY) is a finite-dimensional algebra over ff which contains all scalar matrices. (1.3)
Let
~
be an ff-representation of N where char ff-irreducible if and only if Cff(~) is a division algebra. \:f
Yl(\)l • Then
,?roo!. If iY is ff-irreducible then setting iY iY 1 =:: iY z in (1.2) we see that every nonzero element of C ff(~) has an inverse which is then necessarily in Cff(iY). Thus C ff( iY) is a division algebra. Conversely suppose that It is ff-reducible with underlying vector space '0. By (1.1) '0 = '0 1 ® 'Oz, where '0 1 and 'Oz are nonzero invariant subspaces of '0. Let S be the projection of '0 onto '0 1 , then S is singular and it follows easily that S E: Cff(It). Thus Cff(~) is not a division algebra. (1.4) Let ff be an algebraically closed field and let iY be an \:f -irreducible representation of ~. Then C ff(~) consists of scalar matrices.
Proof. Let S E: Cff(~) and let x be a characteristic value of S. Then x E: ff since ff is algebraically closed. Thus
6
CHARACTERS OF FINITE GROUPS
S - xl E: C3'(t'f) and S xl i~ singular. Hence S xl = 0 by (1.2) and so S xl is a scalar matrix as required. Let X be an extension field of 3'. Then any 3'-representation of ~ may be considered to be a X-representation of ("'. An 3'-representation is absolutely irreducible if it remains irreducible in every extension field of 3'. The next two results are elementary preliminaries. (1.5) Suppose that char 3' YI~I . Let t\l' if2 be 3'-representations of (~~ and let S be a matrtxwith coefficient~. Then ~l (G)S :: S if2 (G) for all G E: ~ if and only if there exists a matrix T with coefficients in 3' such that S Z<M if l (G-I)T Proof. If ~l (G)S S32 (G) for all G (lIT
E: ~
let T
'E iTl (G-I)T ~\ (G) ('"
=
'E iT
l
(HG-I)T 3 2 (GH- I ) ~2 (H)
GE:N
:: 'E 31(G-l)T if2(~J! if2(H) (in
(1.6) Assume that char 3' Y1(\)1 • Let X be an extension field of 3' and let t\ be an 3' - representation of (~ . C3'( iT) consists of sc~matrices if and only if Cx(i'i) consists of scalar matrices. Proof. For any matrix S with coefficients in X let f(S) il(G- 1 ) S 3(G). For x, y E: X and matrices Sand T we have that f(xS + yT) xf(S) + yf(T). Let Eij be the matrix whose (i, j) entry is 1 and in which all other entries are 0 . Thus by (1.5) every matrix in C3'(t'Y}, CX(~) is an 3'-linear, X-linear combination respectively of the matrices f(E ij ). Z(~
REPRESENTATIONS
7
Thus e~(~) or ex(~) consists of scalar matrices if and only if f(E ij ) is a scalar matrix for all i, j . (1.7) If ~ is an absolutely irreducible ~-representation of 6) then e ~( ~) consists of scalar matrices. If char ~YI6)1 and ~(~) consists of scalar matrices for some ~ -representation it of 6) then ~ is absolutely irreducible.
e
Proof. Suppose that ~ is an absolutely irreducible ~ representation of 6). Let X be the algebraic closure of ~. Then ~ is X-irreducible. Thus by (1.4) ex(~) consists of scalar matrices. Hence also ~(~) consists of scalar matrices. Conversely assume that char ~ YI(~I and e~(iJ) consists of scalar matrices. Then by (1.6) x(i\) consists of scalar matrices for any extension field X of ~. Thus x (3) is ring isomorphic to X and by (1.3) iY is X-irreducible.
e
e
e
(1.8) Let (ill be an abeUangroup and let ~ be an ~-irreducible ~-representation of N for some field ~. If i\ is faithful then (~ is cyclic. If ~ is absolutely irreduciblethen ~ has degree 1.
e
.
Proof. Since 6) is abelian it( G) c ~(~) Thus if iY is faithful (~is isomorphic to a finite abelian subgroup of the multiplicative group consisting of nonzero elements of the division ring e ~ (3) . Therefore (\\ is cyclic. If ty is absolutely irreducible then by (1. 7) iJ (G) is a scalar matrix for all G € (\\. The irreducibility of ty thus implies that it has degree 1. The relations in the next theorem are called the Schur relations. They are of fundamental importance for the theory that follows. (1.9) Let 11, ti be ~-irreducible ~-representations of (\) for somefteld ~. Let "(G) == (aij(G» and 5l\(G) == (bij(G» for G € ®. -(i) If ,. is not similar to ~ then ~ ais(G-l )bt/G) N
=0
8
CHARACTERS
OF FINITE GROUPS
for all i, j, s, t . (ii) If ~ is absolutely irreducible and char 5' '"
LJ
N
a, (G -1 )at,(G) -- -I(\;\ 0" 1S J n 1J
I)
S
Y\('41
then
t
where n is the degree of !. Proof. For any matrix S with coefficients in ff let f(S) Then Yl (H)f(S) = f(S)~(H) for all H (: (\;. Let Est be the matrix with 1 in the (s, t) entry and 0 elsewhere. The (i, j) entry of f(E st ) is E('t)ais (G-1) btj (G) . If Yl and ~ are not similar then f(E st ) = 0 by (1.2). This proves the first statement. If n '" ~ is absolutely irreducible and char ::r YINj then by (1:7) f(Est) estI . Thus E(\; Yl(G-1 )S~(G).
e t 6" ;: .6,u a, (G-1) at, (G) s 1] ,'J 1S J
6/Ua, (G) at' (G-l) V)
Consequently eij 0 if i ~ j and en of i. Hence est 6ij '" e 6 t 6". Since s 1J
n
ne '"
6
i=1
1S
=e
]
is independent = I,
n (G-1) n (G)
n
e
= L: L: a . (G-l ) a. ~
i"'l
S1
(G)
I~\
1S
as required. (1.10) Assume that char ff YI('t)j . Then there are only finitely many pairwise nonsimilar absolutely irreducible ffrepresentations ~ l ' ' , . , ~k of (\l • Let X ile an extension field of ff. If 11 s( G) = (arj (G» and 5s has degree ns then {a~j} is a set of linearly independent X-valued functions on ~and
REPRESENTATIONS
9
k
:E
n§ ~ I(~I
s=1
Proof. Let ~l' ••• , ~k be a set of pairwise nonsimilar absolutely irreducible X-representations of ~. Let I\(G) = (afj (G» and let ns be the degree of ~ s. The set of X-valued functions on ~ forms a IG)\-dimensional vector space over 3C. Thus it suffices to show that tafj} is a set of linearly independent X-valued functions on (~. Suppose that ~.. x? a~. (G) = 0 for all G E: Nand s I,J,S IJ IJ some x ij E: X. Then by (1.9)
o
I~l S' x . , .,
n
S
I
J
1
completing the proof. A field 5 is a splitting field of (~ is every 5-irreducible 5 - representation of 6.l is absolutely irreducible. (1.11) Assume that char 5 YI~1 . Then there exists a finite extension field of 5 which is a splitting field of ~ . Proof. Let ~ be the algebraic closure of 5. By (1.4) and (1. 7) ~ is a splitting field of ~. Let ~ 1> •• " ~k be a maximal set of pairwise nonsimilar absolutely irreducible \i-representations of 6.l. Such a set exists by (1.10). Let ~s(G) = (afj (G» for G E: (~\. Let X be the extension field of ~ generated by all afj (G). The finiteness of (~ implies th.,.at Ix: 5] is finite since each afj (G) is algebraic over 5. If I~ is a Xirreducible X- representation of (~~ then in ~, ~ is similar to a direct sum of various ~j' Thus if ~(G) (aij (G» for G E: (~ then all is an 5 linear combination of the a~j . Thus by (1.9) ~ is similar to some ii s in X. Hence X is a splitting field of N.
10
CHARACTERS OF FINITE GROUPS
Assume that char 5=YI("'1 . Let ~ be an 5=-representation of ("'. If G E: <'" then thereexistsarrnite exteilsion field :JCof 5=-such that ~(G) is similar to a diagonal matrix in :JC. (1.12)
Proof. It may be assumed that (~ =
. By (1.11) there exists a finite extension field :JC of 5= which is a splitting field of ~. Thus by (1.1) and (1.8) iJ(G) is similar to a diagonal matrix in :JC.
U. CHARACTERS Let 5= be a field of characteristic O. Let ~ be a group and let ~ be an 5=-representation of ~. For G E: ~ let 9(G) be the trace of il(G). The function 9 is called the character of ~ afforded by the 5=-representation ~. In general a character of ('" is a function 6 which is the character afforded by some 5=-representation of Q;, where 5= is a field of characteristic O. An irreducible character of 6) is a character of ® which is afforded by an absolutely irreducible 5=-representation of (~, where 5= is some field of characteristic O. It should be noted that this definition of a character differs from that used by some authors, for instance Curtis and Reiner. Only if 5= has characteristic 0 does an 5=-representation of ~ afford a character in the sense of the above definition.
(2.1) If 6 is the character afforded by the representation ~ of ('" then 9(1) is the degree of i\. For G E: N, 6(G) is a sum of complex roots of unity, and e(G-l) = eTG}, where
the bar denotes complex conjugation. Thus valued function on ~ .
e is a complex
Proof. Clearly 9 (1) is the degree of iJ. Since ty is an 5=representation of (~where 5= is a field of characteristic 0 it
CHARACTERS
11
is possible to identify the algebraic closure of Q in the algebraic closure of 5= as a subfield of the field of complex numbers e. Since it(G)I~1 I for G e: ~ all the characteristic roots of i1(G) are complex roots of unity and O(G) is their sum. If E is a complex root of unity then ~ € -1 and so 6 (G -1) :; If{G'} for G e: ~. Suppose that char s= ;: O. Then (i) Similar 5=-representations of ~ afford the same character. (ii) If 6 is a character of (j then 6 is a class function. That is to say 6(H-IGH) = 6(G) for G, H e: (t». (iii) If i1 = itt E9 il 2 , (\1 ® ty2 then the character afforded ~ iT is the sum, product respectively of the characters afforded by t;t and il2 . (2.2)
Proof. Statements (0 and (ii) follow from the fact that similar matrices have the same trace. The last statement is a consequence of elementary properties of traces. The principal character 1@ of (» is the character afforded by the unit 5=-representations, where char 5= = O. Thus l~(G) = 1 for G e: ~. The character afforded by a right regular 5= - representation of ~ will be denoted by p (~ • Suppose that 6 is the character afforded by a permutation representation i1 of (~. Then 6 (G) is equal to the number of letters fixed by the permutation il(G). Thus in particular p(~(G) = 0 if G e: ~# and p~(l) = I~I . (2.3)
Proof. If ~(G) = (aij(G» then aij(G) = 1 if iiY(G) = j and aij(G) = 0 otherwise. Thus 6(G) ;: Z aU (G) as required. If 6, 11 are complex valued functions on ~ let 1
(6, 11)~ = I~I ~ 6(G)11(G),
11611(~ = (0,6)('1\
12
CHARACTERS OF FINITE GROUPS
The subscript N will be omitted in case it is clear from context which group is specified. It follows immediately that the hermitian product thus defined has all the usual properties and is positive definite. If 17 is a character of (ill then by (2.1) (fj, 1]) =
I(!I ~ o(G)17 (G-l)
(2.4) Suppose that char 5" 0 . Let 0, 17 be the characters afforded by the 5" -irreducible 5"'- representations ~,~ of (ill • (i) If ~ is not similar to ~ then (0, 17 ) 0. (ii) li ~ is absolutely irredUcible then (0, 0) = 1. Proof. In the notation of (1.9) O(G) = !:i a ii (G) and 17 (G) = !:i b ii (G) for G E: ('~. Thus (0, 11)
':1
1VJ
L;
••
1,J
a 11 .. (G) b ]J .. (G-l)
Hence by (1.9) (0, 1}) :::: 0 if 11 is not similar to ~ = ~ is absolutely irreducible then (0,0)
1
= 0(1) L;
i,j
0 .. 0 .. 1J
IJ
~
and if
=1
(2.5) Suppose that char 5" = O. Let i\ be an 5"-representation of (I,) and assume that ty ~l (!) ••• (!) ~s ' where each 1Ii is an 5"-irreducible 5"-representation of ~. Let 0, 1} be the character afforded by iV, ~l respectively. Then the number of ~i similar to ~l is (0, 1})/(1}, 1}). Proof. This is an immediate consequence of (2.2) and (2.~
(2.6) Suppose that char 5" = O. Two 5"-representations are similar if and only if they afford the same character. Proof. This follows directly from complete reducibility and (2.5).
CHARACTERS
(2.7)
13
If fJ is a character of ~ then (fJ,P{!tj) ;::: fJ(l).
Proof. Clear by (2.3). (2.8) There exists a finite extension If of ~ depending only on (~ such ~!t~Leyery character of N is aff~~~~d by an If -representaJ!on of N. Thus in particular If is a splitting field of N. Proof. By (1.11) there exists a splitting field If of ~ which is a finite extension of ~. Let 9l be an If-representation of (\\ which affords p(\\' Every If-irreducible constituent of 9i is absolutely irreducible. Thus by (2.5) and (2.7) every irreducible character of (~ is afforded by an If-representation and hence every character of {!tj is afforded by an If-representation. If fJ, 1} are characters of (\\ then (J is a constituent of 1} if either 1} = () or 1} fJ + fJ 1 for some character fJ l' If fJ is a constituent of 1} we shall write () C 1}. If fJ is irreducible then «(), 1}) is the multiplicity of 9 as a constituent of 1}. A character 1} is multiplicity free if (fJ, 1}) :; 0 or 1 for every irreducible character (J. The kernel of a character () is the kernel of representation which affords fJ. A character fJ is faithful if its kernel is (1). If If is a field of characteristic 0 and fJ is a character of (;} then If(fJ) is the extension field of If generated by the elements fJ(G) for G E: N. Clearly 1f(9) is a finite extension field of If. If If(fJ) ;::: If we will also write that (} E: If. The field If is a splitting field of the character (} if fJ is afforded by an If -representation of ~. If If is a splitting field of fJ then clearly (} E: If. However it is important to note that If'«(}) need not be a splitting field of (}. It is an immediate consequence of this definition that If is a splitting field of (~ if and only if If is a splitting field of every irreducible character of N. If If is a field of characteristic 0 and IV is an If -representation of @ which affords the character (J the,!l for any isomorphism (j of If into its algebraiC closure If, IVO" is an TF-representation which affords the character eO". Since fJ(G)
14
C H A RAe T ERS 0 F FIN I T E G R0 UPS
is a sum of roots of unity for G E: ~, ~(9) is a normal extension of ~ with an abelian Galois group. Thus if e is a character of (\) and (J is an automorphism of Q(e), ()(J is also a character of (~. If 9 1 = e(J for such an automorphism, e and el are said to be algebraically conjugate characters. Thus in particular e and 11 are algebraically conjugate. It is easily seen that if e is afforded by the representation ~ then "9 is afforded by the contragredient representation ~*. The following notation will be used in the remainder of this section. Xl = lN' X2 , · •• , X is the set of irreducible characters k of ~. For i = 1, ... , k, I. is a representation which affords X. and x. = x.(l). 1 1
.n.1 1
1
1
{t}, .fi 2 ,
,{ll
= {G-1IG E:
"'i(·fU ]
(2.9)
==
is the set of conjugate classes of (\),
•••
St}. 1
I.Sl.1] X1. (G)/x.1 for G
(i) (X., X.)
E:
.Sl.J
6 ...
1]
1]
(ii) ~NX i (G) == I~! 6 h . == ~
(iii) If e -
e = X.1
a. X. then (e, X,) 11--
1
==
a. and
(iv) If e is a character with lie 112 for some i. (v)
p(U .,
;:;:~. x. 1
1
1
==
lie 112
1 then
x1..
(vi) ~. X7 ::: INI. 1
1
Proof. (2.4) implies (0. Let j = 1 in (0 to get (ii). (iii) follows from (0 and (iv) from (iii). (2.7) and (iii) imply (v). By (iii) and (v) 1(\)1 == IIp()12 = ~. x~, proving (vi). ~J
1 1
CHARACTERS
15
(2.10) Let 5' be a field such that II,"" lk are 5'-representations. Let I (G) = (a~.(G». Then any 5'-valued funcs 1) Hon on ~ is a linear combination of the a~. with coefficients in 5'. 1)
Proof. The set 5'-valued functions on ~ is a I("'I-dimensional vector space over 5'. By (1.10) and (2.9) (vi) {a~.} is a basis of this space. 1) ~
(2.11)
(1
·n.
= w.1(.SUI )
1.(G)
J
1
Proof. Let S =
~.)
1.(G). Then 1.(H-1 )SI.(H)
.llj
1
1
S for
1
H E: (~\. Hence by (1.7) S = sI is a scalar matrix. The trace of S is the sum of the traces of alll.(G) for G E: .Sl •• Since 1 J each such matrix has trace X(G J for G. E: .R. we see that J J J x.s = trace of S = I·n.lx.(GJ 1 J 1 J Thus s = w. (.\U as required. 1 J Let G E:.R
(2.12)
pairs G., G. with G.
--
1
Then a..
IJS
) --
and let a..
s
1
IJS
.St., G.
E:
1)
E:
1
1
-
J
S
1,'\1 /) ji' .
Proof. If G
= G.G. then H-1 GH = (H-1 G.H) (H-1 G.H). 1
J
1
The first statement follows. If a" G E: .\l. such that G-1 easily.
.R. such that G. G. = G.
is independent of the choice of G in !l . Further-
more aijl'=
1
be the number of ordered
E:
.n)..
1) 1
J
"'" 0 then there exists
Thus i = j'. The result follows
16
CHARACTERS OF FINITE GROUPS
(2.13)
Let a..
-
1JS
have the same meaning as in (2.12). Then -
(i) wtUl.)w (.~L) ::: z; s a..
t
1
J
1JS
w (.n ). t s
(U) I.R.I!.~l.IXt(G.)Xt(G.) = X 1
G. c 1
.~t, 1
s
1JS
L
1
J
t
Z;
s
a ..
IJS
I·U
S
Ixt(G ),where
s
G. CoR., G C .R • J J s s
Proof. 1: a..
J
L() It(G.) L l' It(G.) = L () <) It(G.G.):: ·Jl. 1 .,lj J ·J\i' .l'j 1 J i <) .11.
S
(0. Multiplying (0
It(G ). Thus (2.11) implies
s
t
by x and expressing (L't in terms of X yields (U). t (2.14)
If G. C
-
1
.n.,1
G.
J
C
.n.Jthen --
Proof. Replace j by jl in (2.13) (ii) and sum over t. By (2.9) (v) and (2.12) this yields that
l.n.II.~L" L) xt(G.) xt(G.) 1
J
t
1
J
L) a .. s
f
1J S
I.R Ip(.,~(G ) S
= a 1J.. 1 1(;')1 = 1
S
I·~U o.. '~1 1
1J
Since I·R." = ISl.I, the result follows. J J (2.9) (i) and (2.14) are called the orthogonality relations for characters. (2.15) G.
J
C
-Let a. iJm be defined..... as in (2.12). If G.1 5L, G C 5l then J m m
C
.n.,1
CHARACTERS
17
a ..
11 m
Proof. Multiply the expression in (2.13) (U) by X (G )/x and sum over t. Thus by (2.14) t m t
E a.. I.~ I E s IJS S t =
X (G ) xt(G )
t
s
m
(\)
I.R' m I a.. I'i 11m.J\
m
The result follows. (2.16) The number of irreducible characters of N e ual to the number of con'u ate classes of 6j. If ff is a field containing ~ X·, ..• , X ) then X.1 is a basis for the 1 k space of ff-valued class functions on N. Proof. Let kl be the number of conjugate classes of ()). By (2.9) (i) {x-J is a set of linearly independent class func1
tions on 6). Thus k < k 1 • Let X = (X. (G.», where G. E: .n .. Thus X is a k x kl 1 1 ] J matrix. Let X' be the transpose of X. By (2.14) x'x = «I @ 1/ l.fli I) 0 .. ) is a nonsingular kl x kl matrix. Hence the 1]
rank of X is at least kl and so kl < k. All statements are proved. (2.17)
Let a. t be defined as in (2.12). Let A.be the
-
IS
matrix whose (s,t) entry is a.IS t' Then -
1
W
t(Jl.) for t ::: 1-
18
CHARACTERS OF FINITE GROUPS
1, ... , k are characteristic roots of Ai' Each Cl.l{fll) is an algebraic integer. Proof. The relation in (2.13) (0 may be rewritten as k
E
s= 1
a .. ) wt(·R ) :::: 0 for i,j:::: 1, ..• , k
(wt(Sl.)o. 1
JS
IJS
S
If i is kept fixed and j is allowed to range over the values
1, ... ,k this yields k homogeneous equations in the k unknowns CI.'{ft 1 ) , ••• , wt(·fl ). Since W{Rl) :::: 1 ~ 0, the dek
terminant of the systems must vanish. Thus for each i the determinant of ""t(·UJI - A. is zero. Hence wt(·uJ is a 1
1
1
characteristic root of A. for t 1
= 1, ... , k.
Since the entries
of A. are all rational integers, wt(SlJ is an algebraic integer. 1
1
(2.18) Proof.
INI x.
1
Let G. J =
I(~I
X.
1
E: .Sf .,
l
Then k
(X., 1
xJ 1
=
~
.njl Xi (G j )
=
x.(G.)
Xi
j= 1
1
J
k E"".(.u.)x.(G.) j=l
1
J
1
J
By (2.1) and (2.17) this implies that 1~ I/Xi is an algebraic integer. Since 1(\)1 and x. are rational the result follows. 1
§3. COMPLEX REPRESENTATIONS A complex representation tv of ('4 is unitary or orthogonal if jj(G) is unitary or real orthogonal respectively for all G t: (~.
COMPLEX REPRESENTATIONS
19
(3.1) A complex representation of ~ is similar to a unitary representation. A real representation of ~ is similar to an orthogonal representation. Proof. Let 8
== ~N
~
be a complex representation of
(~\.
Let
(G) ~(G), where the prime denotes transpose. Then
S' = 8 and 8 is real if
tl' is real. Each
;V'(G);V(G) is non-
negative and ;V' (lHJ(l) == I is positive. Thus 8 is positive definite. The result follows since ;V'(G)8 ~(G) = 8 for G E: N. The result immediately yields another proof of the complete reducibility of complex representations since the orthogonal complement of an invariant subspace of the underlying vector space of a unitary representation of (~ is also invariant. (3.2) If ;V 11 ii2 are similar irreducible unitary representations of (~ then there exists a unitary matrix U such that U-1 ;VI (G)U = ii2 (G) for G E: N. Proof. By assumption 8-1 31(G)8 == tl'2 (G) for some complex matrix 8. Thus ~1 (G)8 :; 8ii2 (G). Taking conjugate transposes this yields that
Thus
Hence 8S' commutes with ;Vl(G) for all G E: (~. Consequently 8S' = cI is a scalar. Therefore c I is positive, and so c is a positive real number. Hence c == S8 for some complex s. Let U == i8. Then U-1 ~ 1 (G)U == ~2(G) for G E: (~, and UU' = I and thus U is unitary.
20
CHARACTERS OF FINITE GROUPS
The results in the remainder of this section are due to Frobenius and Schur. An irreducible complex representation tj is of the first kind if it is similar to a real r~resentation. ~ is of the second kind if ~ is similar to ~ but is not of the first kind. ~ is of the third kind if ~ is not similar to ~. (3.3)
Let;Y be a complex representation of (\j such that f ~(G) ;::: U-1 ij(G)U for some unitary matrix U. Then U ±U. Furthermore U U if and only if iY is of thefirSt kind. U' ;::: -U if and only if ~ is of the second kind. f
;:::
Proof. By assumption ti(G) ;::: U-1 t'J(G)U. Taking complex conjugates this implies that ~(G) ;::: U iY(G) U' -1. Thus f
~(G)
=U
f
ti(G) U'-l
U'U-1 il(G)UU~-l
f
Hence U U-1 ;::: cI is a scalar matrix. Thus U' ;::: cU and U ;::: cU' ;::: c 2 U. Hence c ;::: ± 1. By definition ti is of the first or second kind. Thus it suffices to show that ~ is of the first kind if and only if U' ;::: U. If il is of the first kind then there exists a complex matrix A such that A-I ti(G)A is real for all G C ~. Hence for G C N A-I U-1 ti(G)UA Thus UAA-l ;::: aI is a scalar matrix and so U ;::: aAA-1 • Consequently U' ;::: U-1 ;::: a-I AX-I. Thus if U' ;::: -U then a-I -a or aa ;::: -1 which is impossible. Hence U' ;::: U. Suppose that U' U. Since U is unitary there exists C unitary and D diagonal such that C-1 UC ;: D. Thus CDC-I;::: U ;::: U' ;::: C-l' DC' Hence C'CD ;::: DC'C. There exists a diagonal unitary matrix E such that E2 ;::: D and C'CE ;::: ECfC. Let V CEC-1 • 2 Then V ;::: U and V is unitary since C is. Furthermore
COMPLEX REPRESENTATIONS
V'
= C-l' EC '
21
C-I' EC ' CC-l = C-I/C' CEC-l :;: CEC- I ::: V
Thus V-I :;: V and V-I l'Y(G)V :;: VU-I t'r(G)UV-1
:;:
V-I i:HG)V for G E:
~
The proof is complete. (3.4) If 5 is a complex representation of kind then ~ has even degree.
(iI)
of the second
Proof. Let n be the degree of d. By (3.1) and (3.2) it may be assumed that ~(G) :;: U-1 ~(G)U for G E: ~ and some unitary U. By (3.3) u' :;: -U. Thus det U :;: (_l)n det U. Since det U ;r! 0, n is even. (3.5) For an irreducible character X of N let lAx) = 1/ I(~' I :0 (;,\X (G2). Let X be an irreducible character afforded by the complex representations _ then (i) v (X) 1 if and only if X is of the first kind. (ii) v(X) :;: -1 if and only if X is of the second kind. (iii) v(X) :;: 0 if and only if X is of the third kind. Proof. It may be assumed that ~ is unitary and ~ :;: if ~ is of the first kind. X(G2) is the trace of ~(G) ~(G) ~(G)~(G-l)'. Thus if ~(G) :;: a .. (G) then
ii
1)
v(X) :::
1(\1 6 6 a .. (G) a .. (G-l) ,
i,j
N
1)
1J
If l'Y is of the third kind then v(X> ::: 0 by (1.9). If l'Y is of
the first kind then by (1.9) v(X) :;:
I(~'I., 6.. 6(;" 1,J
I ~I \21
61'
a .. (G) a .. (G-l) 1,J IJ
0(iii a 11.. (G) a 11.. (G-l)
= 1
22
CHARACTERS OF FINITE GROUPS
Suppose that tl is of the second kind. By (3.2) and (3.3) there f exists U unitary such that U :::: -U and U- 1 ~{G)U = ~(G) f 1 for G E: (~. Since UU = -U this implies that if U = (u ) then ij a .. (G-l) :;: -2:; U. a t(G-l) u . t) ~ st , ~ s Therefore by (1.9) v(X):::
-I~'I 2:; VI
;;:: -
t 1,),S,
U.
"
~ u:-:-1)
1(1) X ,.
IS
t)
~ a" (G) (U IJ .,
a t (G-
s
1
)
u ..
1,)
Since U'
u '
1)
U- 1 , ~j Uij Uij ;;:: 1 for all i. Thus v(x)
=
1
as required. All possibilities are exhausted and the proof is complete.
(3.6) For G E: (~ let t(G) be the number of elements H such that H2 :: G. Then
k
t(G) ::: ~ v(X,) X.(G) i=l 1 1 where v(x) is defined as in (3.5) . Proof. Clearly t is a class function on ~. By (2.17) t(G) = ~i ai Xi{G) for G E: N. If !(G) {HIH2 G} then the sets !(G) partition ~. Let G1 , G2 , ••• be a system of representations for the sets l:(G). Then
INTEGRAL REPRESENTATIONS
1
1(\)1 ~
-t(G)
23
1
~ (G) = ~ ~~
I~I ~ ~ (HI!) = v(~)
=
~
as required. (3.7)
If
(I,; ~ontains
t + 1
=
exactly t elements of order 2 then
k .~ v(~)~(1) 1=1
5
k ~ ~(1) i=1
where veX) is defined as in (3.5). Thus in particular t + 1 =
Z~1 ~ (1)
if and only if every irreducible character of N is
afforded by a representation of the first kind. Proof. Set G
1 in (3.6).
There seems to be no analogue of (3.7) if" 2" is replaced by any other integer. This striking result illustrates one of the many ways in which the prime 2 plays a special role in group theory. The symmetric groups are examples of groups in which every irreducible representation is of the first kind, in fact Q is a splitting field for symmetric groups see Curtis and Reiner p. 190. Thus (3.7) yields a simple method for computing the sum of the degrees of the irreducible character of the symmetric groups.
§4. INTEGRAL REPRESENTATIONS Let :.0 be an integral domain with quotient field ff". An representation il of ~ is a :D- representation of (~l if for all G E: (\) the entries of il(G) lie in :0. Such representations of (\\ are sometimes called integral representations. ft -
CHARACTERS OF FINITE GROUPS
24
The theory of integral representations differs markedly from the theory of ff -representations, where g; is a field. We will make no attempt to study integral representations systematically, only proving some elementary results that will be needed in the sequel. (4.1) Let !I) be a principal ideal domain with quotient field ff. Every ff - representation of <'9 is similar to a !D- representation of {'9. Proof. Let 'U be the underlying vector space of the ffrepresentation \J of {'9. Let {v l ' ••• , v n} be a basis of 'U. Let ff[ be the !I)-module generated by {vi rt(G) Ii := 1, ... , n, G E: (\)}. Thus ff[ is a finitely generated torsion free !I)module and so it has a basis {wi}' It is easily seen that {wd is also a basis of 'U and ff[ \J(G) S ff[ for G E: ('9. With respect to this basis all the matrices of the representation have entries in !I) . By (2.8) there exists a finite extension ff of (! which is a splitting field of {'9. Let p be a prime and let 800 be a prime ideal in the ring of algebraic integers of ff such that p E 80 0 • Define !D
{albia, b integers in
ff ,
b
:;t:.
0 (mod
800)}
It is easily seen that !D is an integral domain with quotient
field ff. The elements of !D are called local integers at 80 0 • An element of ff is an algebraic integer if and only if it is local integer at every prime in the ring of integers of ff. Let 1T E: 80 0' 1T ~ 800 2 • Let 80 = (1T) be the ideal of !I) generated by 1T. We will show that every nonzero ideal in !I) is of the form 80 1 for some i. Hence :.0 is a prinCipal ideal domain and 80 is the unique maximal ideal in :.0. Thus :.o/g~ is a finite field of characteristic p. For alb E: :.0, with a, b algeoraic integ~rs let v(a/b) ;:;; if a ;:;;.0 and v(a/b) = i if a == O(mod 800 1 ), a ~ O(mod IPOl+l). Thus an element a of :.0 is a unit if and only if v(a) = O. Also v(ll) 1. If 9 is an ideal of :.0 choose a E: 9 such that v(a) is minimum. Let v(a) i. Then 0()
INTEGRAL REPRESENTATIONS
25
lI(a1T- i } = 0 and so &Oi :::: (a) c 11. If v(b} = j 2: i then b1T- 1 E:: 1> and thus b E:: &Oi. Hence 11 = &Oi as required. Throughout this section we will adhere to the notation just introduced. (4.1) may be used to give an alternative proof of (2.18). We will here generalize that result.
(4.2) Z(",) .
Proof. If suffices to show that 1("':.8l/x(l) is a local integer at every prime p. Choose a prime p. By (4.1) there exists a 1>-representation ~ of (») which affords the character X. Let ~(G) = (aij(G» for G E:: (~i. If Z E:: .8 then ~(Z) = E(Z)I, where E (Z) is a root of unity. Thus for GE::(l},ZE::.8
Let G1 ,
•••
tatives of
.B
,G
m
be a complete system of coset represen-
m
m
2] all (G. Z) all (Z-l G. -1) i=1
.B
in N. Hence for Z E::
1
=E
i=1
1
Therefore by (1.9)
: : I(»):.BI X (1)
all (G.) all (G. -1) 1
1
26
C H A RAe T E RS 0 F FIN I T E G R0 UPS
Since a l l (G) E: [) for G E: (\) this implies that /(\\:.8I/x(l) E: [) completing the proof. For a E: [) let a* denote the image of a in :D/~ == :1)* . Assume that p l I(~ I. Let ~, ~ be :I)-representa(4.3) tions of (\) which are absolutely irreducibie ~-representa tions of 6J. Then ~ * , ~ * are absolutely irreducible '!'* representations of (\). Furthermore ~ is similar to ~ in ~ if and only if ~ * is similar to ~* . Proof. If ~ * is not absolutely irreducible then replacing by an extension field if necessary it may be assumed that 1* is reducible. Hence there exists a matrix T with coeffi~
cients in :D* such that T-l!* (G)T
==
((Sl~G) <S2~G»)
for all
G E: (\). Let S be any matrix with coefficients in !D such that S* = T. Then det S ;: det T ~ O(mod~). Thus S-1 has coefficients in
1).
Hence {S-11 (G)S}*
Thus it may be assumed that 1* (G) G
E: (~.
Let 1 (G)
= (aij (G»
=
= S*-1 I*(G)S* .
(£I~G) <S2~G») for
and let n be the degree of
~
.
By (1.9)
(I~n I) *
= L; a * (G) a * (G) = 0 @
In
nl
Thus 1~l/n ;: O(mod p) contrary to assumption. Hence ~* and ~* are absolutely irreducible. If ~ * is similar to ~* then, as above, it may be assumed that ~* = 5B*. Let >B(G) = (b .. (G» for G E: <'9. Hence by (1.9) ~
L; a ~
In
*(G)b
nl
*(G-l)
= L;a *(G)a *(G-l) 6S In
= e~I)* and so 1 is similar to 58.
nl
~0
CENTER OF GROUP ALGEBRA
27
If tI and ~ are similar in g: they afford the same character X. By (2.9) (i)
EE
i,j ~
a .. (G) b .. (G-l) 11
JJ
=E
X(G) X (G-l) ;: I(~I
('" ~
O(mod p)
thus tI* is similar to 58* by (1.9). (4.4) Assume that p J I('t)1 • If X is any field of characteristic p then a X-representation of 6) is similar to a ~* representation of ~. If 58 is a ~* - representation of ~ then there exists a ~-representation tI of 6} such that tI* = 58. Furthermore tI is absolutely irredUCible if and only if ~ * is absolutely irreducible. If ~l' tl2 are ~-representations of (I,} then til is similar to '4 in g: if and only if til * is similar to tl2 * . Proof. Let II' ... , Ik be a set of pairwise nonsimilar absolutely irreducible g:-representations of 6} such that every absolutely irreducible g:-representation of ('" is similar to some Ii' Assume further that each I i is a~ representation of ~. Let Xi be the degree of Ii' By (4.3) t 1* , ... , Ik* is a set of pairwise nonsimilar absolutely irreducible ~* -representations of @. Since E xi = I~I by (2.9) it follows from (1.10) that any absolutely irreducible X-representation of 6) is similar to a ~* -representation of (I,}. The first three statements now follow from complete reducibility. The last statement is a consequence of (4.3) and complete reducibility.
§5. THE CENTER OF THE GROUP ALGEBRA Let (" be a group and let g: be a field of characteristic is a splitting field of (". Let g:«(I,}) be the group algebra of (;j over g:. Let Zg: (6)) be the center of g:(6J). For any subset tI of (I,} let tI = Etl G E:: g:«(I,}).
o which
C HA RAe TE RS 0 F F' NIT E G R0 UPS
28
Let .il l
,
••• ,
stk be the conjugate classes
of (\). Let
Xl' ..• , X be the irreducible characters of (~. Define k
W'(~ c. ~\ = f c. Wi(M.) j=1 J Y j=1 J J 1
where w. (.n.) J
1
Let a..
IJS
= w. (.R .) = I.R'j I Xi (G) J
1
G
E:
Sf. J
Xi ( 1 ) '
have the same meaning as in (2.12).
The integers a..
~s
and the functions w. have a natural in1
terpretation in terms of ZIT (@) as illustrated by the next result.
.il l' ... , .i k is a basis of ZIT (@). For i,j,s = , k st. i. ~s a.. it s. Each w. is a homomor1 J IJS -- 1
(5.1)
1, ...
:=
phism of ZIT «(\) into IT. Furthermore if A A
=0
if and only if w. (A) 1
:=
E:
ZIT «(~) then
0 for i == 1, •.• , k . -
Proof. For G E: ~, G-l .~. G := .~ •• Thus .it. -J J J for j := 1, ••. , k. Suppose that ~(\) a(G) G E: ZIT for H
Z('1: «(~)
«\).
...
Then
E: 6)
L; a(HGH- 1 ) G
==
L; a(G) H-I GH = L: a(G) G
~
Thus
E:
~
~
6)
a(G) G
k J==
~. 1
~
.
a(G.) .R. , where G. E: JI. for J 1, VI ~ J J J J ... , k. Hence {Sf/iS a basis of Z5(\~) since it is clearly IU
:=
a linearly independent set in 5('1). The definition of a .. IJS
29
CENTER OF GROUP ALGEBRA
yields immediately that
.R.1 i.J
~ a.. s IJS
.R s •
By (2.13) (i)
each w. is a homomorphism of ZIT (N) into IT. Thus clearly 1
w. (0) = 0 for i = 1, ... , k. 1 k ~ Let A = ~j=l c K and suppose that wi (A) j j . " , k. Thus for G E: jt' S
6.
= o for
i = 1,
S
c .l·fi.lx.(G.)x.(G) Jll1IS
1
L:j
= X. (G ) X. (1) lSI
c. w. (.it.) = 0 JIJ
for i = 1, ... , k. Summing over i this implies by (2.14) that
c
s
,I(~I 1(\1\ -- 6C.j.R.11SlIO.
J
j
J
s
JS
= 0
= 0 for s = 1, ... , k and so A = 0 as required. s The next result is quite general and well known.
Hence c
(5.2) WI' •••
Let (l be an algebra with unit over the field IT. Let ,w be distinct nonzero algebra homomorphisms of n
into IT. Then {w.} is a set of linearly independent functiollsfrom - (l to IT.1 (l
Proof. Induction on n. If n = 1 the result is clear. Suppose that ~ a. w. = 0 with some a. ?! O. Then by induction 1
a.
1
a
n
~
1
1
0 for i = 1, ... , n. Hence it may be assumed that 1. Thus
C H A RAe T E RS 0 F FIN IT E G R0 UPS
30
n-1 (A)
W
n
'E
a, w. (B) + W (A) w (B) i=1 l I n n
n-1 ==
'E
0
a. w. (AB)
+
w (AB)
. 1 lIn
1=
ct. Since w. (AB) == w. (A) w. (B) for i == 1, ..• , n I I I n this implies that 'E 1':11 {w (A) - w. (A)} a. w. (B) ;: 0 for
for A, B
E:
-
A, B A
E:
n I l
1
ct. Hence by induction {w (A) - WI (A)} a l == 0 for n ct. Since a l "" 0 this implies that wI ;: Wn contrary to E:
assumption. (5.3) In the notation introduced at the beginning of this section WI' ••• , 'ic are all the nonzero homomorphisms of Zg: (N) into g:. Proof. If w.
1
G
E:
= w., 1
then X. (G)/X. (1) 1
1
= x.1 (G)/X.1 (1)
for
N. Hence
Thus i
== j.
Thus wI' •.• ,
'ic
are k distinct nonzero alge-
bra homomorphisms of Zff (@) into ff. Since Zff (~) has dimension k over ff, the space of functions from Zff (@) into ff is k-dimensional over ff. The result now follows from (5.2). The previous result can be used to characterize the irreducible characters of ~. (5.4)
Let
e be a function from
@
to ff. Then ()
==
cx for
CENTER OF GROUP ALGEBRA
some irreducible character X of @ and some c only if
31 E:
ff if and
(5.5) for G, H
E: @.
Proof. If () satisfies (5.5) then setting G = 1 we see that () is a class function. Assume now that () is any class function with e(1) ;c O. Let G. E: .R. for j = 1, ••• , k. Define J J .
k
~
k
~
a function w on Zcr (@) by w(E. 1 c. R.) = E. 1 c. w(st.) u J= J J J= J J and w(R.) IJLle(G.)/6(1). Then J J 1 A.
E
6(GK-IHK):::;
KE:@
E
e(L-IGLL-IK-IHKL)
KE:@ =
L;
6(L-IGLK-IHK)
KE:@ Therefore
=
1j{~~IJV ~1.R'mlastm 6(Gm )
IGle(1) :::; ISlsIISlt l
~ a stm w(.R'm)
32
C H A RAe T E R5 0 F FIN I T E G R0 UPS
Thus
e satisfies (5.5) if and only if w(Sl)w(.Rt)=L)a
m st m
s
w(.R:)
m
Hence if e = cx then e satisfies (5.5). Suppose that e satisfies (5.5). H e(l) 0 then e = 0 by (5.5). Assume that e(1) ¢ O. Then by (5.3) w = w. for some 1
i. Hence B(G)/B(l) = X.(G)/X.(l) for G E: @. Therefore 1
1
B = e(l)/X (1) ~ as required.
i
§6. SOME PROPERTIES OF CHARACTERS In this section we collect some elementary properties of characters. Several of these are useful in applications of character theory to questions concerning the structure of groups. Let Xl' ~, ... be the irreducible characters of ®. Then for GE: @
(6.1)
1
Xs (G)
_
T@T L) Xs (GH 1) Xt (H) = X (1) @ s
Q
st
Proof. Let Is be a representation of
@
which affords
s X . Let I (G) :::: (a .. (G». Then by (1.9) s s IJ
I ~I \21
L) X (GH-l) xt(H) :::: @
S
I~I \21
: : I~I \21
L) ••
l,l
L a~.11 (GH- l ) a~.(H) ~ JJ
L) 1,J,m ••
x a t (H) jj
a~1m (G) L) asml.(H- l ) IU
\21
SOME PROPERTIES OF CHARACTERS
1 '" = -(1) LJ X . .
s
1,],m
33
s (G) 0 to .. 0 . a.1m s IJ mJ
s t '" LJ a .. (G) s.]]
1 = -(1) 0 X
s
]
X (G) =
s
Xl (1)
0
st
(6.2) Let ~ be a subgroup of ~. If 0 is a character of which vanishes on ~# then I~II 0(1).
Proof. It may be assumed that 0(1) _
1
o(1)/1.t>1
.t>. Then
_ - (0, 19,)c:..
TN - 1ST L; O(H) _ .\>
Thus
(~ =
~
'V
is an integer.
(6.3) If ~ = ~l X ~2 then a character of ~ is irreducible if and only if it is of the form X~, where X is an irreducible character of ~/~2 and ~ is an irreducible character of ~/~l' Proof. Let Xl' X2' ... be all the irreducible characters of ~7~2 and let ~l' ~2 be all the irreducible characters of ~/~l . Then 1 --(X. ~., X ~t) = IIUI L; X. (G) X (G) L; ~.(G) ~t(G) 1] s "-'IU 1 S ~] "-'1
::: O.
IS
2
O't J
Thus by (2.9) the set {Xi ~j} consists of pairwise distinct
34
CHARACTERS
OF FINITE GROUPS
irreducible characters of ®. Since Ei,j Xi(1)2
~j(1)2 =
I®\'
every irreducible character of ® is of this form by (2.9) (vi). Let p be a prime and suppose that P, G e::: @, (6.4) where P is a p-element, G is a p' -element and PG = GP. Let fT be an algebraic numberfield containing the I®I-th roots of unity and let ~ be a prime ideal in the ring of integers of fT such that p e::: ~. If e is a character of ® then e(PG) == 6 (G)(mod ~)
Proof. It suffices to prove the result for every character, and hence for every irreducible character, of the abelian group < P, G >. Let e be an irreducible character of < P, G >. Then 6(1) 1 and 6(PG) = 6 (P)6(G). Furthermore 6 (P) is a pm-th root of unity for some m and so 6 (P) 1(mod ~). The result follows. (Solomon [2]) Let sr l , ... , sr'k be the conjugate classes of ®. Let G . e::: .tt J.• Then E~ 1 X(G.) is a nonJ - - J= J negative rational integer for any irreducible character X of ®. (6.5)
Proof. Let ! be the permutation representation of ® on the elements of @' defined by H!(G) = G-l HG for G, H e::: ®. Let 6 be the character afforded by !. Since H~(G) = H if and only if HG = GH, the number of elements fixed by ~(G) is IC(G)I = 1@1/I·stl, where Sl is the conjugate class of @ containing G. Thus by (2.3) e(G.) . k J I®I/I~.I. Let 6 = E. 1 a. X· where Xl' ... , X are the k J 1= 1 1
irreducible characters of ®. Then each a. is a nonnegative integer and 1 a. = 1
I~I L;6(G.)I·~.lx.(G.) \'Y. J J1J J
=
L; x.(G.) .1J J
s 0 M E PRO PER TIE S 0 F C H A RAe T ERS
35
L. Solomon and J. G. Thompson have pointed out the fact
~hat in contrast to (6.5) 1:;~::::1 xi(G) need not be a nonnegative mteger. A character e of ® is a linear character if linear character is necessarily irreducible. (6.6)
e(1)
1. A
Let X., X·, X be irreducible characters of ®. The 1 J multiplicity of X in X. X. is equal to the multiplicity of X. 1 l 1 in XX.. If furthermore X is linear then the multiplicity of J X in X· X. is 0 or 1 . 1 l-
Proof. The multiplicity of X in Xi X j and the multiplicityof Xi in XXj are both equal to
1/INI
1:;(\\
Xi(G)Xj(G)x(G).
If X is linear then XXj is irreducible. Thus Xi is a con-
stituent of XX j if and only if Xi:::; XX j in which case the multiplicity of Xi in XX j is 1. (6.7) Let ~ be a representation of ® which affords the character e . Let S; be the kernel of eo Then (i) le(G)1 :5 e(1) for G E: ®. (ii) e(G) ;:: e(l) if and only if G E: .po (iii) If le(G)1 e(l) then G.\) is in the center of ®/S). If e is irreducible thenconversely III (G}I :::; e(1) for G.n in the center of ®/~). Thus in particular {Gle(G) :::: e(l)} <1 ® and {Glle(G)1 :::; e(1)} <1 ®. Proof. Let e(1) :::; n. The characteristic roots t I' .. En of ~(G) are complex roots of unity. Thus le(G)1 :::: lEI + ••• + tnl :s n proving (i). If G Ii: ~ then ~(G) ::: I and so e(G) :::; n e(1). If e is irreducible and GS; is in the center of ®/Sj then ~(G) E I by Schur's lemma and so Ill(G)1 :::; InE I : : n. Conversely suppose that Ill(G)1 :::; n. Then "'I .. En :::; E. Hence by (1.12) ~(G) :::: € 1. Thus 0
0
::;
,
36
CHARACTERS OF FINITE GROUPS
G,\) is in the center of @/.\). If furthermore 8(G) Eland G E: s:).
n then
(6.8) (Feit-Thompson [2] Lemma 4.3) Let j) <1 @ and let X be an irreducible character of @ whose kernel does not contain 3). If G E: @ and C(G) n s:) = (1)ihen x(G) ::: O. Proof. Let (1' ~2' ... be all the irreducible characters of @* = @/i). Let Xl' X2' ••. be the remaining irreducible characters of @. If C(G) n ,\) (1), then C(G) is mapped isomorphically into C(G*), where G* is the image of G in @*. Thus by (2.14) ~1(.(G)12 1
=
IC(G*)I ~ IC(G)I
== ~1(.(G)12 I
+ ~lx.(G)12 1
Hence Xi(G) ::: 0 for all i as required. The next result is due to Burnside. The proof was suggested by N. Hamilton several years ago.
(6.9) X(G)
If X is a nonlinear irreducible character of
= o for some element G
E:
6~
then
@.
Proof. If G, H E: @ let G z H if (G) ;;;: (H). It is easily seen that" R:': is an equivalence relation and G R: H if and only if G == HI for some i with (i, I@I) == 1. Let I be a representation of @ which affords X. Let E l ' ••• , (;. n be the characteristic roots of I(H). If (J E: 9 ~ l@l/~ then there exists i with (i, I@I) ::: 1 such that EC!
. 1
]
==
d. ]
Thus
x(H)(J x(H ). Hence if ~( is an equivalence class with respect to : : : then 9 ~ l@'/~ permutes the set {X (H) I H E: ~(}. Suppose that X(G) ¢ O. Let ~( be the equivalence class with respect to : : : which contains G. The arithmetic mean of positive real numbers is at least the geometric mean.
SOME PROPERTIES OF CHARACTERS
37
Since complex conjugation commutes with any element of 9 f.l1@1/ f.l this implies that
J.r; Ix(G)12 I~I
?:
11
~n IX(G)1
2
~11
}
.
1/\111
As x(G) is a nonzero algebraic integer
nll lx(G»2
Therefore :E Ix(G)12 ?: 1111. Thus if X(G) 1I G E: @ then by the orthogonality relations
'B#
I@I ?: X(1)2 +
;;1f:.
1.
?:
0 for all
Ix(G)12 ?: X(1)2 + I@I
1
@
Hence X(1)2 !S 1 as required. The following striking consequence of (6.9) was discovered by Brauer [4] and Wielandt. However, it is a result that seems to be difficult to use. (6.10)
~
~
@' if
and
:E~ l,ft,
is proportional to J= J n, 1 ,\l. in Z ""(@), where the notation of Section 5 is J= J " k
if
= c nk
Proof. Let Xl
j=l
1,~1.1
J
then
Wl(~j=l .u.)J Since
:E~ 1'\!'
we see by (5.1) that only if w,
1
(n~J= l,R,) ]
J=
=
]
is proportional to
0 for i
;;1f:.
n~ 1 ,ii, J=
J
if and
1, or equivalently X,( G) ;: 0 1
for some G in case i ;;1f:. 1. Since a linear character never vanishes this statement is by (6.9) equivalent to the fact that
38
C H A RAe T E RS 0 F FIN I T E G R 0 UPS
Xl is the only linear character of ®, or equivalently ® = as required. The next result due to R. Brauer [5] generalizes a result of Burnside. A similar argument was used by Fong and Gaschiitz [1) in a different context. ~'
(6.11) Let () be a faithful character of ®. Suppose that () takes on a total of r distinct values aI' ... , a r . Then each irreducible character X of ® is a constituent of one of the characters () 0 = 1~' (), .. -:-, () r-l .
= {GIG
Proof. Let 11.
- - ]
E:
®, () (G)
= a.}. Let G. ] ]
E:
11,. ]
If X is not a constituent of () s then
o = 1®1«()s,
X)ttt
= L;es(G.) L; 1
\!y,
x(G)
~j
1
=Ea~"E X(G) .
1
1 ~
j
= 0, ... , r-l then the nonvanishing of the Vandermonde determinant implies that for each j E... X(G) = O. Since e is faithful (6.7) implies that for .n J some j, ~, = {I}. Then X(1) = 0 which is impossible.
If this holds for s
]
(6.12) Let Xl' ... Xk be all the irreducible characters of ~. Let ff be a splitting field of ~ of characteristic 0 and let Xi be afforded by the ff-representation . For i, j = 1, . " .1 k define the matrices Sij by
I.(G) ® I,(G) 1
J
= S~.1 [ IJ
J
Ii (G} 0 l o y (G) . ~ is
SI']'
for all G E: ~. The set of matrices S., determines the IJ - - - - ' - - group ® up to isomorphism.
s 0 M E PRO PER TIE S 0 F C H A RAe T ERS
39
Proof. Let A = (A 1 , ... , A ) be an ordered k-tuple of k nonsingular matrices A. such that A. has degree X.(1) and 1 1 1
A.
1
Let AG
® A.
J
= S:-.l IJ
be the set of all such k-tuples A. If G E: @ let '1(® = {A }. For A E: '1( G let {a.(A)} be the various matrix entries in the set of ma'1(
= (I 1(G), ... , Ik(G)) and let J
trices A1 , ... , Ak , where each subscript j denotes a place in one of these matrices. Thus j = 1, ... , I®I. For G E: @ let a.(A ) = a.(G). By assumption a.(A) a.(A) = ~ c .. a (A) G J J 1 J S IJS S for A E: n, where c.. depends only on the matrices So.. ~s
Thus for A
Es
E: '1(
{c..
I]S
~
and all i, j 0. a.(A)} a (A) JS
1
S
=
0
By (1.10) {as} is a linearly independent set of functions on n® and hence on n. Thus the determinant of the system of equations above vanishes where i is held fixed and j = 1, ... , I@I. Hence ai(A) is a characteristic root of the matrix whose (j, s) entry is c ijs ' Consequently ~ is finite. If A = (Ai)' B = (B i ) are in n then clearly AB :::: (AiBi) (A:-1) are in ~. Thus ~ is a group and ~IU is a and A-:1 1
~
1
subgroup of ~ isomorphic to ®. For each i the mapping which sends A to A. is a representation of '1( which will be 1
denoted by ID.. Since ID. (A ) I. (G) for G E: @, ID. is 1 1 G 1 1 absolutely irreducible. Let ID = ID1 EEl ••• EEl ID k' then ID is a faithful representation of ~(. By assumption every irreducible constituent in any tensor power IDn of ID is similar to some 2) •• Thus by (6.12) every absolutely irreducible 1
40
C H A RAe TE RS 0 F FIN I T E G R0 UPS ~
representation of
is similar to some ID .• Since ID. has 1
1
degree X .(1) it follows from (2.9) (vi) that 1
k
I~I ::;
6
. 1
x.(1)2 ::; I@I 1
1=
Thus @ is isomorphic to ~. As ~ is determined by the matrices S .. the result follows. 1J An algebraic number a is said to require the m-th roots of unity if ~ (a) c ~ , and m is the smallest integer with m this property. Note that if m here is even then 41m. The next result is due to Blichfeldt and Burnside. The proof given here was first published by R. Brauer [51.
(6.13) Let X be an irreducible character of @. Let PI' ... , Pn be distinct primes. Assume that there exist elements GI , ... , Gn E: @ such that X(Gi) requires the Pfi -th roots of unity for some ai> 0, i
1, ... , n. Then
@ con-
tains an element G of order p;l ... pan . n
b· b·+l Proof. Let p.111@1 and p.l 11@1. Then a. --
1
b.-a.h
5. be the field of I@\/p.l 1
1
1
~
b .. Let
1
-th roots of unity over
1
~.
Since X(G ) ¢ 5 i there exists an element 0i E: S ~ 1@1/5 . i 1 such that 6.(x(G.)) :r= X(G.). On the other hand o.(X(G.» 1 1 1 1 J X(G.) for j :r= i. If the result is false then for G E: @, J ci.(x (G» = X (G) for some i depending on G, 1 ~ i ~ n. 1
n
Therefore n =l (1 - 0i)(x(G»
0 for all G E: @. Hence i n~=l (1 - 0i)(x) ::; O. Expanding this product we get that
41
CHARACTER TABLES
x + :6 (\
OJ' (X ) + •••
i<j
:6 0. (X) i
:6
+
i<j
1
o. o. 1
0k(X) + ...
J ~
Hence there exist iI' ... , i. for some J X 0 .... o. (X). Thus 11 1. J X(G. ) 11
= o. . .. o. 11
1j
(X(G. )) 11
= O.
1 such that
(X(G. ))
11
11
contrary to what was shown above. It is an open question, raised by J. Thompson, whether it is possible to find G in (6.13) such that G = HI ... Hn , where H. H. = H. H. for i, j = 1, ... , n and where H. is 1 J J 1 1 conjugate to G. for i = 1, ... , n . 1 Another open question related to (6.13) is the following: Let X be an irreducible character of @ and let m be the smallest integer such that ~ (X) c ~ . Is it true that @ m contains an element of order m? See Brauer [6] problem 41 for comments related to this question.
§7. CHARACTER TABLES Let Xl' ••. , Xk be the irreducible characters of let .\1 1 ,
••• ,
@
and
.\lk be the conjugate class es of @. Let G. E: .\1. J J
for j = 1, ... , k. The matrix X
= (X .(G.)) is a character 1
J
table of @. Any character table of @ can be transformed into any other by permuting characters and conjugate classes. If .\', <1 @ then .,) is the kernel of p S;' Thus .\) is the
(w
intersection of the kernels of some characters. Hence if X is given it is possible to find the lattice of normal subgroups
C H A RAe T ERS 0 F FIN I T E G R0 UPS
42
of ® by using (6.7) and to determine which classes lie in a given normal subgroup. Thus in particular it is possible to determine whether .\l. {t}. Hence X.(1) can be read off J 1 from X. By (2.9) X determines I®I. Then by (2.14) X determines I·Q.I for each j. Thus IZ (~)I is determined by X. J Since ®' is the intersection of the kernels of all the linear characters I®'I can also be determined. There exist two nonisomorphic nonabelian groups of order 8. We will show that they have the same character table. Let ® be any nonabelian group of order 8. Then (W = Z{®) and I®'I = 2. Thus ® has exactly 4 linear characters Xi' i = 1, ... ,4. Since ~~ 1 X.(1)2 = 8 it follows that ® has exactly 1=
1
one other irreducible character Xs with Xs (i) ::0 2. If Z E: Z(®)# then by (6.7) IXlI{Z)I = 2. Thus Xs(Z) :::: ±2 since Z2 ::: 1. Since Z is in the kernel of Xi for i = 1, ... ,4 it cannot be in the kernel of XS' Thus by (6.7)Xs{Z) -2. Hence XS(1)2 + XlI(Z)2 8 and so XlI(G) = 0 foz:G ¢ Z(®). It is now easy to show that a character table of ~ is 1 1 1 -1 1 1 1 -1 0 2
=
1 1 1 -1 1 1 -1 -1 1 -1 1 1 0 0 -2
Dade (1j has shown that if in addition to X it is known when a conjugate class consists of m-th powers of the elements in another conjugate class, for all integers m, then this information is still not sufficient to determine 0; up to isomorphism. It is an open problem to find reasonable conditions which together with a character table will characterize a group @. For a discussion of this question see Brauer
[6, §4]. (7.1)
Let a..
-
I1S
have the same meaning as in (2.12). A know-
ledge of all the a.. is equivalent to a knowledge of X. ~s -
43
INDUCED REPRESENTATIONS Proof. Given X then as remarked above, x.(l) can be determined. Thus by (2.15) a..
~s
1
for all i,j,s. Suppose that all a..
matrix (w.(.\l.)) is determined. JL J
1
J
is determined
are given. By (5.1) and (5.3) the
IJS
,
I@I, 1·~tI and
= 6 ... Thus by IJ Hence also I@I = 2:. I"LI all i a ...
JJI
J
(2.12)
= {t} if and only
if for
l.n.1 is determined for all J
j.
can be found. Since w. (.n.) :: J 1 J 1·\1..1 X.(G.)/X.(l) we see that X.(G.)/x.(l) is determined. J IJ 1 IJ 1 Thus by (2.9) it is possible to determine J
"
-\(Gj )
' - X.(l) j
1
w.(.n.) 1
J
= I@! X .(1)2 1
Hence x.(l) can be computed and thus also x.(G.). 1
1
J
§8. INDUCED REPRESENTATIONS Let ,\) be a subgroup of @. Let ,\)Gu
...
,,\)G m be all the
distinct right cosets of ,(,') in @. If ~ is an tr-representation of 3) of degree n let ~(G) be the zero matrix of degree n in case G E: @ - s:..,. Define ~*(G) = (~(G.GG:l)) for G E: @, 1 J where each pair of indices denotes a sub matrix of degree n. (8.1) If ~ is an tr-representation of .f) of degree n then F* is an tr-representation of @ of degree 1@:3)1 n. Proof. For G,H Then iV*(G) ~*(H)
E: @
let B .. = (2:
= (B..). IJ
IJ
i'V(G.GG-l) i'V(G HG:1 )).
SIS
S
1
Given i there exists a unique t
44
CHARACTERS OF FINITE GROUPS
such that G.G 1
E:
~Gt or G.GG-tl -
1
E:
~. Thus B.. = ~(G.GG-tl) -
IJ
1
lY(GtHG:1 ). G.GHG:l E: .\1 if and only if G.GG-1 E: ~G.H-IG-tl t J 1 J 1 - J 1 and this is the case if and only if GtHG: E: ~. Thus B.. ::: 0 J IJ if G.GHG:l Et ~ and 1 J B.. :::: ~(G.GG-tl) iY (GtHG:l) = ~(G.GHG:l) IJ
1
J
1
J
if G.GHG:l E: ~. Hence ~*(G) ty*(H) ::: ~*(GH), ~*(1) = I 1 J and thus ~ *(G) is nonsingular for G E: ®.
The representation ~* defined above is said to be induced by ~, An 5-representation ~ of ® is monomial if there exists a permutation 5-representation lYo of ~ such that the underlying vector space '0 of iY has a basis {v.} with the property 1
that v.5(G)::: A.(G)v.5o (G) for G E: ~,where A.(G) E: 5, 1 1 1 1 ~o is called the associated permutation representation of ~. If ~o is transitive then ~ is transitive. A character of ® is a monomial character or a permutation character if it is afforded by a monomial representation or a permutation representation respectively.
(8.2) The tensor product of monomial ff -representations or permutation ff-representations of ® is a monomial ffrepresentation or a permutation ff-representation respectively. Proof.
This follows directly from the definition.
Let .\) be a subgroup of ®. Let ~ be a representa(8.3) tion of of degree 1 and let ~o pe the unit representation of ~). Then ~* is a transitive monomial representa!ion of ~ and ~ci is its associated permutation representation. Conversely if 11 is a transitive monomial representation of
45
INDUCED CHARACTERS
with associated permutation representation ~o and if .\') is a subgroup of ® consisting of all elements G such that i'fo (G) leaves a given object fixed then ~ is similar to and ~1) ::: ~3', where ~ is a representation of degree 1 of .\) and ~o is the unit representation of ~). @
Proof. .\) in
Let .\')
= ,\)GI , ... , ,\)G m
be all the right cosets of
Let '0 be a vector space with basis VI"'" v • m The definition of i'Y* implies that vi ~*(G) = ~(GiGG't)vt' @.
where GiGGt E: .\'l. This proves the first statement and also shows that ~o ;;; ~3'. Let ~(G)
(a .. (G)) for G E: @. IJ
Since ~ is the associated permutation representation we see that a .. (G) = 0 if G.GG:-I ¢ .\). Let GI :::: 1. Let a .(G,) 11 1 IJ 1 J c., let ~(G) (c. 0 . .) iIf(G) (C-;-I 0 . .) and let ~(G) = (b .. (G)). 1
1 IJ
Then b .(G.) 11 1
1
= b.11(G:-I). 1
1
1)
IJ
Thus if G.GG-;-I E: .\) then 1 J
Since ~ is a representation of ® it follows that ~ = b u is a representation of .\) of degree 1. Thus ~ = ~ * as required. It is not true in (8.3) that ~ = ~* for some representation of degree 1 of ~). For example let @:;; (G) be the group of order 2. Let iIf(G) any field of characteristic sentation of degree 1 of ,\)
§9.
(_~
==
~
=
2 and (1).
~). ~o
Then i'f
~ ~3'
in
is the unique repre-
INDUCED CHARACTERS
Let ~) be a subgroup of @ and let () be a complex valued class function on ,\1. Define 9(G) 0 for G E: ® -~) and
=
46
CHARACTE,RS OF FINITE GROUPS
for G E: 6}. The function 0* is said to be induced by (). Clearly 0* is a complex valued class function on 6} and «()l
+ ( 2 )* =
()l*
+
()2*'
(9.1) Let ~ be a subgroup of 6} and let () be the character afforded by the representation ;V of .\). Then ()* is the character afforded by ;V*. Any two representations induced by are similar. m
Proof. Let 11 be the character afforded by iV *. By definition 11 (G) = :0.(G.GG-;-l), where {G.} is a complete system 1
1
1
1
of right coset representations of £1 in ~. Since function on S), 9(HG.GG-;-lH-l) = (J(G.GG-;-l) for H 1
1
1
1
(J
E:
is a class i). Thus
The last statement now follows from (2.6). An important consequence of (9.1) is the fact that the class function induced by a character of a subgroup is again a character. This is probably the most important method for constructing characters of groups. Frequently we will be considering various subgroups of a given group ~. In such situations (J* will always denote the class function of ~ induced by the class function 0 on any subgroup of ~. (9.2) Let ,\) c \1 C 6} where ,\) and \' ~re subgroups of Let e be a complex valued class function on ,\) and let 7J be the class function of \' induced by 9. Then e* 7J*.
@.
Proof. Let 9(G) = 0 for G E: ~ -,\). For ME:\,
0' (M)
47
INDUCED CHARACTERS
(1/1~1) ~t' e(LML- 1 ). Let 9(G) ::: 0 for G fore e(M) = (11 I,\) I) ~QJ e(LML-1 ) for M nition
for M
E: QJ -i'.
E: QJ.
There-
Hence by defi-
E: QJ.
If .\) is a subgroup of OS and 8 is a complex valued class
function on (~ then the restriction of 8 to S:" denoted by e1r:' is clearly a complex valued class function of ~.,. ,1 (9.3) Let ~ be a subgroup of QJ. Let 8,1] be complex valued class functions of .\i,QJ !espectively. Then 9*1] = (81] I~)* Proof. Let 9(G)
= 0 for G
(01]1';)* (M) = ~
I.£)I
'V
E QJ
1] (M) =
E: QJ -~.
For M
E: QJ
0 (GMG-1 ) 1] (GMG-l)
I~I ~ 0 (GMG-l)
1]O*(M)
(9.4) (Frobenius reciprocity theorem) Let S) be a subgroup of QJ. Let 0,1] be complex valued class functions of ~,QJ respectively. Then -
Proof. By definition
48
C H A RAe T ERS 0 F FIN I T E G R0 UPS
1
(T},e*)@
2:;
I@I 1.f>·1 =
=
= =0
Since O(G)
1
(T},O*)@ ==
G,H E:
1
1
T} (HGH-1 ) e (HGW 1 )
G,H E:
1
@
2:;
I@I l.pl
T} (G) tJ (G)
G,H E: ®
I~I ~
T} (G) O(G)
E: ~
for G
@
2:;
I@I 1.f>1 1
T} (G) e (HGH-1 )
I~I E £)
- .\) this implies that
T} (G) e (G)
(T} l.p' e),!)
Let .\) be a subgroup of ®. Let Xl"'" X be all k the irreducible characters of GJ and let ~ 1, ••• , ~ m be all (9.5)
the irreducible characters of let ~~ ::: J
2:;.
1
~).
Let X.
I
1 3;
= 2:;. a .. ~. and
J
1J J -
b.o X.. Then a .. = b .. for all i and j. 1J
1 --
1J
1J
Proof. By (9.4)
a1J..
=
(X. ,~.)\; = (x.,~:Lu = boo I,.\) J •. 1 J ~ 1J J
(9.6) Let (J .!?~ a permutation character afforded by the permutation representation ~ of ®. The number of domains of transitivity of ~ is equal to 19,1 )®. GJ
(j
Proof. If ~ has m domains of transitivity then by (8.3) + ••• + 1 c; * , where '\;u"" ~)m are subgroups of
= 1,-
*
~)l
~)m
®. By (9.5) (1~,1~\~ sult follows.
1 for any subgroup
S)
of 6;. The re-
49
INDUCED CHARACTERS
H Sj is a subgroup of ~l, G €: @ and () is a complex valued class function on .\.) define the complex valued class function OG on f>G by (;I G(M) = (;I (GMG-I ) for M
{tp}H = eGH . H
e is a
S:·P.
€:
Thus
character of ,\) then (;IG is a charac-
ter of s:",G. The following elegant result was discovered by Gallagher [1]. However it is a result that seems to be difficult to apply. (9.7) The following conditions are equivalent. (0 H GI ,G 2 , ••• are elements of @ whose orders are pairwise relatively prime then n. G. = 1 if and only if G. = 1 . 1 1 1 f or eac h 1. H X is a nonprincipal irreducible character of @ then there exists a prime p and a S -group '13 such that (:X'li3,I )
= o.
i3
p
Proof. dividing
Let PI' ... ,p be the set of all distinct primes . Let '131' be ~ S -group and let e. = 1~ = Pi
z;. c .. X., where X J IJ J ters of @. Let
=
I
1
""i
11U , X , ... are the irreducible charac\Y
2
6
s
G I , · · · , Gn G1 G2 ••• Gn =1
Observe that s for (G 1 ,
••• ,
>
G )
1 and s
;t!
=1
if and only if
n,19.(G.) 11
= 0
(1, ... ,1). Since O. vanishes on p~ -ele-
n
I
l
ments it may be assumed that in the definition of s, G.
1
ranges over the p. -elements of @. Since 11. (G) ) 0 for any 1
p. - element G. it follows easily that s 1
1
1
=1
if and only if
condition (i) is satisfied. For any n-tuple (G , ... ,G ) define HI' G ... G. for 1 n I l i 1, ..• , n - 1. Thus
C H A RAe T ERS 0 F FIN I T E G R0 UPS
50
1
x
6
H , ... ,H 1
n-l
X. (H)··· X. (H- l ) J 1 J n-l 1 n
Applying (6.0 n - 1 times this yields that
1
By the Frobenius reciprocity theorem s = 1 if and only if for j
;;t!
n1;:l
c h ::; 1. Thus
1 there exists i such that c .. = O. 1J
In other words s = 1 if and only if condition (ii) is satisfied. Thus the statement that s = 1 is equivalent to both statements (i) and (ii). It was first pointed out by P. Hall that condition (i) of (9.7) is satisfied in case @ is solvable. This can be proved by induction in a straightforward manner. He also conjectured that
INDUCED
51
CHARACTERS
the converse is true. The proof of this conjecture has recently been given by J. Thompson. However, it is far beyond the scope of these notes. (9.8) (Mackey) Let Xl,tI be subgroups of ®. Let 0,11 be complex valued class functions on ~), tI respectively. Let ()G l .\!; •.• ,'£)G m tI be all the (,u, tI) double cosets in ® and let G)i :: ~)Gi for i :: 1, ... , m. Then
n \'
«() *,11 *)®
m ( G'1
:: . ~
1-1
()
)
1®.,1111U 1
\'Y.
1
<M
i
Proof. By definition (f)
*,11*)Ci}
1 11M
1-';1
m I®I
1
1 1 [\II I®I
l.u! If MN-l
HGi L for H E:
and
E
f)MN"l (G)11 (G)
.E
E
N E: G) M,G E: ®
E
N"l (G)l1(G)
f)M
M,N,GE:®
,u, L E: tI then OHGiL ::
=
GE:®
E
G L f) i (G)l1(G)
GE:®
E
1
f)G i (G)l1 L - (G)
GE:®
E
f)G i (G)l1(G)
GE:®
=
E G E: G)i
N
(G)11 (G)
f)
f)G i (G)11(G)
f)
Gi L
52
C H A RAe T ERS 0 F FIN IT E G R0 UPS
Thus
m
=
1 11 - E E IRII i\il IGI i = 1 M E: ()~
E
NE:@ MN-1 E: £'>G. \1 1
The result now follows since
1.\'1 Ittl =
I~Gi ~I
I®/
(9.9) Let e be the character afforded by a transitive permutation representation ty of ev. Let ~) J:>e the subgroup of ()~ consisting of all elements G such that ~(G) leaves a given object fixed. Then 11911@ is equal to the number of domains of transitivity of ~). Thus in particular iT is doubly transitive if and only if lIell@ 2. Proof. By (8.3)
* Let Sj = \? and is).
(J
(J
= 1]
=
1,\, in
(9.8). Then in the notation of (9.8)
Thus He II@ is the number of double cosets of ,\) in implies the result. Suppose that ,\' <j
@
@
which
Let {} be a complex valued class
INDUCED CHARACTERS
53
function on .\). Then OG is also a class function on ~) for G €: ®. Define the inertial group J(£1) of IJ by ",(e) ::: {G IG
Clearly
.("1
€: @,
£1 G ::: £1}
C ...'(9) and £1 G ::: £1 H if and only if GH-1
€: ...'(9).
(9.10) Let~)
ible character of @ then there exists an irreducible character 9 o,f ~ and a positive integer e = e~ (X) such that X IS) = e
Li 9
Gi
-
where {G i} is a complete system of coset representatives of j(lJ) in @.
Proof. The first statement follows directly from the definition of £1*. Let X be an irreducible character of @ and let 9 be an irreducible constituent of X IS)' Then X c £1* by the Frobenius reciprocity theorem. Since - (X !'£;' e) S) ::: (XI~Y £1G),p for G €: @ XI.\) has the required form. .. The integer eSj(x) appearing in (9.10) is the index of ramification of X with respect to Sj. (9.11) Let X be an irreducible character of (};. Let ~)
.'."1
irreducible constituent of XI3(£1) such that 9 C ?; IS)" Then X :::
t*
and e. (X) ::: e _ (?;), 5,) .))
Proof. By the Frobenius reciprocity theorem XC?; * . Hence it suffices to show that ?;* is irreducible. Let {G.} be a complete system of right coset representatives of Jt9) in ®. As Sj
C
,,1(11)
n
,,1(£1)
G·
1.
As tl~ ::: eS)(t) IJ,
54
CHARACTERS OF FINITE GROUPS
G·
Gi
t I.\~ = es,(t) 8
G· and (0,8 \ )
0 for Gi ~ 3(8) we see that G G· (t I~i' ~ I@i)@i = 0 for G i ~ ,..1(0), where @i = 3(0) n J(O) 1 Thus by (9.8)
which shows that t* is irreducible. Furthermore
which implies that e~ (t) =
eg;/x)'
(9.12) Let X be an irreducible character of @. Let ~
cyclic then e4) (x)
1. If 3(9)/.\) is solvable then
Proof. Assume that J(8)/.\1 = (G£» is cyclic. In showing that e ;:;: ef,(X) = 1 it may be assumed by (9.11) that @ 3(0). Let II be a complex representation of ~ which ~fords the character 9. Define ;yG(H) i}(GHG-l). Since 8 = e there exists a matrix C such that
ci ~(H)C-i
=
ty Gi(H) for H
E: S).
Thus if n = I@:~I, C-n~(Gn) is a scalar matrix by Schur's lemma. Hence there exists a scalar matrix S so that Sn = C- n B(Gn ). Define iV(GiH) = siciiJ(H) for H E: S;1, i 0, i .•. , n - 1. Thus i}(GiH) = SiC ~(H) for all i, H E: S) and if Hu H2 E: .\') then
INDUCED CHARACTERS
iY{GiHI ) ~(GjHa)
==
==
55
sid ij (HJ sje j iY(Ha) Si + jei + j iHG-jH I Gj) i}{Ha) Si + je i + j il (G- jHI GjH ) a
= i}(GiHIGjH a)
Thus ij is in this way extended to a representation of @. Let Xl be the character afforded by this representation. Then e* == X1P@!.\'Y· Since @/,p is cyclic p@/,p = LiAi' where A. ranges over all the distinct linear characters of 1
@/,p. Thus 8* = Li \ Ai' As Ai is linear X/'i is irreducible and XI AI" (1)
==
X1 (1)
stituent of 8*, X e,p(x)
==
e(l}. Since X is an irreducible con-
XIAi for some i. Thus X(1)
==
8(1) and so
1.
==
Suppose now that 3(e)/,p is solvable. It remains to show that e,p(X} 113(lJ):,pI. The proof is by induction on I@:,p[. It may be assumed by (9.11) that @ =3(9). Thus XI,p where e
== eO
ef,{X)' If @ = .,p the result is trivial. Suppose
that @ ~,p. Since @/S; is solvable there exists @l
..• ,
e._ (1; J '\")
1
t n are distinct irreducible characters of @l with ==
er.; (t: J. By induction e", {1;.} 110; :,pl. Since e",. (X) = ",")
J
.v 1
I
-v
ne~{t:l) and nip this implies that eS;(x)l p eS;(t: 1 ) and so
e._ (X) 11@:f,1 as required. ,,)
It can be shown that if ,p
56
CHARACTERS OF FINITE GROUPS
general eS}X) 113(9): ~)I. However a proof of this result involves the theory of projective representations. See Curtis and Reiner p. 364. We will content ourselves with another special case, (9.13) (Ito [1]) Let X be an irreducible character of @. Let .\)<1 @ and let e be an irreducible constituent of XI.\)' Assume that .\) is abelian. Then e9) (X) I L..1(9):.\)L Thus in particular X(1) II@:.\)I.
-
Proof. By (9.11) it may be assumed that J(e) ::= @ and X is faithful. Thus XI.\) ::= ee, where e ::= e.\)(x) and e is a faithful character of 9). Since 9) is abelian 1X1) = 1 and so ,S) is cyclic. As gG == 8 for G E: @ it follows that ,S) c Z(@). Thus I@:Z(@)II I@:'s)I. Hence by (4.2) X(l) 11@:'s)I. The ~esult follows since X(l) = ee(l) = e. There is no known general criterion which ensures that e9)"X) == 1. The next result gives a sufficient condition for this to happen and is a slight generalization of part of Lemma 4.5 in Feit-Thompson [2J. We will here and elsewhere make use of the following theorem due to Schur and Zassenhaus. See for instance Scott p. 224 and p. 227 or Zassenhaus p. 162. If .\)<1 @ and S) is an S-subgroup of ® !hen there exists a subgroup .\' of @ such that ,S) l! = @ and .t> Ii l! = 1. If furthermore either ~ or @/ ~ is solvable and l! 1 is another subgroup of @ such that .~~ 1 = @ and .t> Ii \! 1 = 1 then ~ is conjugate to ~ 1 in @. -The group l! in the theorem above is said to be a complement of .S) in @. It is clear that ~ is isomorphic to @AP. (9.14) Assume that .S)
e
e
INDUCED CHARACTERS
57
the irreducible characters of 3(e)h:;, each 1; Ai is irreducible and tAi 1;Aj if and only if i = j. If furtherm().I'~ X is irreducible constituent of e* then X I~- ::: L;J' eGj where .J
{G/ is ac::omplete system of coset representatives of 3(e) in
@.
Proof. By the Schur Zassenhaus theorem there exists an abelian complement ~ of s., in 3(e). Let ~ be a maximal subgroup of 3(e) such that ~ has a character S with t I~ ::: e. If ~ 'I! 3(e) then there exists ~ 1 C 3Un such that
1\1 1 : \II = p is a prime and \I <j ~lsince 3(e)/.\'> is abelian. Let ~ '" ~ n ~'~1 ::: \11 n ~. Let be the character of \I induced bye. Then (j ::: sP\I/S)' Thus 9 ::: :Ei SAp where
e
A. ranges over the irreducible characters of ~/~. If B 1
then
eB ::: e.
E:
581
Thus sB = sAi for some i. Let q be a prime,
q II~:~I and let Q be a q-element, Q
E: ~.
Since s(1) ::: 0(1)
and e(1) 11.s;1 we get that to) ;t!l 0 (mod q) as S) is a S-subgroup of J(e). Thus by (6.4) r(Q) 'I! O. Since ~1 is abelian
Therefore Ai(Q) ::: 1. Thus the kernel of Ail~ contains every element of prime power order and so ~ is in the kernel of Ai' Thus t B t. Since ~/.(I is cyclic it now follows from (9.12) that there exists a character t 1 of \1 1 such that tIl \I ::: S and so r 11 ,Q ::: e contrary to the maximality of ~. We have shown the existence of a character t of 3(e) such that 1; IS)
O. Therefore
e : : tPS(e)/.\)
::: :E tAp where \
ranges over the irreducible characters of 3 (e)/ fl, Each tA. is irreducible as A. is linear. By the Frobenius reci1 1 procity theorem
58
C H A RAe TE RS 0 F FIN IT E G R0 UPS
Thus rAi
;r.
If
rAj if i
;r.
j.
x is an irreducible constituent of e*
L:. (t A.)* 1
1
then X is an irreducible constituent of some (t Ai)*' Since (t Ai)
*I,f)
= L: j () G·J,
where { Gj } is a complete system of
coset representatives of 3(e) in @ the result follows.
§lO.
M-GROUPS
An M-group is a group @ all of whose irreducible characters are monomiaL A complete classification of M-groups has not been given. However there exist large classes of groups which are M-groups. First a result which restricts the nature of M-groups. (10.1)
(Taketa) Every M-group is solvable.
Proof. Let @(O)
= @ and
@(i + 1)
=
[@(i), ~(i)J. Let @
be an M-group. If @ is not solvable then @(i) = @(i+l) '" (1), for some i. Let X be an irreducible character of 6; of minimum degree which does not contain @(i) in its kerneL By assumption X A* where A is a linear character of some subgroup ~) of @. Let 11 = 1 ~y By the Frobenius reciprocity theorem
l(~i
"
c 11 *. Thus if
~
is
a~
irreducible
constituent of J.L* then.f (1) < X(1) and so ~(1) is in the kernel of t. Hence @(1 is in the kernel of J.L. This im.
/U(i)
c:.
(i)
/U(i + 1)
•I
phes that \'Y C "-'. Consequently @ = \'Y C~) and thus @W is in the kernel of A. Since @(i)
M-GROUPS
(10.2) ~~group
59
Assume that S>
Proof. The proof is by induction on I@I. Let X be an irreducible character of ®. If X(1) = 1 then X is monomial. Suppose that X(1) > 1. Let \' = {A.} be the group of all l linear characters of ® such that XA. = X. By (6.6) XX = J ~ \,Aj + (), where e is a sum of nonlinear irreducible characters of ®. By (9.13) the degree of every irreducible character of ~ is a power of p. Since XX (1) = I\'I + () (1) this implies that pi I \' I. Hence there exists A E: ).1# such that AP = 1~' Let ~o be the kernel of A. Thus I~:®ol = p and (XX I®o' 1~o)
>
2. Hence XI®o is reducible. Let I; be an
irreducible constituent of X I~o' Then I; (1) < p~(1)<
xU)
and so
xU)
since 1;(1) and X(1) are powers of p. By the Frobenius reciprocity theorem X C 1;* and so X(1) < 1;*(1) = pl;(1). Consequently xU) = pt (1) and therefore X ~ t*. By induction I; is monomial and so X is monomial. (10.3) The direct product of M-groups is an M-group. Every nilpotent group is an M-group. x .t1n where each '\)i is an M-group. Let X be an irreducible character of ~. By (6.3) X = \ ... Xn' where Xi is an irreducible character of ili Proof. Let ® = .\:)1
X •.•
for i = 1, ... ,n. By assumption each X. is monomial. The 1 first statement follows from (8.2). The second statement follows from the first and (10.2). For generalizations of (10.2) and (10.3) see references in Curtis and Reiner p. 357.
CHAPTER II
§ 11.
THE SCHUR INDEX
(11.1) Let fF be a field of characteristic 0 and let X be a finite extension field of fF. If e is a character of which is affqrded by a X-representation of then Tr3<1fF (0) is a character of (~ which is afforded by an fF-representation of N. Proof. Let ('V(G) ;:: (aij (G)) for G E: (~. Let m be the regular representation of X over fF. Define ~r(G) ;:: Wl(aij(G)) for G E: N, where each m(aij (G) ) is a submatrix of degree [X: fF] . Then '1(, is an fF-representation of Nand TrX/ fF (9) is the character afforded by ~. (11.2) Let X be an irreducible character of ~ and let fF be a field of characteristic O. Then there exists a positive integer m such that mx is afforded by an fF(X )-representation of (I,}. Proof. By (1.11) there exists a finite extension field X of fF(X which is a splitting field of X. Since TrX/fF(x) (X) :::: LX:fF J X the result follows from (11.1). Let X be an irreducible character of 6) and let fF be a field of characteristic O. The Schur index mfF (X) is defined to be the smallest positive integer such that mfF (X) X is the character afforded by an fFed-representation of @.
(11.3) Let )( be an irreducible character of W and let fF be a field of characteristic O. 61
62
C H A RAe T E RS 0 F FIN IT E G R0 UPS
(i) If mx is the character afforded by some ;:r(x)-representation of ® then m;:r (x) I m. (ii) If X finite extension field of ;:r which is a splitting field of X then mt;= ) I lx: 5' (X . Proof. (i) It may be assumed that mx is afforded by an ;:r(x)irreducible ;:r(X )-representation ~ of ®. The minimality of mt;=(X) implies that m;:r(X)X is afforded by an t;=(x)-irreducible ;:r(x)-representation W of (\l. By (1.9) and (2.9) ~ and rt are similar. Thus m = m;:r (X). (ii) By (11.1) [X:;:r (X) ] X is afforded by an ;:r (x)-representation of ®. The result follows from (i). It can be shown by using the theory of algebras that an irreducible character X always has a splitting field X such that m ff (X) = [X:;:r (X) ]. See Curtis and Reiner § 70. (11.4) character afforded by some ;:r-irreducible ;:r-representation of ®. Then there exists an irreducible character X of N such that () = m;:r(X) 2; Xa where a ranges over 9 ;:r(x)/;- and ()
is the unique character afforded by an ;:r-irreducible ;:r-representation of N which has X as a constituent. Thus if 11 is .the character afforded by some ;:r - representation of ® and X is a constituent off] then m;:r(X) I (X,11). Proof. Let 9 = g;:r(x)/g:' Let X be an irreducible character of ()) such that (fJ, X) = m'" O. Since g:(fJ) = ;:r, «(),X a ) :::: a (tl,X) for a E: g. Thus fJ =: m 2;g X + eo, where (eo,xa ) = o for all a E: g. By (11.1) m;:r(X) ~gxa is afforded by some g:-representation of 6t Thus by (1.9) and (2.9) mg:(X) l;gxa tl + 91 , where ()l :::: 0 or e1 is a character. Consequently tl = m ~gxa and m ::s m;:r(X). Since e and mg:(X)x are both afforded by 5'(X )-representations of ® we see that e mg: (X) X + ()2' where e2 0 or e2 is a character. Hence m;:r (X)::s m as required. The uniqueness of from (1.9) and (2.9).
e foliows directly
THE SCHUR INDEX
(11.5) m;; (x.)
1
63
Let X. be an irreducible character of (~. Then ;; is any field of characteristic O.
x. <1:) where
Proof. P(~) is afforded by an !f-representation m of ~. Let iY be an ;;-irreducible constituent of 9l which affords a character e with (e,x.) ;>! O. By (11.4) (e, X) = m5'(X) and p(®) = ae+ l} where (1], x.) 0 and a is a positive integer. Consequently
as required. Let n be an integer, n:::::: 1. Define the dihedral group 1'n by X'n ==
n
It follows directly from the definition that I!' nl ::: IOn'
2 +l and each of !'n and ();n has a normal cyclic subgroup of index 2. If n 1 then !'n is the noncyclic group of order 4. -n~
= B2 is the unique involution n in C n while :t'n has 2 + 1 involutions. Furthermore every proper homomorphic image of X'n or.o n of order at least 4 is a dihedral group. It is easily verified that A2
(11.6) Let (~be a 2-group such that (A)
64
CHARACTERS OF FINITE GROUPS
Proof. By (9.10) X(1) :::; 2. If (\j is nonabelian and X is faithful then X(l) = 2. Let B E: (\j - (Ai. Hence XI(A) = A + AB and X = A* by the Frobenius reciprocity theorem. If @ = 1:'n or On then B- 1 AB = A-I and so AB(Ai) = A(A- i) = A(Ai) Thus X (Ai) =
E
i +
E
-i as required.
In Section 7 it was observed that 1:'2 and O 2 have the same character tables. By making use of (11.6) it can be shown that for n::::: 2 1:'n and.Qn have the same character tables. However the next two results show that!'n and C n differ in properties concerning the Schur index. (11.7) Let g: be a field of characteristic O. Let ® be a 2-group sUCh that <.A) <1@,I(\j:(A)I=2 and thereexist-sB E: ~ - (A) such that B2 = 1. Then m g: (X) = 1 for eve-'!:'y irreducible character X of 6L Thus in particular mg: (X) = 1 for every irreducible character X of '!)n' Proof. It may be assumed that X is faithful. If X(1) = 1 the result is clear. If X(1) ~ 1 then by (11.6) X(1) = 2 and X(G) = 0 for G E: ® - (A). Hence
* X) ~ ( l(B),
1 l(B) * (1) X(l) = 1 = ~
* is afforded by a ~-representation of @ (11.4) Since l(B) implies that mg: (X) = 1. (11.8) Let X be a faithful irreducible character_ ()~ C n and let g: be a subfield of the field of real numbers. Then mg: (X) =
2. Proof. By (11.5) and (11.6) mg:(X) = 1 or 2. Hence it suffices to show that mg:(X) = 2 for go equal to the field of real nl!-mbers. By (11.6) X E: g:. J?very el.ement in en is of the form Al or BAi. Furthermore (Al)2 = NI and
65
THE SCHUR INDEX
{BAi )2
:::
B2 B-1 Ai BAi ::: B2 A-iAi
B2
Therefore by (11.6)
L ('»
2
n X (Ni) + 2n X (B2)
~
X (G2)
1
i 2
n
::: L::
(€2i +
=:
1
Thus by (3.5) mff{x)
;r.
i
E- 2 i)
2n +l
1 and so mff(x)
== 2
as required.
Let X be a faithful irreducible character of 0 n' By using the theory of algebras it is quite easy to describe all the splitting fields of X over Q. We will only need the following result. (11.9)
Q(
Let X be an irreducible character of Cu. Then , X) and Q (v- 3, X) are both splitting fields of X.
Proof. If X is not faithful then X is a character of a dihedral group and the result follows from (11. 7). Suppose that X is faithful. By (11.6) Q(X) ::: Q (10 + 10-1 ) is the maximal real subfield of Q2 n. Since Q2 n we get that Q{X, r-t) Q2 n. Since X = A * where A is a linear character of a suhgroup and A E: Q'2 n it follows immediately that Q(X, r-t ) is a splitting field for X. Let ff = Q{v-3, X) .We will prove that ff is a splitting field of X by exhibiting a group of nonsingular matrices with coefficients in ff which is isomorphic to .c~nd whose trace function coincides with X. Let w = (- 1 + v- 3)/2 E: ff . Then w is a primitive cube root of unity and u.\2 + w + 1 ::: O. Let J
(1.1;2 u.,2 ). w -u.,
It is easily verified that the character-
istic roots of J are Rand -R. Thus J is diagonable in some extension field of ff. Let € = a + b Fi where a, b are reaL Thus a, b E: ff. Define
66
CHARACTERS OF FINITE GROUPS
Thus Ao has the characteristic values E and £. == c 1 , Since J is diagonable in some extension field so is .AJ for every i. Hence
~n
==
I and
~n-l
"" (- ~
B~1 AoBo = (~ ~) -
bJ
_~) ==
B~.
Furthermore
1\;1
Hence the mapping sending BiAj to B~ ~ is a homomorphism of On onto \Ao, BO>. Since ienl = 2n+l = HAo, Bo> I, this mapping is an isomorphism. For G E:
§12. A COMBINATORIAL RESULT AND SOME CONSEQUENCES The following result due to Brauer [1, Lemma useful.
IJ
is very
(12.1) Let A = (aij) be a nonsingular matrix of degree k .with complex entries. If (J is a permutation of the k2 ordered pairs (i,j) i, j = 1, ... , k define A(J = (a(J(i j»' Suppose that IDl is a group of permutations on the k2 orde~ed pairs (i, j) ~1:1ch that for (J E: IDl A(J can be derived from A by permuting ~he rows of A and A(J can also be derived from A by permuting the columns of A. Then the number of orbits under the action of IDl consisting of rows is equal to the number of orbits under the action ID1 consisting of columns. If furthermore 9R is cyclic then the number of rows left fixed by 9R is equal to the number pf columns left fixed by ID1Proof. If (J E: 9R then by assumption there exist permutation matrices ~(((J), Sl\«(J) such that ~«(J)A A(J == A5{\((J).
A COMBINATORIAL RESULT
67
1(0') and Sl,\(a) are uniquely determined since A is nonsingular. Then 1(0') I( r) = I( TO') and 58(0') 58( T) = 58(aT). Thus I' and 58 are both permutation representations of IDl, where ' denotes transpose. Since A is nonsingular A-l "(0') A ;;; Sl,\(a) and so 1'(0') and 58(0') have the same trace. Thus I' and 58 afford the same character (). The first statement now follows from (9.6). If IDl ;;; (0') is cyclic then the number of rows or columns left fixed by IDl is ()(a) by (2.3). In applications of (12.1) A will generally be the character table of some group@. We will first introduce some terminology. Let 5 be a field of characteristic O. Let € be a primitive 1~lth root of unity and let 9 ;;; 95(€)/5' Every element in 9 is of the form as for some integer s with (s, I(~') = 1 and 0< s < I(~)I, where €as €s. Two elements G, H E.: (\~ are 5conjugate if G is conjugate to HS for some s with as E.: 9. It is clear that 5-conjugation is an equivalence relation. The equivalence classes are the 5-conjugate classes of~. If € E.: 5 then G and Hare 5-conjugate if and only if they are conjugate. At the other extreme G is ~-conjugate to if if and only if (G) is conjugate to (H). (12.2)
characteristic O. 5-conjugate classes of
on
Proof. Let € be a primitive I~I th root of unity. Thus 5(€). If as E.: 9 (€)/5 then ()as (G) () (G s ) for G E.: (\1. 5 The result follows. The next result was probably known to Schur.
(J E.:
(12.3) Let 5 be a field of characteristic O. The maximal number of pairwise nonsimilar 5-representations of ~ is equal to the number of 5-conjugate classes of~. If furfuermore 9 5 (€)/5 is cyclic, where € is a primitive I@I th root .of unity the number of irreducible characters of @ whose values lie in ~. is equal to the number of conjugate classes of ® which are 5-conjugate classes.
68
CHARACTERS OF FINITE GROUPS
Proof. Let X
9 = 9 .5'{E)/if·
If as
(Xi (G j )) be a character table of (19, Let 9 then the mapping sending Xi to
E::
as Xi permutes the irreducible character of N. The number of orbits under the action of 9 equals the number of characa ters of the form ~a Xi ' where a ranges over 6)5(Xi)/ ff • By (11.4) thip is the maximal number of pairwise nonsimilar 5-representations of N. The number of irreducible characters fixed by all the elements of 9 is the number of irreducible characters whose values all lie in 5. The mapping sending G into GS for as E:: 9 permutes the conjugate classes of (19. The number of orbits being the number of 5-conjugate classes of (~. A conjugate class is fixed under the action of 9 if and only if it is an 5-conjugate class. Since X.as(G.) = s 1 J Xi (Gj) and X is nonsingular the result follows from (12.1). A conjugate class of (19 is a real conjugate class if it is an 5-conjugate class where 5 is the field of real numbers. An element G is a real element if it is contained in a real conjugate class. The next result is of fundamental importance for many applications. (12.4) (Burnside) An element G is real if and only if G is conjugate to G-1 • The number of real valued irreducible characters of ~ is equal to the number of real conjugate classes of 6), If INI is odd then {1} is the only real conjugate class and 1N is the only real valued irreducible character of (~). Proof. The first two statements follow immediately from (12.3). Suppose that I(~I is odd and G is real. Then there exists H E:: (~ such that H-1 GH = G-l. Taking inverses we get that H-1G-1H = G. Thus H- 2GH 2 = G and so H2 E:: C(G). Since I(~I is odd H is a power of H2 and so H E:: C(G). Thus G = G-1 or G2 = 1. Hence G 1. Therefore {l} is the only real conjugate class in N. Consequently by the first part of the theorem 6) has only one real valued irreducible character which must necessarily be 1(~.
RAT ION A L V A L U ED C H A RAe TE RS
69
US. RATIONAL VALUED CHARACTERS The results in this section have applications in number theory. (13.1)
Let 1{ be a cyclic group. For A a~ (A)
=
I~I
if ~{
o
if ~ ~ (A)
E: 't(
define
= (A)
Then Q''t( is a rational integral combination oJ permutation characters of ~. Proof. The proof is by induction on IIJ. If Iftl ::: 1 then Q'I 11' Suppose that III> 1. Let S8 be a subgroup of I with >l\ ~ I. If ~ = (B) then Q'~ (B) = III and if >8 ~ (B) then o~ (B) O. Thus if {3 = .t Q'~ where S8 ranges over all subgroups of 9{ with 5B ~ 'tl, then {3 (A) l'tl I if (A) ;It 't( and (3(A) 0 if (A) = ~. Hence Q'! III 11 - {3 and the result follows by induction. (13.2) (Artin) Every rational valued character of 6} is a rational linear combination of characters of the form where !( ranges over the cyclic subgroups of ~.
1i'
Proof. Let 'l) be the vector space over Q consisting of all rational linear combinations of rational valued characters of ~ and let n be the dimension of 'l). By (12.3) there exist n pairwise nonconjugate cyclic subgroups 11"'" In of ~. Let Q'i::: a ~l i be defined as in (13.1). Then Q't vanishes outside the Q-conjugate class of (~; which contains a generator of Ii. Thus Q'~ is a basis of 'l) as required.
at .... ,
(13.3) (R. Brauer) There exist cyclic subgroups ~j of (~ and nonprincipal linear characters of ~lj such that p(\~ = 1 N + Z; aj t'j, where each aj is a rational number.
C H A RAe T ERS 0 F FIN I T E G R0 UPS
70
Proof. For a cyclic group ~ let a be defined as in 1l (13.1). Let 131l = qJ(l1l:l)p ~ - 01l:' where (p is the Euler fupcti~m. If ~ is an irreducible character of 1l: (A) then ~(Al) E l~ where E is an I~tlth root of unity. Therefore :0 ~(Ai) a~ i (AI) = I~I Tr~ 11ll/tJ (E). Thus I(~, O''ft)'' I ~ (p (11l1). This implies that
(t, 13,,) ~l
(~,
= qJ (l1l/ )
0' ~)11
~ 0
Since (p®, 1®) = 1 it now suffices to show that P(<») - 1(®) :;
(1/ / (~ I ):0 13;, where" ranges over all cyclic subgroups 11 of ®. Let X be an irreducible character of N. Then (X, p(Y, - 1(f,,) X(1) - (X, 1(})}. On the other hand the Frobenius reciprocity theorem implies that
=
1
1(\\/
qJ (I~I)
X(I)
Since X was an arbitrary irreducible character PN - IN :; (II 1<» / ) :013~ as required.
~-ELEMENTARY
GROUPS
71
§14. 5=-ELEMENTARY GROUPS Let 5= be a field of characteristic 0. A group ~ is 5=elementary with respect to the prime p if (0 ~) = 11~, where 11 <1 ~, 11 is a cyclic p' -group and ~ is a p-group. (ii) If two elements of ~ are conjugate in .\) then they are 5=-conjugate in 11. Let E be a primitive I10 th root of unity. The second condition is eQuivalent to . (ii)' Let ~ = (A). If Ai is conjugated to AJ in .\) then there exists a E: 9 ~(E)/~ such that (Ei)a = Ej The group ~) is said to be 5=-elementary if it is 5=-elementary with respect to some prime p. S) is elementary if it is 5=elementary, where 5= is a field containing a primitive 1':'( Ith root of unity. It is easily seen that an 5= - elementary group .\) is elementary if and only if S) = ':'( x 'P. A group .s; is Qelementary if and only if S) = ~~, where ~ <J S), ~ is cyclic and ~ is a p-group. It will be shown later that questions concerning Schur indices of characters of an arbitrary group N can be reduced to the corresponding questions for 5=-elementary subgroups of (~. These questions are however far from being answered for tJ-elementary groups. The results in this section will be needed later. Fong [1] first proved (14.1) by making use of the theory of modular characters. The remaining results in this section may be found in Solomon [1], though (14.3) is a slight generalization of the corresponding result there and the proof of (14.2) is here simplified. Some of the results in this section have also been obtained by Berman and Witt by different methods (see Curtis and Reiner for references). Roquette [2J first published a proof of (14.5). His proof was based on the theory of algebras. All the material in this section is closely related to the work of R. Brauer [2]. Throughout this section 5= is a field of characteristic 0, .'J) is an tJ-elementary group and ~ = (A), ~ and p have the same meaning as in the definition of 5=- elementary groups.
72
C H A RAe T ERS 0 F FIN I T E G R0 UPS
(14.1) Let q be a prime, q q )' ho, and let X = (!ho. If X s.) then mx<x) 1.
:;t:
p. Let I.\))
= qa11o , where character
Proof. It may be assumed that X is faithful. By (9.13)X(l) is a power of p. By (10.2) X = A*, where A is a linear character of some subgroup \' of ,f). Thus I~): \' I is a power of p and so 'n ~ \'. Let C be the Sq-group of W. Since A is linear \' c li = CC. Let f; be the character of li induced by A. Then li <j ~) and X ::: t*. Furthermore li = 0 x SB for some group SB and so t ::: 111], where 11 is a linear character of Q and 11 is an irreducible character of ~. Let XI be an irreducible constituent of the character of ~'l3 induced by 11. Since 1~'i311 110 (10.2) implies that Xl' and thus also xt, is afforded by a X-representation of ~'l3, ~ respectively . .l.)/li is cyclic since ,\)/<5 is a q' -group of automorphisms of the cyclic q-group Q Thus by (9.12) XII ~ = ~i 11i' where 11 ::: 11 1 , 11 2' • •• are distinct irreducible characters of ~. Hence xt lli P,o ~i 11i' By the Frobenius reciprocity theorem
Therefore by (11.4) mx(x)
1 as was to be shown.
(14.2) Let E be a primitive ifll th root of unity. For any !.I!teger t let At (A1 ) = EU and let 'Vt ::: C 'l3 (At). Define
= Tr5=(E)/5= (At (Ai)). Let
::: 0
if
P
E:
'l3
'V t
Then 9t is a character afforded by some 5=-irreducible 5=representation of .\).
ff'- E LEMEN TA RY G R0 UPS Proof. There exists a linear character J.L of that tlt is in the kernel of Then Trff'(x)/(f(X)
8
r
J.L
73
Wt\
such
and J.LI! = At. Let X ::: J.L*.
Furthermore X is an irreducible
character of ~. Since X I~
Ptl/i3t
we see that (1~, X) Sj :::
(l , XItl)tl :; 1. As 1; is afforded by an (f-representation of tl S) (11.4) implies that m(f(X) = 1. Thus by (11.4) e is afforded t
by an (f-irreducible (f-representation of S;. The following well known fact is required for the proof of the next result. See M. Hall p. 189. If tl is a p-group which contains a unique I:;ubgroup of order p then either tl is cyclic or p = 2 and tl is a quaternion group.
Let (f ::: ~. Let 1T(~[) ::: {Pi}' Let 3C be a field of characteristic 0 which contains a primitive n Pi th root of unity. Suppose that one of the following hypotheses is satisfied (14.3)
(ii)
p::: 2 andH
(ui)
p = 2 and tl does not contain a cyclic subgroup of index 2.
E:
3C.
Then if X is a nonlinear irreducible faithful character of there exists S)"O
~
an irreducible character of ~ and a ranges over S3C(it)/3C(x)
74
CHARACTERS OF FINITE GROUPS
and 19 :JC(A)/3C (x) I ::: l.p: ~ I . Furthermore AI<S ::: ~f = 1 /li' where Ill' ... , /lp are distinct irreducible characters 0 f li and 3C(/l1) ::: 3C(A) for i ::: 1, ... , p. Let ~ be a minimal subgroup of ~) such that XI ~ ~a
cpa, where cp is an irreducible character of (i, a ranges
over 9(cp) ::: 9 3C (cp)/:JC(x) and 19(cp) I ::: l.p:~ I. If {G i } is a complete set. of coset representatives of ~ in ~) then XI ~ == G· a' Gi • Since (cp G)a ::: ~i cp 1. Let ai be defined by cp 1 ::: cp
(cpa)G we see that
Thus the mapping sending G. into a. is an isomorphism of 1 1 s)/ ~ onto 9(cp). By the Frobenius reciprocity theorem X c cp*. Since X{1) ::: I~):~I cp(1) ::: cp*{1) this implies that X ;- cp*. Thus by (9.13) I.\):~I is a power of p. Assume first that cp is nonlinear. By (10.2) there exists a subgroup i\ of ~ with I~: i\1 ::: p such that cp J is induced by the character w of ~. Let !' be the intersection of all conjugates of i\ in ,\:-'. Since ~ c i\ and 11
E:
:;t:
/lp are
3C(cp) for
9 3C( 11 i)/3C( cp) such that /li
a
i~plies
Ej aj E lli , where (E li) = l!/li. Thus Ili = ll i This a that X lli ::: ~ 11 ,where /l /ll and a ranges over
9 X (/L)/X(X) contrary to the minimality of li. Thus X(/li)
~-ELEMENTARY
GROUPS
75
x(cp) for i 1, ... , p. Now let Ii ~ and cp :::: A. (*) is proved in this case. Assume next that cp is linear. Since cp and cpu have the same kernel this implies that cp is faithful and so (.! is cyclic. Let (.! ;; (AP), where P c 'P. Let cp(A) = Ct.' and cp(p) = E. Thus w C X(cp). Since [X(w) :x] Ii'll and g(cp} is a p-group this implies that w C X(X). m If 'P contains a cyclic subgroup of index 2 and order 2 then hypothesis (ii) holds. Thus F1 C :JC. If X(p) = cp*(p) ::: E + c l or X(p} = E + E-1+2 m - 1 then X(X) = X(E) ::: X(cp) contrary to hypothesis. Thus it may be assumed that X(p) =
m 1 E + E1 + 2 - . Hence in particular 'l3 is not a quaternion group and so there exists an involution T C 'P - (Pi. Thus m T-l P'I' = p1 + 2 -1. If p = 2 and 1'l3: (P) 1;r 2 then there also exists an element T C 'l3 - (P) such that T-l PT ::: p 1 +2 m - l and once again it may be assumed that T is an involution. Suppose now that 'l3/(P) is cyclic. Let I (P)I :::: pm and 1'l3: (p) I = pn. Thus 'l3 :::: (p, B) where T can be chosen of n-l 1+ m-n order p so that BP T. Furthermore B-1 PB P P ~D ~D(1 +pm-l) ~D with m > n 2: 1. Then T-1.P" T = .P" = .P". Thus
(T, pP) is an abelian group. Since I(T, P) : (T, pPi I
= p,
(T, PP)
76
CHARACTERS OF FINITE GROUPS
p ::; 2 and l(p)1 = 2m
8. There exist elements B, S E: ~ m-n such that S-1 PS = p-l, B-1 PB ::; pI + 2 where I(B, p)1 = m n 2 + , m - 1 > n ?': 1 and I ~I = 2m +n+ 1. Since (B, P) is not a quaternion group there exists an element T of order 2 ?':
n- 1
in (B, P) (P). Thus it may be assumed that B 2 = T. Let m cp(P) = E. Thus E is a primitive 2 th root of unity. Hence .\l(cp) = .ll(W, d. Since ~/(P) is abelian \ T, p)
i3
q;
of "(T, P) induced by tp. Thus q;(P) = Tr'Q(E)/ Q(E2) (E) = O. This implies that ';p vanishes outside "(T, p2). Therefore 'q) I~(T, P2) is reducible. Hence by (9.11) -;p I"(T, p2) = 1/1 1 + 1/12' where Wu l/J2 are distinct irreducible characters of fl(T, p2). Since the order of any element in (T, p2) is at most 2m - 1 we see that X(I/Ii) c X(w, E2). Let \£ ~C~,(T) and let A be the character of \£ induced by 1/11' Then (X\lbt):X(A)] = 2n and so (X(w, E): X(A) J ?': 2n+1. Since X c X(A) this implies that X(X) ::; X(A). (*) now follows by setting ~ = -U. This completes the proof of (*) in aU cases. By (*) XI\£ ~~=::1 ~al1t, where a ranges over 9X(lll)!X(X) and no two of Ili' Ilj are algebraically conjugate. Let ~;o
be the subgroup of ,(') consisting of all elements G such that 1l1G ::; 1l 1CJ for some CJ. Let t be the character of ~;o induced by 111' Then I'\;:~;ol = p, t* = X and X(X) X(t) as required. Let ff = ~. Let 1T(:'f) = {Pi}' Let X be a field of characteristic 0 which contains a primitive IIi Pi th root of unity. If p = 2 assume that also A E: X. If X is an irre(14.4)
\T-ELEMENTARY GROUPS
77
ducible character of .\) then there exists a subgroup ~ and a linear character A of I.! such that A* == X and JC(X) ::: Thus in particular mJC(x) = 1. -
Proof. The proof is by induction on ,.\) I. If X(1) = 1 the result is clear. If X(1):> 1 then by (14.3) there exists a subgroup '\)0 of .\) with .(10 ;z! .\1 and an irreducible character t of ~)o such that X t* and JC(x) = JC(t). Since '\)0 is ~ elementary the result follows by induction.
(14.5) Suppose that .\1 is a nilpotent group and X is an irreducible character of ~. Then there exists a subgroup I.! of ~) and a linear character A of I.! such that A* = X andeither ~(A) ::: ~(X) or 1·\)1 is even and ~(A) ;;:; ~(x.,';::"'1). Furthermore either m~ (X) = 1 or I~)I is even and m~(Ff) (x) m~(J:3) (X) ::: 1.
Proof. Let f, = '1.\ x· .. x '13n , where 'l3i is a pi-group and Pl' ••. ,Pn is a set of primes. Thus X rr~;;:;1 Xi and ~(X) ::: ~(XH ••. ,X n ), where Xi is an irreducible character
of
'1.\.
If Xi
Ai then X (rr~l Ai)* and t:l (rrbl Ai ) = t:l (AI' ...
An)" Thus it may be assumed that N == '13 is a p-group. The first statement follows from (14.4) with JC ::: ~(x> or ~(F1, X) since X(p) = € X(l) with € a primitive p th root of 1 for P an element of order p in Z(N). This implies that m~(x) ;;:; 1 if p;z! 2 and m~(vCT) (X) ;;:; 1 if p = 2. Assume now that p ::: 2. We will prove by induction on 1'131 that m~(J:3) (X) ::: 1. If x(1) ;;:; 1 this is clear. If '13 has a cyclic subgroup of index 2 this follows from (11.7) and (11.9). Otherwise by (14.3) there exists a subgroup '13 0 of '13 and a character t of '130 such that t* ;;:; X and ~(v'-3,x) ~(,t=3, t). Thus m~(..c3) (X) :5 m~(yC3) (t) = 1 by induction.
78
C H A RAe TE RS 0 F FIN I T E G R0 UPS
§16. THE CHARACTER RING Throughout this section the following notation will be used. (®) is a group and 5" is a field of characteristic O. ~ is an integral domain whose quotient field has characteristic O. X is an extension field of the quotient field of ~ which contains subfields isomorphic to 5" and to ~INI' Thus by choosing suitable isomorphisms it may be assumed that 5".:=. X and ~ I®I c X. Thus any character is a X-valued function on Nand a'i>:" linear combination of characters is well defined. e{~;rP"~' 5") is the ring of all ~-linear combinations of characters afforded by 5"-representations of (~. eft~« ..~, 5")' is the ring of all ~-linear combinations of characters e E: 5". 'U..~«\~, 5"), 'U..~«\~, 5")' is the ring of all X-valued class functions 8 on N such that
e1,\,
E:
e{~;rP)' 5"), eft;r}~)' 5")' respec-
tively for every g:-elementary subgroup .\' of N. Observe that, as the notation indicates, 'U..~(N, g:) and 'U..~«")' g:)' do not depend on X. 'O~(N,
g:), 'O1)«(\)' g:)' is the ring of all :I)-linear combina-
tions of functions
e*,
where 8
E:
e "';n(:!j), g:), en:I)(:'), g:)' re-
spectively for some g:-elementary subgroup ,'ij of N. If p is a rational prime then '0 :I)«\\' g:, p) is the ring of all :I)-linear combinations of functions 8* where () E: e
f6:I)(~),
g:)
and ,\, is a subgroup of N which is g:-elementary with respect to p. In all these cases if :I) is the domain of rational integers the subscript :D will be omitted. In case g: is a splitting field of ~ we write
The ring e f~«(\) is the character ring of N. The elements of eft«\) are called generalized characters of N. Thus a
THE CHARACTER RING
79
generalized character of (\~ is a rational integral linear combination of irreducible characters of N. Or equivalently a generalized character of (~ is a difference of two characters of N. The following results might well be called the fundamental theorems of character theory.
These results have applications in number theory and group theory. The most important special case occurs when :D is the ring of rational integers and ff is a splitting field of N. This special case was first discovered by Brauer who was also the first to realize the importance of results of this type. Roquette [1J simplified Brauer's proof. Berman, Witt and Brauer all independently generalized the original result. In case ff Q Swan also obtained (15.1) independently. See Curtis and Reiner p. 301 for references. The original result was given a new and elegant proof by Brauer and Tate (1j who made use of Roquette's simplifications. Using their methods Solomon [1] gave a proof of (15.1) and (15.2) and also of a related result of Brauer [2]. He also obtained a new result concerning splitting fields. This last approach will be followed here. Before proving (15.1) and (15.2) we will deduce a result that will be of use later. (15.3) Let e be a complex valued function on (~'. Then () is a generalized character of N if and only if the fOlloWing conditions are satisfied. (i)
(J
is a class function on N.
(ii) For every elementary subgroup .\) generalized character of ~).
Q! (\\, elf .J
is a
C H A RAe TE RS 0 F FIN I TE G R0 UPS
80
Proof. This is a restatement of the fact that 'U.«\\, (f) =
ek:N, (f), where (f is the field of complex numbers. (15.4) Let () be a complex valued function on N. Then () is an irreducible character of N if and only if () satisfies conditions (i), (ii) of (15.3) as well as
(iv)
(J
(1)
>
O.
Proof. If () is an irreducible character of N then (i) - (iv) are clearly satisfied. Assume that (i) (iv) are satisfied. By (15.3) (J :E a i X., where each X. is an irreducible charl l acter of (\\ and each a i is a rational integer. By (iii) 1 = ~ ai 2 •
Thus there is exactly one nonzero a = ±1. Hence i for some i. By (iv) 0 = X. as required.
(J
= ± Xi
1
We will prove several lemmas and then use them to give a proof of (15.1) and {15.2L Throughout the remainder of this section n denotes the exponent of Nand
(15.5)
To prove (15.1) and (15.2)Jt suffices to show that
1(~ E: 'U(N, (f).
Proof. By definition
By (9.3) 'U 5)«"" (f), 'U '.O(N, (f)' is an ideal of 'U. '.0«(\\' (f), 'U.'.O«\\' (f)' respectively. The result follows.
(15.6) Let ~"'l = tl'l3 be an 5-el~.rl?:entary group with respec!j:g p where ~ = ,A) is a p' -group and ~ is a p-group.
THE CHARACTER RING
81
Let a !":IfI and let S be the ring of integ,ers in ~ a' Then there exists 11 E: C!Jl>s("\ ff) such that 1J (AI) = a or 0 according to whether Ai is ff-conjugate to A or not. Proof, Let ff a be the extension field of ff generated by a primitive a th root of unity. Define the f~nction t on ":II by I::(A 1) a or 0 according to whether Al is ff-.conjugate to A or not. Thus I:: = Z; ~-Ol c. A., where A. = AJ A is a J=
J J
J
faithful irreducible character of ~ and the Cj are complex numbers. By the orthogonality relations a-I
= -1 L:
c.
a i=O
J
Thus c
j
S and I::
E:
"
t (A l) =
A, (A I ) J
==
Tr ff aI ff
(A . (A» J
Z;t Ct Tr ffalff (At)' where t ranges
over a suitable index set. Let
et
be the function defined in
ef£l':"'"
(14.2). Let 1J = Z; t c t et' Then 1] E: ff) and 1] (Ai) t(A i ) for i = 1, , .. , a. The proof is complete. Let G E: (ill the ff-normalizer Nff(G) is the set of all elements H E: (~ such that H-1 GH :;:: Gi, where G is ff-conjugate to Gi in (G>. The definition implies that N~(G) == N«G) ) and Nff(G) :;:: C(G) if ff contains a primitive I(~I th root of unity. If A is a pi - element of N and 'l.~ is a Spgroup of Nff(A) then (A) '.l3 is an ff-elementary group with respect to p. Let A E: N and let p be a prime. There exists (15.7) l/J E: 'V m(N, ff, p) ~uch that w(G) :;:: IN ff(A) I or 0 according to whether G is ff.::-conjugate to A or not. Proof. The cyclic group (A) is ff -e lementary with respect to p. Let a :;:: a'l be the function defined in (13.1). Thus a*
E:
"m
«(f,}, (F,
that If; :;:: a
*
p). The definition of a implies directly
has the required properties,
82
C H A RAe TE R5 0 F FIN I T E G R0 UPS
(15.8) Let p be a rational prime and let 8" be a prime divisor of pin (fl.. Let A be a p' -element in N of order a. Then there eXists If-E:'O(R(N, g:,p) such thate (G) = 0 if G is a pi -element of N WhICh is not g:-conjugate to A and e (A) '" 1 (mod ll). Proof.
Let tl
= (A).
Let ~ be a Sp-group of Ng:(A) and
let .,) = ~~ •. Thus .,) is g:- elementary. Let S be the ring of integers in ~a and let 'I be the function defined in (15.6). Since S S (fl., 17 * E: '0 (fl.("l, g:, p). The only p' -elements of .') are in 1( thus the definition of 17 * yields directly that 17 *(G) = o if G is a p' -element which is not g:-conjugate to A and 17*(A) Since
=
I.~)I i:(\\
17(G- 1 AG)
=
r~1
INg:(A)I
=
I Ng:(A):'1.~1
'l3 is a Sp -group of Ng:(A), pl 17 * (A). Hence there
exists a rational integer b such that b 17 *(A) '" 1 (mod p). Then e = bl} * has the required properties. (15.9) Let p be a rational prime and let ~ be a prime divisor of p in
Let INI = pCb, where (p,b) = 1. Then b1(\\ g:, p).-
E:
Proof. Let I; be the function constructed in (15.9). By
s
induction I;(G)P = 1 (mod 8"s) for all s and all G E: <"t The construction of I; was independent of the choice of the
THE CHARACTER RING
83
s prime divisor ~ of p. Thus for suitable s t (G)p _ 1(mod pC) s for all G E: N. Clearly b!; P E: '0
N
sps)
E:
'0
Let AI' A 2 , ••• be a complete system of representatives of the IF-conjugate classes of N. Let lb i be the function defined in (15.7) corresponding to Ai' Then
Since ~p
s :=
1 (mod pc) it follows that
Thus by (15.7) b(1
N
tP
s
)
E:
'0
(15.11) 'O«\~,
IF, p).
Proof. Let € be a primitive nth root of unity. Then {€ iT 0 ~ i < ({'(n)} is a basis of
w.1
"
1
1
1
is a character of some subgroup which is 5=-element-
ary with respect to p. Let a :.:; E. a .. € j, where a.. is a i J IJ IJ rational integer for all i, j. Thus b1 {\' E .. a .. € j 1iJ'!'. Let ,
Xl
= 1N,
1,J IJ
1
X2' ..• be all the irreducible characters of N. Then
84
CHARACTERS OF FINITE GROUPS
This implies that 2;.
a.
1 10
1/!~ 1
2; i
a
(wi, Xs)
io proving the result.
:= bo sl' Thus bl (~ =
The proof of (15.1) and (15.2) is now very simple. Let {PJ be the set of distinct primes dividing INI. For each i 1 c· let I~I := Pi Ib , where PiYbi' By (15.11) b 1(,; E: "O(N, 0:). i i There exist rational integers a. such that 2; a.b. = 1. Thus 111
1m :=
:E. a.b.l(\! 1 1
E: "O((~,
0:)
1
By (15.5) the proof is complete. The following lemma is of use in computing Schur indices. Let X be an irreducible character of N. Let .c be a field sUCh that o:(X) c .c c 0:(£), where £ is a pllmitive nth root of unity and 1 0:( € ):.c J is a power of p. Then there exists a subgroup .\) of N which is .c-elementary with respect to p and an irreducible character ~ of 3) such that ~ E: .c and PY (X, ~*). (15.12)
Proof. Let INI := pCb, where (p, b) 1. By (15.11) there exist subgroups .\). of (\\ which are .c-elementary with reJ spect to p such that bllU:= 2; a. cpr-, where cpo is a character ~ J J J of ~. and cp. E: .c. By (9.3) this implies that XI ~_
.)j
J
bX
:=
6
a. X cp~ J
J
:=
E
J
a. (XI ,-"
"'j
For any irreducible character ~ of
J
'\)j
let T(~) = Tr.c (O/.c (~).
SCHUR INDICES AND SPLITTING FIELDS
85
Thus there exists subgroups ~). of (~\ which are .e-elementary with respect to p and irredud6le characters ~.1 of ~.1 such that bX 2: d. T(~.)*. Hence 1
1
b
Since (b, p)
:=
l.e(~):.el
1 this implies that for some ~ (X,
~
*)
~
= ~i and
~
~i
0 (mod p)
Since ~ E: ff (E) and Iff (E) : .e I is a power of p this implies that .e(~) = .e. Hence ~ E: .e and (X, ~*) ~ 0 (mod p) as required.
§la.
SCHUR INDICES AND SPLITTING FIELDS
By combining the results of the last two sections we can now derive some information concerning Schur indices and splitting fields. (Brauer (2]) Let ff be a field of characteristic O. Let X be an irreducible character of N. For each rational prime p there exists a finite extension field (£ of ff, a subgroup .\:, of N which is .e-elementarywith respect to p and aIliITeductble character ~ of ~ with ~ E: .e such that i f pa Imff(x) then palmff(~)' (16.1)
Proof. Let n be the exponent of N. Let ff c .e c ff(€) such that (ff(€):.e] is a power of p and p! I.e: ffl. Let ~ be the character constructed in (15.12). Then mff(~) ~ * is afforded by an .e-representation of N. Hence by (11.4) m.e (X) I (X, mff(~H*). Since mff(x) I m.e (X) I.e: ffl and p Y (X, ~*) x I.e: ffl this yields that palmff(~) as was asserted. The following special case of (16.1) is useful.
86
CHARACTERS OF FINITE GROUPS
(16.2) Let 5 be a field of characteristic O. Assume for any 5_:elementary subgroup .(l QL N and allY irreducible character cp of "), m (cp) = 1. Then m (x) = 1 _fo_r~_ 5 5 irreducible character X of N. The next result was an open question for about 40 years. For this result the argument can be considerably simplified. See Brauer and Tate [1J and Curtis and Reiner p.294. (16.3) (Brauer) Let n be the exponent of N. Then ~n is a splitting field of N. Proof. By (16.2) it suffices to prove the result in case is n-elementary. Since ~n-elementary groups are nilpotent every character of (~ is monomial. The result follows. (16.3) has been strengthened as follows.
(\l
{PJ
(Solomon [1]) Let 1T(N) == Let 5 be a field of (16.4) characteristic 0 which contains a primitive IIPith root of is even assume also that R E: 5. Then unity. m (x) - 1 for any irreducible character X of N.TIiiis in 5 particular m,q (X)I2 IT. (p. 1) and if INI ip odd then "
m,q(x)l II.(p. "
1
1
1
__
1).
1
Proof. By (16.2) it may be assumed that N is 5-elementary. The result follows from (14.4) and (11.3). (16.5) (Fang (1]) Let p be a prime and let INI = pCb, where (p,b) == 1. Let 5 be a field of characteristic 0 which contains a primitiv-e-bth root of unity. If p == 2 assume that either FI E: 5 or F3 E: 5. Then m (x) = 5 irreducible character X of N. Proof. By (16.2) it may be assumed that (~; is 5-elementary with respect to some prime q. If q ~ p the result follows from (14.1). Suppose that p q. Thus N is nilpotent since ~b c 5 and the result follows from (14.5). The special case below which motivated (16.5) was first
EQUATIONS IN GROUPS
87
proved by using the theory of algebras and parts of the class field theory. It was of great importance in applications of the theory of modular characters. (16.6) (Brauer) Let p be a prime. There exists an algebraic number field s: in which p does not ramify such that ms:(x) = 1 for every irreducible character X of N.
§17. EQUATIONS IN GROUPS The next result is due to Frobenius. The proof given here is due to Brauer L3]. (17.1) Let II be an integral domain which contains a primitive 1(\;1 th root of unity and whose quotient field has characteristic O. Let n be a positive integer. For any subset .\1 ~ N which is a union conjugate classes define () ~I (G) = e~ \1 (G:N) for G E: @ ~ . .
INI ( INI , n)
if
Proof. If .\) is a subgroup of N then the quotient of INI/ONI, n) by I~)I/(I.\)I, n) is a rational integer c and JI is a union of conjugate classes of ~:~. Thus for H E: S:.,
n
.(~
If the result has been proved for elementary groups then since () .\1 is clearly a class function the result follows in general
from (15.1). Thus it may be assumed that group.
(\~
is an elementary
88
CHARACTERS OF FINITE GROUPS
Let .\1 1 of
= {I},
~. If .\1
then
e Jl
= ~
.11 2
"",
.\lk
be all the conjugate classes
U .\1 i' where i ranges over some index set
e.~I i'
Thus it may be assumed that.n
.\1 i
for some i since the result is trivial in case .\1 is empty. Furthermore k
E
i=1
=
INI (INI, n)
Thus it suffices to prove the result in case .11 i
1. If N
~
= .~l.1
with
("\ X N2 with (IN11, IN21 ) = 1 then every element in .11 is 9.f the form K1 ~, where .\1-1 ranges over a conjugate class .\I~i) of Ni for i 1,2. Thus for G i E: Ni , i = 1,2
Since (\\ is elementary it thus may be assumed that N is a p-group for some prime p. Let e = e.(t' We need to show that if X. is any irreducible character of (\\ then (9,x.) E: S. By (10.2) x. = A* where A is a linear character of some subgroup .\) of N. By the Frobenius reciprocity theorem (9,., A). = (9, x.). Let s ~ ,. . e (H) x .\.') .\.') '.' A (H-1). It remains to show that s/I.£) I E: S. Let INI pa. It may be assumed that n
= pb
is a power of p. If b ;::: a 1 for H E: ~) and so 9 (H) = 0 for H E: .\';. then Hn Suppose that b < a. Then s = pa-b ~ A(H- 1 ) where H ranges over these elements in ~) such that Hn E: .\!. Distribute these elements into pairwise disjoint sets as follows. H1 and H2 are in the same set if and only if each is a power of the other and Hr = HP E: .n. It suffices to show that for any such
det ~ ~~ Am-I) ;;; O(mod pb). If A
E: :I(
and A has order
EQUATIONS
IN GROUPS
89
pd then d > b since .n #- {I}, Furthermore Aj E: ;I( if and only if jpb == pb (mod pd), that is j 1 (mod pd-b), Hence d-b if A(AP ) = E pb_ 1 A(A-l)
L;
Ei
i=o
Since E is a pb -th root of unity this implies that E;I(A(H- 1 ) o or A(A-l)pb, Hence in any case E~A(H-l) =: 0 (mod pb) as required, . Let e.~l be defined as in (17.1). Then e{I} is a (17.2) generalized character of N, Proof. Let {xi} be the set of irreducible characters of N. Let () = e{I} = Ea Xi' By (17.1) each a i is an algebraic i C integer in (J Inl' Let a E: HA / A' Then E a = E for some
\
"INI "
c with (c, 1(\;1 ) 1 where E is a primitive INI -th root of unity, Since 9(G) = e(G C ) for G E: N this implies that
Thus a.
1
(17.3)
=
a.
E:
(J as required,
1
Let n be a positive integer and Ie! .\1 be a union of
90
CHARACTERS OF FINITE GROUPS
conjugate classes of (1,\. Let 'M .(l {G I Gn E: .(l}. IT X is a character of @ then [1/( I @ I , n)] ~'M.q X(G) is an algebraic integer. Furthermore (I OJ I, n) I' Proof.
'iII.n ,.
Let O.U be defined as in (17.1). It suffices to prove
the result in case X is an irreducible character. Then
The first statement follows from (17,1), The last statement follows by setting X = IN' By using different methods Solomon {3j has shown that in case .~l
= {I}
in (17,3) then even
is an algebraic integer, Frobenius has conjectured that if (INI, n) = 1~{1} I then ~
(17.4) Let INI = glga with (gH ga) ::: 1. Let X be an irreducible character of N. IT gl X. (1) then X. (G) ::: 0 unless Gga ::: 1. Conversely if X(G) ::: 0 whenever Ggl then gllx.(1).
==
1, G
;c
Proof. Suppose first that X(G) ::: 0 whenever Ggl ::: 1, G ;c 1. Let p be a prime such that plgl and let il3 be a
1
91
EQUATIONS IN GROUPS
S -group of (~. Thus X(G) :::: 0 for G p
E: ~:#.
Hence by (6.2)
1~llx(l). Therefore g l lx(1). Assume now that g l lx(1). Let W :::: {GIGg2 :::: 1}. By (17.3) g21~~lx(G)12. Let W :::: U .\ where each .~li is a conjugate class of ('~ and let G i
E: ,(t •• Thus ~ .. lx(G)12 1.1
:::: ~il.'lilx(G.)x 1
x(G/ Hence by (2.17) x(1)1 ~~lx(G)12. Therefore
As x(1)
> 0,
Ix(G)/2
>0
for all G
E: ~
and ~~lx(G)12
is a rational integer, this yields that
Therefore ~N_~lx(G)l2 :::: 0 and so X(G):::: 0 for G
E:
N
11
as was to be shown. As an immediate consequence of (17.4) we get (17.5) Let INI gIg2 with (glg2) :::: 1. Let X be an irreducible character of ~. If X(G) :::: 0 whenever 1, G ¢ 1 then X(G) :::: 0 whenever -~2 ¢ 1.
CHAPTER III § 18.
CRITERIA FOR SOLVABILITY
Throughout the rest of these notes we will to a large extent be concerned with applying the theory of character to questions concerning the structure of finite groups. We begin with one of the earliest and most elegant results of this kind due to Burnside which also forms the starting point of P. Hall's theory of solvable groups. Only elementary properties of characters will be used in this section. The first four results are due to Burnside. The remaining results in this section are due to P. Hall. (18.1) Suppose that (I.n I, X(1» == 1 for some conjugate class of JI of @ and some irreducible character X of @. Then for GE: .\1 ~ither X(G) == 0 or Ix(G)1 ;;: X(1). Proof. There exist rational integers s, t such that sl.{11 + tx(1) == 1. Hence sl.~llx(G) + tx(1) X(G) ;;: X(G). Thus by (2.17) a == X(G)/X(l) is an algebraic integer. Let a 1 ;;: a, all, ... , be aU the algebraic conjugates of a. Since every algebraic conjugate of a root of unity is again a root of unity it follows that each a i is a sum of X(1) roots of unity divided by X(l). Thus lail < 1 for all i. Therefore lITail < 1. Since TIai is a rational integer this yields that ! TIai I == 0 or 1. In the first case a ;;: 0 and thus X(G) ;;: O. In the second case I a I == 1 and so I X(G) I = X(l). (18.2) Suppose that I.SII is a power of a prime for some conjugate class .SI ~ {t} of @~ Then @ is not a noncyclic _Simple group. 93
94
C H A RAe T E RS 0 F FIN I T E G R0 UPS
Proof. Let I·~ll = pC where p is a prime. Let G E: .~!. = then G €: Z(®) and the result follows. Suppose that c > 0 and ® is a noncyclic simple group. Then every nonprincipal irreducible character of ® is faithful. Let Xl = 1~, X2 , ••• be all the irreducible characters of ®. Then
If c
1 +
"E
i7" 1
X.(1) X.(G) 1
0
1
Thus there exists i ~ 1 such that Xi(l) Xi(G) ~ 0 (mod pl. Hence (I.'ll, Xi(l» = 1 and Xi(G) ~ O. Thus by (18.1) 1Xi(G) I = Xi(l). Therefore by (6.7) G E: z(~) contrary to the fact that ® is a noncyclic simple group. (18.3) If I ® I solvable.
= paqb
where p, q are primes then l~ is
Proof. The proof is by induction on !®I. It may be assumed that p ~ q. If a 0 or b = 0 the result is clear. Assume that ab ~ O. Let ~ be a Sp-group of ®. Let P E: Z(~)# and let .\l be the conjugate class of ® with P £ .n. Since ~ c C(P) it follows that I·~ll = I~: C(p) I is a power of q. Henceby (18.2) there exists.\'>
Proof. Suppose that ~ is simple. Thus X is faithful. If nonabelian then XI'll is irreducible. Thus by (6.7)
IX(G)I = X(1) for G E: Z('ll)# and so Z('ll) c Z(~) contrary to the simplicity of ~. Thus 'll is abelian. Hence if P E: ~# and .'l is the conjugate class of ® with P E: ~l we see that (I·\ll,x(l) = (1~:C(P)I,p) = 1. Thus by (18.1) X(P) 0 or Ix(P)1 = xU). By (6.7) the latter case implies that P E: Z(®)
CRITERIA FOR SOLVABILITY
95
contrary to the simplicity of ®. Therefore X(P) :::: 0 for P E: .\l#. Hence 1~llx(l) by (6.2) which implies that I~I
: : p. The next three results are elementary preliminaries.
(18.5) If ® is solvable and 16)1) 1 then there exists a prime p and a p-group ~ such that (I) ;z! ~ <J (\). Proof. Induction on 1®I. If 16)1 is a prime let ~ :::: ®. Suppose that (~ is not a prime. Since ® is solvable, there exists ,'> <1 ® such that I®:.\') 1 is a prime. By induction S) contains a normal subgroup distinct from (I) which is a pgroup for some prime p. Let ~ be a maximal normal psubgroup of ~). Thus ~;z! (I). Since ~ is the intersection of all the Sp-subgroups of ,\) it follows that ~ is characteristic in ~) and thus ~ <1 ~. (18.6) Let I(\}I = II p~i, where {PJ = 1T(®). Suppose that l ",'\ and ~)2 are subgroups of 6} such that for each i, c· c· p. 111,\)11 or p.IIl~)21. Then 6} :::: S)1.\)2 and 1S)1 n S)21 = I
I
---
--
(hu ~).
Proof. Since 3)1 ~)2 is a union of cosets of .\)1 and also a union of cosets of S)2 it follows that for each i, Pi Ci 11.11)1 S)21. Hence I ® II ! S) 1'\)21 and so ® .\) 1'\)2' By counting cosets it is easily shown that !3)111S)21 = 1'\)1~;'21I.11;'1 n .\)21. Since I(~I 1'\)1.\'\1 is the least common multiple of 1~)11 and 1S)21 the result follows.
(18.7) Let 1® I == IIf:::: 1 p1 i where 1T(®) :::: {PI"'" ps}' Suppose that for each i = 1, ••• , s, C~_ .contains a subgrouJ! ,\\ with 1~ :s\ 1 pfi. Let !' j = .\)j n ... n S)s for j :::: 1, j-1 c· ... ,s. Then I X'J' 1 :::: 11'-1 p.1 1 . 1-
Proof. The proof is by induction on s- j. If s j 0 then 1'. ::::,\) and the result follows. Suppose now that J s
96
C H A RAe TE RS 0 F FIN I T E G R0 UPS
It'.
)+1
j c· I = II. 1 p.I. Since F
1
l). =
J
X' .
)+1
~.
J
the result
follows from (18.6). Let r~1 = p';l p~2 m, where Pl and P2 are primes (18.8) not dividing m. Suppose that ~ contains a subgroup t' with I'!'l = PIC1 p;2 and ~ contains subgroups iJi for i ;: 1 such that I~:€\l is a power of Pi and 1@:3)il > 1. Then ® is not simple. Proof. By (18.3) l' is solvable. Hence by (18.5) there exists a p-group 'l3 with (1) ~ '<J !' and p = Pl or P2' Without loss of generality it may be assumed that p = P2' Replacing .pI by a conjugate if necessary it may be assumed that some Sp-group of '!' is contained in -Pl' Thus , c .\'>1 since' is contained in every Sp-group of !'. By (18.6) t'.pl ;: <M. Thus every conjugate of -S:>l is of the form D-l ,~\ D for some D E: 1". Therefore D-l,D = , since' <J '1:'. Consequently , en @ G-l.pl G <J <M. The group n<MG-I ,~\ G is a proper normal subgroup of <M and so <M is not simple.
(18.9)
s 1 PiCi where 1T(<M) Let I<M\ ::: IIi:::
{ PI' •.•} , Ps •
Suppose that for each i = 1, ... , s, G) contains a subgroup .p. c· 1 with I<M:~ I = Pi l . Then <M is solvable. Proof. Induction on I<MI. If s < 1 the result is clear. Suppose that s) 1. Let t' n~ 3 .\'>.• By (18.7) IX'I c c 1::: I PI 1 P2 2. Thus by (18.8) @ is not simple. Let 1 ~!(
CRITERIA FOR SOLVABILITY
97
Let 1T be a set of primes. P. Hall has introduced the following propositions: E1T
<» contains
C 1T
® satisfies
D1T
a
S1T
-subgroup. and any two
S1T - subgroups of @ are conjugate. ~ satisfies C 1T and any 1T-subgroup of ~ is contained in some S1T - subgroup of ~. E1T
The next result which includes the converse of (18.9) is the beginning of a systematic investigation of solvable groups. (18.10) Every solvable group satisfies D for every set 1T of primes 1T. Proof. Let ~ be a solvable group and let 11 be a set of primes. The proof is by induction on I~I. If I~I is a p-group the result is trivial. Suppose that the result has been proved for solvable groups of order strictly less than I~I. By (18.5) there exists a prime p and a p-group 'U such that (1) ¢ 'V <j @. By induction ~/ '.l3 satisfies C 1T • Let ~): be a S1T - subgroup of ~/ '.l3 and let ~)o be the inverse image of s:>: in ~. Suppose first that p E: 1T. Thus S) = S)O is a S1T - subgroup of ~ and so ~ satisfies E1T. Let " be any 1T-subgroup of ® and let ~* be the image of ~l in ~/'.l3. By induction ~* is conjugate to a subgroup of .~\j. Thus ~ '.l3 is conjugate to a subgroup of ~). Since ~ c !I~ this implies !I is conjugate to a subgroup of ~ and so ~ satisfies D 1T • Assume next that p t 1T. By the Schur-Zassenhaus theorem .\)0 contains a subgroup ") such that S)O = ~ '.l3 and S) n $ = (1). Thus ~ satisfies E . Let !I be a 1T-subgroup 1T of ~ and let !I* be the image of !l in ~/$. By induction ! * is conjugate to a subgroup of S)t. Without loss of generality it may be assumed that !I * c S:)t. Thus 11 'V c ~)o. Since ~$ = ("$ n .N~ it follows from the Schur-Zassenhaus theorem that ~ is conjugate to ~ ~ n S) in ~(~. Thus ~ satisfies D in this case also. 1T
98
§
C H A RAe T E RS 0 F FIN I TE
G R 0 UPS
19. QUOTIENT GROUPS
Let .s; be a subgroup of ~ and suppose that ~o is a S1T- subgroup of ~ for a set of primes 1T then a normal complement of .\) in ~ is called a normal1T-complement in ~. In these definitions the phrase "in~" will often be omitted when the context determines the group ~. There are many theorems by many authors which assert that under various conditions on ,\) and '\')0 there exists a normal complement of .u over '\)0 in 6.S. We will here follow a paper of Brauer [7] which has several of these theorems as consequences. The proofs given here make use of the results of Section 15, in particular (15.4). In most of the applications these results can be avoided. If ,\)/~o is abelian then the normal complement of .u over .\")0 can often be constructed by more explicit methods which do not make use of the character theory. If ~V.s;o is solvable then generally a routine induction argument may be used to reduce the problems to the case that .\")'/'\)0 is abelian. If G E: ~ and 1T is a set of primes then G has a unique decomposition in the form G ::; G 1T G1T I = GIG where 1T 1T' G1T , G1T , is respectively a 1T-element, 1T'-element. The element G 1T , G 1T ' respectively is the 1T-factor, w'-factor of G. If 1T = {p J let Gp ::; G 1T • Throughout this section the following notation will be used: £")0
C ~,
where ,\) is a subgroup of 6.S and
1T ::; 1T(a;/~\).
'\)1' ••. '~)n are the inverse images in ,\:) of the conjugate classes of ($)/£")0)#'
For i = 1, ... , n, ~i is the set of G E: ()) such that Gll is conjugate to an element of ~\. ~o ~ - U ~i' The sets 3\ are pairwise disjoint. This is not the case
P=l
99
QUOTIENT GROUPS
in general for the ~i' It is clear that each @i is a union of conjugate classes of ~. We will be concerned with groups in which the following hypotheses are satisfied. (I~:~)I, I~:.\\I) = 1
(19.1)
(19.2) If HI, H2 E: S) and HI is conjugate to H2 in ~ then HI ·\?o and H2 '\)J are conjugate in S)/S)o' (19.3) If H is a 11-element of that p IT(H)I. Then
£1-.\)0
and if p --
E:
11 such
p j' IC@(H): C-\,(H) I I~il
(19.4)
1~:-\11
I.\)il for i
=
0, ... , n.
The main purpose of this section is to prove the following result. (19.5) Assume that (19.1) (19.4) are satisfied. Then ~o is the unique normal complement of .\) over -\)0 in ~.
The proof of (19.5) will be given in a series of short steps. Throughout this section it will be assumed that (19.1) - (19.4) are satisfied. If p E: 11" then by (19.1) -\' contains a Sp-subgroup ~ of @. Hence by the Sylow theorems every p-element G of @ is conjugate to an element H E: ~ c .\). G is of the first kind or the second kind according to whether H E: S)-S)o or not. By (19.2) this definition is independent of the choice of H. For any G E: ~ let a (G) be the product of all factors Gp with p E: 11" which are of the first kind and let f3(G) be the product of all factors Gp with p E: 11 which are of the second kind. Thus G a (G) j3(G) G11 '. The sets
(19.6) S)i
~i
n
()ji
are pairwise disjoint. For i
0, ... , n,
S).
Proof. Suppose that ~i n ~j is non empty . If i = 0 then j = 0 by definition. Suppose that i, j > O. If G E:
100
CHARACTERS OF FINITE GROUPS
@i n @j let .fi be the conjugate class of @ with GlI E: .R. Thus .R .pi and ~. i.Yj are both nonempty. Let Hi E: ~n ~i,Hj E: .Rn ~j' By (19.2) Hi.no is conjugate to Hj .po in S;;I ~0 and so i = j. The first statement is proved. By definition .\)i C @i ~ for i > O. If H E: @i with i > 0 then H1T is conjugate in @ to some element of .ni. Thus by (19.2) H1T E: .\1i' Since .\1/~0 is a 1T-group Hll' E: .po and so H = H1T H1T , E: ~i.\)o = ~;i' This shows that ~)i = @i ~ for i> O. Therefore
n
n
n
n .\)
n
n c:. .O 1'\)i:: 4'0
1=
The proof is complete. (19.7)
SJ -
If a(G)
~)o'
¢
1 then G1T is conjugate to an element of
Proof. If G 1T is a p- element for some prime p E: i this follows from the definition. The proof is by induction on the number of distinct primes dividing the order of G 1T • If G 1T is not a p-element then G 1T HGp where H is the p' -factor of G1T and where p may be chosen so that o(H) ¢ 1. Thus by induction H is c'onjugate to an element of SJ - £1 0 , Replacing G by a conjugate it may be assumed that H E: ~) .po. By (19.3) C.p (H) contains aSp-subgroup '.130 of C@ (H). Since Gp E: C@ (H) it is conjugate in C@ (H) to an element of '.130 c .p and so G1T = HGp is conjugate to an element K E: .p. If K E: .\)0 so is every power of K and thus o(G) = 1 contrary to assumption. The proof is complete. (19.8)
(i) (ii) o(G) lie (iii) (iv)
G E: @o if and only if o(G) :: 1. If G E: @ then the elements G, G 1T and in the same . If G E: @o then every power of G is in @o. All 1T1 -elements of @ lie in @o'
Proof. If a(G) ¢ 1 then G Ej: @o by (19.7). Suppose that G E: @i with i ¢ O. Then G 1T is conjugate to some H E: .\)i'
QUOTIENT GROUPS
101
Thus a(G) is conjugate to a(H) and (3(G) is conjugate to p(H). Let Hp be the p-factor of p(H) for some prime p. By definition Hp is conjugate in @ to an element of .\)0' Thus by (19.2) Hp E: .\)0' This yields that p(H) E: .po. Hence H = a (H) p~H) E: a(H).po and so a(H) E: .\)i iYo = .pi' Thus a (H) '#:- 1 and so a (G) '#:- 1. This proves (i). If G E: @{ with i > 0 then we have just shown that a(G) is conjugate to an element a (H) E: .pi' Since a (G) = (a(G))1[ we see that a(G) E: @i' If G E: 6\0 then a(G) E: @o by (0. By definition G and G 1T lie in the same @i' (ii) is proved. Statements (iii) and (iv) are immediate consequences of (i). (19.9) Suppose that G ~ x 'l3 is a 1T-group where 'l3 is ~ p-group for some p E: 1T tlnd W = (A) is a cyclic p' -group. If a (A) '#:- 1 then (i is conjugate to a subgroup of ~.
Proof. By (19.7) A is conjugate to an element of ,\). Thus it may be assumed that A E: .\;"'. By (19.3) Cg,(A) contains a
Sp-group of C@(A). Thus 'l3 is conjugate in C@(A) to a subgroup of .\) and so ~ is conjugate to a subgroup of 5). (19.10) Let ~ be an irreducible character of ",/5)0' Define the function e := e ~ on @ !2.Y e (G) = ~ (H) if G E: ~i and H E: '\"'i' Then (J is an irreducible character of @.
"'i
Proof. Since ~ is constant on each we remark that by (19.6) G is well defined. Since each @i is a union of conjugate classes of 6\, e is a class function on @. The result will be proved by showing that e satisfies conditions (i)-(iv) of (15.4). We have just observed that condition (i) is satisfied. Since e (1) = ?; (1) > 1 so is condition (iv). We next verify condition (ii). Let
C H A RAe T E RS 0 F FIN I T E G R0 UPS
102
(lo is conjugate to a subgroup of .p. Hence it may be assumed that (.Yo c ,u, thus e ,~ = ~'(.Y and so G is a generalized character of (.Yo' 'AssumeOthat a(A) = f. T.hus by (t9.8) (iii) a(AJ} = 1 for all j and if P E: ~ then a(AJP} = a(P}.
8,
Thus by (19.8) (ii) () (AjP) = e (P). Hence it suffices to show that eI ~ is a generalized character of ~. By the Sylow theorems and (19.1) ~ is conjugate to a subgroup of £"r. Hence it may be assumed that 'l3 c ,u. Therefore BI'l3 = ~ I'l3 which shows that condition (ii) is satisfied. Let Hi E: ,ui for i = 0, •.. , n. Then by (19.4) 1
n
II e II® = ffiff i~ I @i I It (Hi W 1
n
120
= ,.\)'
==
I S\ 11~(Hi} 12
II t II~ = 1
Thus condition (iii) is verified and by (15.4) ducible character of ®.
e is an irre-
®o is a normal complement of .\1 over .\)o in ®.
(19.11)
Proof. Let 1:; l ' 1:; 2' ••• be all the irreducible characters of . Let 8 i = e~i be defined as in (19.10). Then the definition of e i yields that Bi (G) == e i (1) for all i if and only if G E: @o. Hence by (19.10) ®o
~;:®o
I
= 1 '\):'\)0 I
this yields that 11
=
=
I
®:~
I
®o as was to be shown.
NON SIMPLICITY CRITERIA
103
120. NONSIMPLICITY CRITERIA The results in this section are all consequences of (19.5). We begin with a theorem of Wielandt [1] which generalizes a classical result of Frobenius. (20.1) Let ft be a subgroup of N, let !o ,l) = N(~) (!~Assume that (l,l): ~ I , II :-!o 1 )
<1
~ and let 1 and
! n IG c 10 for G E:: ('II - .\), Then (;'1 and ,l) each have at most one normal complement of ~ over !o. N has a normal complement No of ~r over! 0 if and only if ,l) has a normal complement .\}O of ! over 11 0 , In that case (~\o
b~ - U();(,l) - ,l)o)G . Proof. Suppose that No exists. Let .po
=
bl o n .p. Then
Thus .p ::: !.po' The uniqueness of .po follows directly from the fact that (I.\}o: 110 I, 1 ~: !o I) :::: 1, G Assume that .po exists. Suppose that H E:: ,l) n.p with G E:: (,' - S). Let a :::: I.)):!I. Thus ~ E: ! n !G c !o c .po' Since (a, l.p:,l)o I) :::: 1 this implies that H E: Con-
S);.
sequently ,l) n .\) G c .\)0 for G E: 61 - ,l). Let 11, Ni,.pi be defined as in section 19. Suppose that conditions (19.1) - (19.4) have been verified. Then (19.5) implies that 610 is the unique complement of ,l) over .\}o' If G E: N then by (19.i) G'll is conjugate to an element of 4), Thus U(\j(.\'> - .\}o)G =
U~1l)\
and l'lo has the required form.
n remains to verify conditions (19.1) - (19.4).
104
CHARACTERS OF FINITE GROUPS
Let
~
be a subset of G-l
~,
~o.
f
)8
~G c ~
If
for G e:: 6;
then )8 S ~ n ~ ~o and so G e:: ~. If ~ is a Sp-subgroup of ~ for some p e:: 11 this implies that Nl~(~) S ~ and so )8
~
= {H}
C~.
is a Sp-subgroup of
This verifies (19.1). If
~o we get that C(~\(H) :::: C~(H).
for some He::.\)
Furthermore if H' e:: ~ and H' is conjugate to H in 6\ then H' is conjugate to H in ~. Thus in particular (19.2) and (19.3) are proved. Since
16\1 : : ~?1= 0 1(;'\1. 1 and I~I = ~?1= 0 I'1.'1 ..... '
show that I~il
= IN: ~11~il
for i
>0
it suffices to
in verifying (19.4).
If .\1 is a conjugate class of (~\ with .\1 c Ni for some
n
i > 0 then we have just shown that .\1 ~ is a conjugate class and .\1' ~ ~ ~Q' Thus for H e:: .11
n s
1·111 = 16\:C~(H)1 = Il~i:~II~:C.\1(H)1 :::: IN:~II·l1n ~I This implies that I(~\il = IN: ~IINil for i > 0 and completes the proof. Due to its importance in many applications we state a special case of (20.1) explicitly. As will be seen later more direct proofs can be given of this theorem. (20.2)
(Frobenius) Let
~
that ~ ~G = \ 1) for G e:: anormal complementOt ~ in
n
be a subgroup of @ -
~
(,i.
(~\.
Assume
Then there exists .
As an application of (20.1) we will prove a special case of the Frobenius conjecture mentioned in section 17. See Feit [1], Wielandt [1] . (20.3)
IW :10 I
W be a subgroup of N and let W1)
Let
G e:: (~\ - NN (~) and (a, m) = 1. Let m G = 1} . If Im21 = m then ml
9.n
=
{GIG e::-
('I, -
NONSIMPLICITY CRITERIA
105
N(\\ ('I) and let9Jlo ::: ~ n rol. Since (1'1 0 I, 1'1l: ~o I) ::: 1 and ~o <1 'I it follows that ~o <1 ~. Furthermore ~/!o <1 ~/~o' By the Schur- Zassenhaus theorem there exists a complement of !/!o in ~/~lo' Thus there exists a subgroup ~'() of ~ such that ~o ~o and I~: ~'() I ::: a. Suppose we know that ~'() <1 .\) then by (20.1) there exists a normal complement (\10 of ! over ~o' Since 1(~\:Nol ::: a we get that INol = m and so No 9Jl <1 N. It remains to show that ~o <1 .\). This will be done by showing that ~o = 9Jl o ' Since I~'O II m, ~'O C 9Jl o ' Let I~: ~II ::: h and let I!o I ::: a o ' Thus I~I ::: haao and 19J1 o I > hao since .\)J c \Dl o . It suffices to show that lrolo I hao . Let G E: (\\ - 9Jl. Then some power H of G has prime order p, where p %m. Let' be a Sp-subgroup of (\\ with HE:'. Let be a Sp-subgroup of ~. Then '.13 is a Spsubgroup of N since p 1 1(\\: ~ll . Thus , ::: A for some A E: ~. Hence ~ ==
Proof. Let
'0
0
'0
1 1 HA - = HGA - E: ('oAn 'oAG) A-I
::: m
n
m
Thus AGA -1 E:
~
by assumption. Consequently ®
'+'0
'+'0
AGA-l c ~ln ~AGA-l .
9Jl::;
G
U® (~ - rol o ) . The number of distinct conjugates of .\) is at most I(i,\:.\)1 and I~ rol o l:s hao (a - 1). Therefore m(a - 1) ::: 1('1 - roll :s
IN:~IIS)
- rol o I
==
m(a - 1)
Thus we must have equality and so IS? - 9Jl o I ::: hao (a - 1). Therefore 19J1 o I ::: hao as was to be shown. If H is a ll-element of N, the lI-section ~N' 11 (H) is the set of all elements G E: N such that GlI is conjugate to H in N. Either or both of the subscripts will be omitted if (\j and/or 11 is specified. Let r ll (.\'» ::: I~k (1)1 for any .y , 11
group
~.
The following two lemmas will be needed.
106
CHARACTERS OF FINITE GROUPS
(20.4) Proof. If G
E: ~(H)
there are 16}: CN(H)I possibilities
for Gll • If H' ;;: G1f then G = H' G1f , where G1f , may be any 1f' -element in C (H' ). Since C(~) (H) is isomorphic to N C (H') the result follows. N (20.5) Let p be a prime. Then any group ~.
rp(~) 'if-
0 (mod p) for
Proof. Let '13 be a Sjl-subgroup of ~. Distribute the p'elements of ~ into equIvalence classes where H is equivalent to H' if and only if p-I HP = H' for some P E: \.13. The number of elements in any equivalence class is a power of p and {H} forms an equivalence class if and only if H E: C N (\.I3). Thus rp(~) F r p (C.p(\.I3» (mod p). Thus it may be assumed that.p C~ ('D). This yields that for P E: \.13, @5(p) consists of all elements PG where G is any p' -element of .p. Thus I@5(p)1 = rp(,f) for P E: \.13. Since @5(P I ) is disjoint from @:i(P2 ) for PI:f. P 2 and ~ U p €\.13 e(p) we get that l.pl ;;: 1\.I3lrp(.p) which implies the result. Let .p be a subgroup of (~\ let .po <1 .p and let Assume that if ~ is any abelian 1f-subgroup of .p (including ! ;;: (1», every conjugate class of C(~l {~O consisting of 1f-elements meets .p in a conjugate (20.6)
1T = 1T (.p/~-o)'
class of C.p (!). Then there exists a normal complement of .p over .po in N. Proof. If H is a 1T-element of .p let and
~*
C.p (H). Then .p* = (\1*
n
(~*
C (~(H)
.p. If ! * is an abelian
1T-subgroup of ~* then W ;;: (H, !*) is an abelian subgroup of ~ and
NONSIMPLICITY CRITERIA
=
107
C 6y* (~*) (\.\) = C~*(1(*)
If .fl* is a conjugate class of C(,~*(~*) = C ; (~) consist-
6
ing of 1I'-elements then Jl*
C
6\* and so
Then by assumption .\l* (\ ~* is a conjugate class of Hence the pair N*, ~* satisfy the same assump-
C~*(W*).
tion on (,~ and ~ for the same 11'. Observe also that this assumption is satisfied for any subset 1T 1 of 1T. Let Ho = 1, Hl , • • • , Hm be a complete set of representatives of the conjugate classes of ~ which consist of 'IT 1 elements for 1T 1 C 11'. We next prove by induction on !611 that r «\1) r (~) 1T 1 _ 1!'.1 (20.7) I(~I - -71.6="-JBy letting ~
= ( 1) we see that (\; = U~O ~(~ (Hi) and ~
=
u~Oeco.(H.), where the sets e",(H.) are pairwise disjoint. 1=
4'
1
",
1
Thus by (20.4)
If H. Ef: Z (®) then since ®* = Cru(H.), ~* = C (H.) sat1
isfy the same assumptions as get by induction that
V}
~
and
1
.\)
~
1
and l®* I < 16)1 we
C H A RAe T ERS 0 F FIN IT E G R0 UPS
108
r
(C@(H i 111 IC@(HiH
»
Thus if exactly k of the H. lie in Z(o) it follows that 1
k
r
(03)
r 1Tt
1031
= k
lT
(,p)
I~l
which implies (20.7)
The result will be proved by showing that (19.1) - (19.4) are satisfied and then using (19.5) Let p E: 1T and let H be a 11-element in ,p. Let 1T 1 = {p}. Applying (20.5) and (20.7) to the pair C@(H), C,p(H) yields that p ,rlC@ (H)I:
C.p (~). Thus
(19.3) is satisfied. Letting
H = 1 yields (19.1). (19.2) follows from letting !I = <- 1) . It remains to verify (19.4). Each ~i is a union of 11-sections. Thus it suffice!;> to show that ISO; (Hi) 1 = 103: .pll~,p (Hi)] for i = 0, ... , m and this follows from (20.4) and (20.7) applied to the pair C(}/H ), C.p (Hi) . i The next result due to Brauer [7] is a generalization of earlier results of Grlin and P. Hall. If 'n c .\) c 0) where 'll and ~ are subgroups of (Ilj then ~ is weakly-closed in ,p with respect to (~ if ~G c ,p for some G E: (l) implies that = !I.
,.G
(20.8) . Let 11 be a nonempty set of primes. Assume that the following conditions are satisfied. (i) ~ is a S11-subgroup of (l). u ~ Z(~) and U is weakly closed in ~ with respect to 0\. (ii) If !I is any abelian 11-subgroup of .p = NCl}(U)' then
D11 holds for C(l} (~) . Let ~/.po be the maximal 11-quotient group of ,p. Then there exists a normal complement of .p over ,po in ~. Proof. Let 'It: be an abelian 11-subgroup of .p and set 0)* = i>* C,p(~) bi*n i>. Thus <-~, u) is a 11-subgroup
C~(!I),
NONSIMPLICITY CRITERIA
of
('i
and since D'IT holds for
@
109
= C O;(l)' <~, (S) S lB G for
some G E: (1,). Thus (S = (SG c Z(~) as IS is weakly closed in lB. Since ~l c ~G this implies that IS S CO; (~) (~* .
By (ii) there exists a S -subgroup ~* of (1,)* with (£ c ~*. As lB* c lBH for soml H E: ('I; we conclude that ij c- Z(lB*). Clearly-ij is weakly closed in 58*. Thus (i) is satisfied if (1,;' 58, (£ are replaced by (1,)*, ~* , (l . H ~* is an abelian lI-subgroup of .p* = NO)* (£) then ~I = (~, ~*)
is an abelian lI-subgroup of .p and
N* n
c('Ii. (~*)
=
C@* (1*)
Thus D'IT holds for C@* (~*) by (ii). Since N(~* (l) (I,)*n.p = .p* we see that also (ii) is satisfied if N, ~, ij is replaced by (~;*, ~*, (l . We will next show that (i) and (ii) imply that if .fi is a conjugate class of N consisting of 'IT-elements, then .~l n .p is a conjugate class of .p. By D'IT' ~l n ~ is nonempty. Hence .n n .p is nonempty. Let HI' H2 E: ,(l n ~ and let 1f = S 1\. Let G- I HI G H2 . Then G-I T'1 G and 1)2 are both S'IT-groups of C(I,) (H2 ). Hence by D'IT in C (H2 ) it may be assumed that G-I!\ G = 1"2 . Thus N G (IS, IS ) c 1"2' As T' 2 c t\H for some H E: (,; the weak closure of IS in t\ implies that IS = ISG. Therefore G E: .p. Thus .ft n .p is a conjugate class of .p. In view of the first part of the proof this may be applied to (,i* = C (i,} (~) and ~* C.p (I) for any abelian subgroup ~ c .p. Thus the hypothesis of (20.6) is satisfied and the reSUlt follows from (20.6).
110
CHARACTERS OF FINITE GROUPS
The next result is essentially due to Griin and P. Hall. See M. Hall, Theorem 14.4.6. (20.9) Let p be a prime and let ~ be a Sp-subgroup of (\). Let IS be a subgroup of Z(~) which is weakly closed in ~. Let ~ = N®(IS) and let tJ/tJ o be the maximal p-quotient
group of ~. Then there exists a normal complement of tJ over ~o in (\\. Proof. This is an immediate consequence of (20.8) with {pI since (20.8) (ii) is satisfied by the Sylow theorems.
=
11
(20.10) ~
of (\\
Let p be a prime. Assume that aSp-subgroup abelian. Let ~ = NG) (~) and let tJ/~o be the
maximal p-quotient group of tJ. Then there exists a complement of tJ over tJo in G). Proof. Let
~
= (£ in (20.9).
Let p be a prime. Assume that the Sp-~ of tJ is abelian and ~ ). Then there exists a subgroup ~o of ~SuCh that
(20.11) ~
~ = ~o x
'0
3.
n:
Proof. Let HI' ... , Hm be a complete system of coset representatives of ~ in tJ. For P E: ~ let t(P} 1
H;-I PH .. Since 'U is abelian t{P) is independent of the 1
1
choice of the H.. Furthermore t(P l P:a) 1
=
t(P1 ) t(P 2 )
,
t(P 1 ) := t(P 2 ) in case PI is conjugate to P 2 in tJ and t('U) c 8. Let 'Uo be the kernel of t. If Z E: 'Uo n 8 then 1 = t(Z) = Zm. Hence Z 1 since (p, m) = 1. Thus == <. 1) and so 'Uo 8 = 'Uo x 8 S 'U. Thus
\\0 n 8
l'Uo1181
S I'UI
=
l\\ollt('U)1
< 1'130 1181
NONSIMPLICITY CRITERIA
Hence we must have equality and so ~ ~. The proof is complete.
~o
x .8
111
= ~.
Clearly
'!Po
(20.12) Let p be a prime. Suppose that aSp-subgroup '!P of 1M is abelian. Let .8 = '!P n Z{N@('!P)). Then there exists a normal complement of .8 in 1M. Proof. Let
~ =
N@('!P). Let '!Po be defined as in (20.11).
Then ~/'!Po is isomorphic to ~o /'!P o x '!P/'!Po' where ~o is a subgroup of ~ whose Sp -subgroup is '!Po' Hence ~o is. a normal complement of B in ~. It is easily seen that ~/.\1o is the maximal p-quotient group of .f>. Thus by (20.10) there exists a normal complement IMo of S> over ~o' Then
<.\lon.B = @o n
~n.B
= ~o n.8 = ( 1)
Since I@: @o I = l~: ~o 1 = 1.81 this implies that @0.8 @. The result is proved. As a special case of (20.12) we get one of the earliest results of this type. (20.13) (Burnside) Let p be a prime and let '!P be a Spsubgroup of @. Assume that ~ S Z{N@('!P». Then there exists a normal p-complement in @. (20.14) Let p be the smallest prime in n{(\\). Assume that a Sp-subgroup of @ is cyclic. Then (Ib has a normal p-complement. Proof. Let '!P be a Sp-subgroup of ~ and let ~ N@('!P)/C'!P('l3). Thus ~ is a pi -group of automorphisms of the cyclic p-group 'l3. Hence Is>!l(p-l, I@I). As p is the smallest prime in n(@) this implies that I~! = 1. Hence N~('l3) = C(~('l3) and the result follows from (20.13). A group in which every Sylow subgroup is cyclic is called a Z-group.
112
(20.15) ticular
CHARACTERS OF FINITE GROUPS
@
Let @ be a Z-group. Then is solvable.
@'
is cyclic. In par-
Proof. By repeated application of (20.14) one gets a normal series of @ in which every factor is a Sylow group. Hence @ is solvable. Since every abelian subgroup of @ is cyclic it suffices to show that @" = (1). Assume that @" '" (1). Then by solvability @" '" @III • Hence by factoring out @III it may be assumed that @/I is abelian and therefore cyclic. Let @" = (G). Since @/C@(G) is an automorphism group of the cyclic group (G) it is abelian. Hence
@'
c C@(G).
Choose H E: @' such that (H@II) :::; @' /@" Thus @' :::; (H, G) is abelian and so @/I = (1) contrary to assumption. (20.16) Let ~ be a nonabelian Z-group. There exist integers m, n, s all greater than 1 such that sn 1 (mod m), ((s-1) n, m) = 1, mn :::; I@I where @ :::; (A, B) with Am = 1 :::; Bn, B-1 AB = AS and @' :::; (A). Conversely given positive integers m, n, S satisfying these conditions then the group generated by two elements satisfying these relations is a nonabelian Z-group. Proof. The converse is immediate since (I@:@'I , I@'I) 1 and @' and ®/(W are cyclic. Suppose that @ is a Z-group. By (20.15)~' (A) for some A E: @. Choose B E: @ such that @/@' :::; (B@'). Then @ :::; (A, B) and B-1 AB = AS for some positive integer s. Thus B- n AB n Asn:::; A and so sn :: 1 (mod m). a(1-s b ) b a b a Since B - A - B A A , every commutator is a power of B-1 A-I BA :::; A1-S . Thus (s 1, m) :::; 1. n n Since B E: @' there exists k such that B = Ak. Hence k _ k sk A B 1A B A . Consequently k(s -1) :: 0 (mod m) n and so k .:: 0 (mod m). Therefore B :::; 1. If a prime p n p divides both m and n then (A m/p , B / ) is a noncyclic
NORMAL 1I"-COMPLEMENTS
113
p-group contrary to assumption. Therefore (m, n) 1 and the proof is complete. It was observed by Frobenius that (20.13) can be used to prove the following special case of the conjecture mentioned in Section 17. Let I ® 1 ::; m l m 2 with (mlJ m 2) = 1. Let 9Jl i {G I G 1 = I} for i :; 1, 2. Assume that 1 9J1 i I :; mi for i :; 1, 2. Then 9Jl i
m'
Proof. By symmetry it suffices to show that 9J1 2
(~j
Proof. If @ contains a normal1T-complement ®o then (i) and (iii) follow from the Schur-Zassenhaus theorem. Let HI' H2 E: .p and G- 1 HI G ::; H2 . Then G ::; HGo with
114
CHARACTERS OF FINITE GROUPS
H E:: ~ and Go E:: ®o. Replacing Hl by H-l Hl H it may be assumed that Gol HlGO ::: H2 • Thus
and so Hil H2 E:: ~ n ®o ::: (1). Hence Hl ::: H2 and (ii) is satisfied. Conversely suppose that (i), (ii) and (iii) are satisfied. We will verify that (19.1)-(19.4) hold with .po == (1). The result will then follow from (19.5). Clearly (19.1) is satisfied and (ii) yields (19.2). Let H E:: ~# and let 13 be a Sp-subgroup of C® (H) for some prime p E:: 1f such that p Y I (H) I. Thus l! ::: (H) x i3 is elementary and by (iii) l!G ~ ~ for some G E:: ®. Hence HG E:: ~ and so by (ii) there exists A E:: fj with HGA H. Since 13GA ~ C.p (H) (19.3) is satisfied. To verify (19.4) observe that since ~!) ::: (1), each ~i is a conjugate class of .p. Let Hi E:: .p- and let cr 1f,(C®(H i Then INil:::: I(\}: CN(Hi)tci' Let Ib®(Hi)1 ai bi where ai ::: I Cl~ (Hi) 11f' By (17.3) ci 2::: bi and so Il~\1 2::: I~I/(aibi)bi::: I~I/ai' By (ii) I~il :::: I.\'): C~(Hi)1 and by (19.3) I C.\')(Hi) 1 : : ai' Hence I ~i 1 2::: I~: ~ II.pi I for
»·
i ;:: 0, .•. , n. Since E~=O 16\ i I :::: 16\ I and l':~:::: 01 ~i I I ~ I this implies that I (\ I :::: I(~\: ~ II.\')i I for i = 0, ... , n which verifies (19.4) and completes the proof. (21.2) (Suzuki [3]) Let .t) be a S1f-subgroup of ®. Then ® has a normal1f-complement if and only if the following conditions are satisfied. (0 Every elementary 1f-subgroup of N is conjugate to a subgroup of ~. (ii) There exists a set ~ of coset representatives of ~
in ~ such that ~H ;; ~ for H
E::
£>.
Proof. If ®o is a normal 1f -complement in ~ then (i) follows from (21.1) and (ii) is satisfied for 6J o = ~. Sup-
NORMAL 1T-COMPLEMENTS
115
pose (0 and (ii) are satisfied. Thus (21.1) (i) and (21.1) (iii) are satisfied. ~ence by (21.1) it suffices to verify (21.1) (ii). Let Hl1 H2 E: ~ and G-I HI G = H2 . Let G = HA where A E: 11 and H E: ~. Replacing Hl by H-l Hl H it may be assumed that A-I HI A = H2 and A E: 11. Thus H1 AH-l = H2 H-1 1 A 1 Therefore HI A
:=
E:
11
n
fJA = {A} -
AHI and so Hl ::: H2 as required.
(Frobenius) Let p be a prime and let 'l.' be a Sp-subgroup of @. @ contains a normal p-complementif and only if whenever two elements in '13 ~re conjugate in @ they are conjugate in '13. (21.3)
Proof. This is an immediate consequence of (21.1) as conditions (i) and (iii) are automatically satisfied by the Sylow theorem. The argument used in (21.6) is essentially due to Burnside. Our primary object is (21.8) below. This proof is due to G. Higman and was communicated to the author by J. Alperin. The following property of groups will be considered. Let p be a prime. For any two Sp-subgroups '13 1 and '132 of @ there exists G E: C@ ('13 1 n '132 ) with '13P = 'l.l'2 . (21.4)
(21.5) Let p be a prime. If @ contains a normal p-complement then.- @ satisfies (21.4).
Proof. Let 9Jl be a normal p-complement in ®. There exists G E: 'lJl such that '13 1G = ~2' If P E: ~l n ~2 then {p, pG} ~ ~1G = '13 2 ' Thus [p, Gl E: '132 n ~Jl = (I). He"lce G E: CG ('13 1 n '132 ), (21.6) Let p be a prime. Suppose that @ satisfies (21.4). Assume further that for any I1-group .\) in @, N@ (.\)/C@ (~) is a p-group. Then (\\ hasanormal p-com-
plement.
116
CHARACTERS OF FINITE GROUPS
Proof. Let
~ b~ Sp-subgroup
of
~
and suppose that
for some H E: ~, {p, pH} E: ~. Then pH E: ~ n ~H. By (21.4) there exists G E: C~ (~n '.PH) such that ~H : : : ~G. G-1 HG-l H Hence ~ H ;;: ~ and P = P . Let HG-l = G1 PI = PI G1 , where P l is a p-element and GI is a p'-element. Since G1 PI E: N~(~) this implies that PI E: ~ and by assumption GI E: C~ (~) ~ C(\~ (P). Therefore pH ;; p HG- 1
= -pG I PI
= PP 1
The result now follows from (21.3). (21.7) Let p be a prime. Suppose that @ contains no normal p-subgroup distinct from (I). Assume that every subgroup 4> of ~ with .t> "* @ satisfies (21.4). Then ~ satisfies (21.4). Proof. Let ~l and ~2 be Sp-subgroups of (~I. The proof is by induction on 1~1: ~l n ~21. If 1i3 1 : i31 n '13 21 : : : 1 then ~l ::::; ~2 and (21.4) is satisfied with G 1. Given ~l and '.\3 2 let l' = ~l n ~2' If '1:' : : ; (I) the result is trivial. Suppose that (1) "* l' ~1' Thus N{\j(!') 1'. For i = 1, 2, let ('Ii = ~i n N(\) (1'), let 9l i be a Sp-subgroup of N(\\ ('1:') with (~\ ~ iRi and let ~i be a Sp -subgroup of N with 9li ~ ~i' Since Ni S; 'l3i n ~i it follows by induction ~hat there exist Gi E: C(~ (~i n ~i) ~ C(~ ('t» such that ~~l ::::: ~i for i = 1, 2. By assumption there exists H E: C(;l ('1:') such that 9l~ : : : 9l z • Since 9l z f: ~1 H n ~2' induction yields the existence of K E: C(\l (~IH (') ~z) f: CN (1') such that ~1 HK ~2' Consequently
'*
(21.8)
(Frobenius) Let p be a prime and let '13 be a
'*
THOMPSON SUBGROUPS
117
Sp-subgroup of 0), Assume that for every subgroup ~ of '.f3, NO} (S»/C(~l (~) is a p-group. Then (~} contains a normal p -complement. Proof. The proof is by induction on 1@ I. If I @! = 1 there is nothing to prove. The assumption of the theorem is clearly satisfied by every subgroup of @. Hence by induction it may be assumed that @ has no p-factor group ;r. {I}. If (I) ~
*
'TT'-group? If the answer to this question is in the negative then it
can be shown by using (20.1) that there exists a simple group @ such that 'TTl and 'TT z are disjoint nonempty subsets of 'TT(@) with 'TT(@) = 'TTl U 'TT z and such that every proper subgroup of @ is either a 'TTl -group or a 'TT z -group. Baer [1].
§22. THOMPSON SUBGROUPS In this section a generalization of (21.8) for odd primes due to Thompson [3] will be proved and some consequences of it will be deduced. The proof of the result is not too long, however a fuller appreciation of the subtlety of the statement of the result may be obtained by studying the difficulties involved in the original proof of a weaker statement (Thompson [2]).
118
CHARACTERS OF FINITE GROUPS
The following elementary result will be needed. We state it here without proof. (22.1) Let p be an odd prime. Let ® == SL(2, p), Then the only abelian p' -subgroups of ® which are normalized by a Sp-subgroup of ® lie in the center of ®. For any group ® let m(~) denote the minimal number of generators of ®. Define d(@1 == max m(!() where !( ranges over all the abelian subgroups of ®. The Thompson subgroup T(®) is defined to be the subgroup of ® generated by all abelian subgroups ft of ® with m(.!) : : : d(®). Clearly if .p is a subgroup of 6} with T(~) S; .p S; ®, then T(®) is a characteristic subgroup of ~. (22.2) (Thompson) Let p be an odd prime and let ~ be a Sp-subgroup of ®. If C®(Z('ll» and N®(T(~» have normal p-complements, so does ~. Proof. Induction on 1® I. Suppose that @ has no normal p-complement. Let :IC be the set of nonidentity p-subgroups .p of ~ such that N@ (.p) has no normal p-complement. By (21.8) :IC is nonempty. Define a partial ordering on X as follows. If .p1' .pa C X then .\)1 < .t>-a if and only if one of the following holds: (i) I N® (~l ) Ip < 1 N® (~a) Ip . (ii) 1 N@ (.pI) Ip == I N6} (S)a ) Ip and 1.p1 I < l.pal. (iii) .p1 == ~3' Let .p be a maximal element of X with respect to -< and let ~l = N@(",). It may be assumed that ~ S; ~* S; 'll, where ~* is a Sp-subgroup of 91. If .p :; ~ then 91 s; N® (T(iJ» and so has a normal p-complement by assumption contrary to the choice of ~. Thus .p 7- ~. It is easily verified that the maximality of ~ with respect to < implies that 91/~ satisfies the assumptions of the theorem. Thus by induction 9l/f> .has a normal p-complement .(l/~. Since ~* s; t' it follows that Z(~) s; Z(iJ*) and so
THOMPSON SUBGROUPS
119
C'91 (Z(13*» has a normal p-complement. If N'91 (T(13*» does not have a normal p-complement then the maximality of ,p implies that 13 = 13* since T(tl *) is thelargest normal p-subgroup of ®. Let \I /,p be a chief factor of ('0 with \I S; .IL If \.1 '" .R then \.11J3 '" ~. Hence by induction \.113 has a normal p-complement t'. Then t' is characteristic in .\l and so ® = 13·ll ~ NO) (t') contrary to (22.3). Thus .f!/,p is a chief factor of ®. Since .{l and IJ3 normalize C$ (.fl/,p) it follows from (22.4) that (22.5)
C$ (.Il/,p) = ,p
Let ti = CSl (,p) so that ti <1 ~. If (i .9; ,p then .\l = ,p(i since Jl/,p is a chief factor of ®. As (i Ii ,p Z(£) is a Sp-subgroup of ti it follows by (20.13) or (21.8) that (S contains a normal p-complement which is thus normal in @ contrary to (22.3). Hence (S ,p. As .Il contains all pi -elements of 0) this implies that C® (,p) is a p-group and so by (22.4) CO) (,p) = Z(,p). Thus in particular (22.6)
Z('l.\)
S;
Z(,p)
=
C® (,p)
Let 9)l be a maximal subgroup of (\l with $ ~ fll. By induction 9Jl contains a normal p-complement which necessarily centralizes ,p. Thus by (22.6) $ = 9)1 is a maximal subgroup of ®. Let q be a prime divisor of 1·ll: S)1. The number of Sq-subgroups of ofl/,p is prime to p, so 'l3/,p normalizes some Sq-subgroup O/,p of .n/,p. Thus $D is a group and so $C = ® and .Il C by the maximality of $. Since off/4' is a chief factor of 0) this implies that .\i/ £>
120
CHARACTERS OF FINITE GROUPS
is an elementary abelian q -group. Since N('I\ (T('l3» has a normal p-complement, it follows that T('V) 1 .\). Let d ::: d('V) and choose an abelian subgroup 'l( of 'V with m(~) ::: d and 'ft 1 ~. Let ~'o = Yl n .\). Let 'V o ::: .\)Yl. By (20.11) .~~/~ = ~ll/.\) x .fi 2 /.\) where .~li/~ ::: Yl/ft o . Let
THOMPSON SUBGROUPS
121
G E: .R - 4:1. If m(5l.~) ;:::: 3 then m(C5l.\ (A, G-l AG» ;:::: 1. Thus there exists V E: $ such that (A, G-l AG) ~ C(,; (V). Hence (4:1, A, G-l AG) .s;: Ct~; (V). Since ~ = (~, A) the maximality of ~ implies that
'*
(" = (4:1,
A, G-l AG)
= C6; (V)
This contradicts the fact that C~ (Co) = (1). Hence m(5g) '!5 2. Since @/4:I is represented faithfully as a group of linear transformations on ~ this implies that ~/.v is isomorphic to a subgroup of GL(2, p). Since 6;/!t> is generated by its p-elements it is isomorphic to a subgroup of SL(2, p). Therefore by (22.0 .n/!t> £ Z(@/.v) contrary to (22.5). This contradiction completes the proof. The following corollary of (22.2) had been conjectured for some time however no direct proof of it is known. One major difficulty in attempting a direct proof arises from the fact that unlike (21.8) the hypothesis is not readily amenable to an inductive argument. (22.7) (Thompson [2], [3]) Let p be an odd prime and let ~ be a Sp-subg!:.oup.of ~,. Suppose that for every characteristic subgroup ~ of tl, N@(,p)/C@(.v) is a p-group. Then @ contains a normal p-complement. Proof. Induction on 16; \. If !t> is a characteristic subgroup of ~, !t> '" (1) then by induction N@ (!t»/~ has a normal p-complement 9)l/,p. By assumption every p' -element in 9Jl centralizes !t>. Thus by (21.8) 9)l has a normal p-complement and so also N@ (!t» has a normal p-complement. The result now follows from (22.2). (22.8) (Thompson [1]) Suppose that @ contains a maximal subgroup 9)l which is nilpotent of odd order. Then @ is solvable. Proof. Induction on \~; I. Let 1f ;: 1f(ml). If (1) '" 91 ~ ~m andm<j ~ then by induction @/':l1 is solvable. Thus ~ is solvable since ':l1 is nilpotent. Assume that no nonidentity
122
C H A RAe T E RS 0 F FIN I T E G R0 UP S
subgroup of Wl is normal in ();. Let p E: 7r and let 'l3 be a Sp-subgroup of IDL Since ID1 is maximal N61 (13) = IDl. Thus 13 is a Sp-subgroup of ();. The maximality of 'lJl also implies that 9)1 = N(~ (T('U)) and 9)1 = C61 (Z('l3)). Thus by (22.2) ® has a normal p-complement. Applying this to every prime in 7r we see that ® Has a normal 7r-complement ~. Let q be a prime divisor of I S) \. The number of Sq - subgroups of .p is a divisor of S). Hence by the Schur-Zassenhaus theorem there exists a Sq-subgroup C of S) with IDl ~ N® (C). Thus ~mc is a group and so ~mc = ~) by the maximality of 9)1. Therefore S) C is solvable. Since ~)1 is solvable this implies that (\j is solvable as was to be shown. It should be mentioned that (22.2), (22.7), and (22.8) are all false for p = 2. The group PGL2 (7) provides a counter example to all three statements.
CHAPTER IV §23. T. I. SETS A subset ~ of @ is a trivial intersection set in @ or a T. I. set in @ if ~( ~ N @ (~) and ~ G (1 ~ b (1) for G E: 6) N6) (~). If ~ is aT. I. set in @ we will also say that !{ has the T. I. property. The importance of this property anq of the following result was first realized by Brauer and independently by Suzuki [1]. The remainder of these notes will to a large extent be concerned with deriving consequences of (23.1) and its generalizations. (23.1) Let ~( be a T.1. set in 6) and let 91 = N6)(~). Let a, (3 be complex valued class functions on 91 such that a vanishes on 91 - ~ and (3 vanishes on {U ~ ~ G n ~l }---=-! . Then a *(A) = a (A) and (3*(A) = (3(A) for A E: !( #. If furtli""ermore 0' (1) = 0 then -
Proof. The first statement follows directly from the definition of a * and (3*. The number of distinct conjugates of !{ in ® is I 6) : 911. Thus the T.1. property and the first statement imply that (a*,
(3*)~
=
I~I ~
a*(G){3*(G)
1
= 191 I ~
0
(A) f3(A)
123
=
(a, (3) 91
124
CHARACTERS OF FIN I T E GROUPS
It should be observed that if ~ is a S-subgroup of N@ (t) in (23.1) then (3 automatically satisfies the required assumptions. As a first illustration of how (23.1) is used we give here an alternative proof of Frobenius' theorem (20.2).
(23.2) Suppose that (! = N<»«(!) and lj is a T.!. set in Then ~ has a normal complement in @.
~.
Proof. Let {~i} be the set of nonprincipal irreducible characters of <:. Let ai = ~i (1)1<: - ~i' By (23.1) Ila{112 = ~i (1)2 + 1. By the Frobenius reciprocity theorem (a 1(~) ;::; ~i (1). Since (1) ;::; 0 this implies that = \i(1)I~ - Xi for some irreducible character Xi of @. If G E: ~ and G is not conjugate to an element of <: then Xi (G) = \i (1) Xi (1) and so G E: ID1i by (6.7) where ID1i is the kernel of Xi' Let e = 1@ + !;i Xi (I)Xi and let ID1 ;::; n i "Ul i . Then e(G) ;::; I ~ I = 0(1) if G E: ID1, while e(G) 0 if G ¢ IDt Thus IDl <1 <» and 0 P @/IDl. Furthermore ID1 n
at
at
t,
The following elementary result also due to Brauer and Suzuki is very useful when used in conjunction with (23.1). Let n ~ 2. Let {Oij 11 ~ i, j ~ n} be a collec(23.3) tion of generalized characters of ~ such that 0ij (1) = 0 for 1 ~ i, j ~ nand
1
~
i, j, s, t
~
n
Then there exists a sign € = ±1 and irreducible characters Xh' •. , Xn of @ such that eij = dXi - Xj) for 1 ~ i, j ~ n. If n > 2 then € is uniquely determined.
CHARACTERS OF RELATIVELY SMALL DEGREE 125
Proof. Since II e ij 112 : : 2 - 26 ij and e ij (1) = 0 it follows that eii 0 and e ij is the difference of two irreducible characters if i j. Since (e ij , eji) = 26 ij - 2 it follows that eij = -e ji • If n == 2 this implies the result.
'*
Suppose that n > 2. Since (e 12' e13) = 1 we see that e 12 = E(Xl - X2) 'and e 13 = dXl X3)' where Xu X2' X3 are pairwise distinct irreducible characters and E ±1 is uniquely determined. Let j > 2. If (Xl' eIj) :::: 0 then (Xu e1j) : : : -E(e 12 , e 1 j) = -E. Similarly (X3' e1j) :::: -E. Thus e1j = -dX2 + X3) contrary to e 1 j (1) ::::: O. Thus e 1 j : : : E1 (Xl - Xj) for some irreducible character Xj Xl and El : : : ±1. Since (e l21 e1 ") == 1 we get thatE I : : : E. Consequently elj dXI - Xj~ and Xl Xj for 2 :5 j :5 n. Since (e 1 j' e lt ) 1 - 0jt it follows that Xj =I Xt for j t. If 1 < j t :5 n then II (elj - e1t ) 112 ;: 2 and (ejt, e1te1j) = 2. Therefore ejt = elt - elj : ;: E(Xj - Xt) as required.
'*
'*
'*
'*
§24. CHARACTERS OF RELATIVELY SMALL DEGREE In this section it will be shown that if ® has a faithful character whose degree is small relative to some prime p in 1T(®) then the Sp -subgroup of ® is normal in ~. Groups satisfying the following assumptions will be studied. The next two results are from Feit [5]. They are implicit in Feit-Thompson [1]. (24.1) (i) 5) is an abelian subgroup of @, Z(@), 191: 5).81 ::: t and
91 : :
N@(4),
.8 =
91 0
}
U
C®(H)} - .B
tHE:£)1f
(ii) ,\; and 9l o are T.1. sets in @ and N@(91o ) (iii) 91 =I @.
91.
126
C H A RAe T ERS OF FIN I T E , GROUPS
(24.2) Assume that (24.1) is satisfied. Let I be an irirreducible character of @ which does not have £1 in its. kernel. Then
Proof: Let 119? = (l + f3, where f3 has S) in its kernel and no irreducible constituent of a has S) in its kernel. Let b = f3(1). Then b = (X 1,\)' l~J),\)' By (6.7) I f3(G) I = b for G E: ,\J.8. Hence (6.8) and (24.1) imply that
1=
I ~ I :::
II X 112 >
~ I x(G) 12 o
?
_111 {-L;IX(G)12 + 91.8
+
I~I ~
~) .8
1£11
By assumption a
la(G) + f3(G) 12}
1f3(G) 12 b2
x(1)2 = - t
L;
91 oU,8
+
'*
II a 1191 +
T
0 and so II a 1191
?
1. Thus
2
X(1)2 b -,-
tl.\)l./
t
which implies the required result. (24.3) Suppose that (24.1) is satisfied. Assume further that .\) is not contained in any proper normal subgroup of ®. Let X be a nonlinear irreducible character of @. Then X(l)+ 1 > 1 S) 1112
CHARACTERS OF RELATIVELY SMALL DEGREE 127
Proof. Let Xo = 1@, Xl' ... ~e all the irreducible characters of @. Define ai = (Xi' XX), b i = (Xii f), 1~) for all i. The Frobenius reciprocity theorem implies that b i = (Xi' 1~ ). Thus by the Frobenius reciprocity theorem
Since.\) is abelian X(1) ~ (24.2) implies that
!1 XI f) II~.
As a o
= bo = 1 = X(1)2 - 1
1~ 11/2 Since X(1)
'*
1 this yields the required result.
(24.4) Suppose that for some prime p the Sp-group 'V of @ is abelian and a T.1. set in @. Assume that @ has a faithful character X with X(1) ~ p1/2 - 1. Then 'V
'*
'*
'*
91 0 n 91 ~. Let 'VI' 'V 2 be Sp - subgroups of C'V (H), C'V G(H) respectively. By definition 'Vi (1) for i = 1, 2 and C®(H) ~. Thus by induction the Sp-subgroup of C® (H) is normal in C® (H). Hence 'V 2 ~ C® ('VI) S; 91. Thus l\2 ~ 'V and so 'V 2 S; ~ n 'V G = (1). This contradiction establishes the result. H
E:
'*
'*
128
C H A RAe TE RS 0 F FIN I T E G R0 UPS
(24.5) that Z(O) Z (0) I nonlinear
Let q be a prime and let 0 be a q-group such == 0' is cyclic and 0 /0' has exponent q. Then q, \ 0: 0' I == q2n for some integer n and every irreducible character of Q has degree qn.
Proof. It is easily verified that in any group [G,HHG,K]H
==
[G,KHl
Since ,0' == Z(O) this implies that [G, Hi] == [G, H]i for G, H E: C. Since ~ E: C' == Z(C) we see that [G, H]q :::: [G, ~] == 1 for G, H E: C. Thus I Z(C) I == q as Z(C) is cyclic. Let to: 0' 1 == qm and let {P-i} be all the linear characters of C. Then for Q E: C, Ei I J.Li (Q) 12 = qm. If Q E: .c -C' then IC: Cc (Q) \ ::::: q and so by the orthogonality relations ICc (Q) \ == qm = Ei I Ili (Q) \2. Thus if X is a nonlinear irreducible character of C then X(Q) O. Thus
Therefore X(1)2
==
qm completing the proof.
(24.6) (Ito [2}) Suppose that C is solvable. Let i3 be a Sp-subgroup of ~ for some prime p. Assume that ~ has a faithful character X with x(l) < p 1. Then ~ is abelian and i3
Proof. Induction on I ~ I. Since every irreducible constituent of X1i3 is linear i3 is abelian. Assume that X is a faithful character of minimum degree. Let Xl be a nonlinear irreducible constituent of X with kernel J!. If .\/ *- (1) then i3.11
CHARACTERS OF RELATIVELY SMALL DEGREE 129
Suppose that 1T(@) = {p, qll ... , qn} with n > 1. By (18.10) there exist Sp,qi -subgroups @i of @ with ~ ~ @i for i 1, •.. , n. By induction @i ~ N@ (~) for each i. Hence 1 @I I \ N@ (~)! and so @ N@('V) as required. Thus it may be assumed that 1T(@) {p, q}. If ~ is contained in a proper normal subgroup of @ the result follows by induction. Hence I @: @' I is a power of p. Let @o @ with 1 @: @o 1 p and let 'V o be a Sp-subgroup of ~o' Thus '13 0
e
If 00 is a characteristic subgroup of 0. with .0 :;..0 0 then Co
(24.7) Suppose that @ has a normal p-complement 9R for some prime p. Assume that @ has a faithful irreducible character X with x(l) < p - 1. Then @ = ~ x IDl, where 'l3 is a Sp-subgraup of @. --
130
C H A RAe T ERS 0 F FIN I TE G R0 UPS
Proof. Let q E: 1T(IDl). By the Sylow theorems there exists a Sq-subgroup C such that ~ ~ N(C). Hence ~C is a solvable group and by (24.6) C ~ N@ (~). Thus! IDlIIIN@ (~>I and so ~ . By (20.12) this implies that ~ (1 Z(@) = (1). Assume that l' :;; ~ n ~G '" (1) for some G E: @. Since {~, ~ G} <; C@('l:') '4- @ we get that ~ == ~G by induction. Thus ~ is a T.1. set in @ and the result follows from (24.4). The next result is due to Isaacs and Passman.
'*
CHARACTERS OF RELATIVELY SMALL DEGREE 131
(24.9) Suppose that X(1) :::: P - 1 for every irreducible character X of @ where p is a prime. Let 'l3 be a Sp-subgroup of @. Then ~@. --Proof. Induction on I @ I. Since XI'l3 is a sum of linear characters of 'l3 for every irreducible character X of @ it follows that 'l3 is abelian. If l' = 'l3 'l3 G '" (1) for some G ) G E: @then ('l3, 'l3 ~ C@Cr').1f C@(!') = (\j then induction applied to (\j IT' yields the result. If C (\j (1'. *- (\j then '1)
n
n
Let S be the set of all nonprincipal irreducible characters X of (\j with X(J = X. By the previous paragraph each X in S is faithful. Since (;,;# contains p' -elements S is nonempty by (12.3). If XE: S let XI'l3 = :E~i' Since X(G) E: (J for G E: 'l3 it follows that the Ai are all algebraically conjugate. Thus the kernel of XI'l3 is equal to the kernel of each ~i and so each Ai is faithful. Hence 'l3 is cyclic. Let 'l3 = (P) then pP = 1 since X(P) E: (J and X(1) :::: P - 1. Thus I 'l.~ I = p, X(1) = p - 1 and XI 'l.~ = :Ef~} Ai for some irreducible character A of 'l3. Suppose that C(\j ('l.~) *- 'l.~. Let A be a p' -element in C(\j (P)# and letW = (P) x (A). Then XIQI' = 2;~-1 Ai IJ. 1=1 1 n
132
C H A RAe T E RS 0 F FIN I TE
G R0 UPS
where each Ili is a linear character of (A). Reading subscripts mod p we get that p-l
z:;
, Al (P) Ili (G) :: X(PG)
X(PG)cr-l
i=1 p-1 ,
::: L:
;\1(P)ilit(G) i=1 p I ' Therefore ~i:l {Iii (G) - ilit (G)}A 1 (P) ::: O. Hence ili (G) ::: ill (G) for all i and G E: (A). Hence ili ::: IJ. for all i. Consequently X(A) ::: X{l) IJ.{A) contrary to the fact that Z{(\;) ::: (1). Therefore C(~j('13) :::: '13. Hence every element of b\ is either a p' -element or is conjugate to a power of P. Let k be the number of conjugate classes of (\j which contain elements in '13#. By (12.3) this implies that there is at most one orbit J with I :J I > 1 under the action of (cr) on the irreducible characters of (\\ and I J I ::: k. Let each character in J have degree d. Then k I (p - 1) since I (cr) I ::: p - 1 and I N I ::: 1 + kd2 + I s I (p - 1). Reading mod k yields that 1 =: 0 (mod k). Thus k = 1. Hence :J is empty and I (\1 I = 1 + I s I (p - 1). Since x{1)11 (~; I for X E: Sand x(1) =p 1 this implies that 0 =: 1 (mod p - 1). Thus p 2. Hence by assumption every irreducible character of (\I is linear and the result follows. The next result is due to Blichfeldt and is of a slightly different type. (24.10)
Let X be a faithful character of ('II, Let 1f = 1}. Then ('1\ contains an abelian S1f-sub-
{P I X(1) < p group,
Proof. Induction on INt. Any 1f -subgroup .£1 of (;,\ is abelian since XI ~ is a sum of linear characters. It remains to show that (;,j has an S1f-subgroup, Suppose first that there exists (~\o
FROBENIUS GROUPS
133
16\: ~ I ::: p and p E: 1f. By induction i> contains an abelian S1f-subgroup ~. Let '13 0 be a Sp-subgroup of ~ and let '.1J be a Sp-subgroup of ('I with '.1Jo £ '13. Thus (~, '13) £ CN ('13 0 ), Hence by induction it may be assumed that C{'Ij('.l3o ) ::: (i,; and so by (20.13) ('II has a normal p-complemente The result now follows from (24.7). Therefore ('.I/N' is a 1f' -group: Hence by (20.13) Z«(i,l) is a 1f' -group. It may be assumed that X is a faithful character of minimum degree. Let Xl be an irreducible constituent of X and let .~ be the kernel of Xl' The minimality of X(1) implies that Xl 1N and so .R N. If .\f (1) then by induction .R £ ~ where ~/.fl' is a S1f-subgroup of ('II/Sf. Since N/('I;' is a 1f' -group ~l N. Thus ~ contains a S1f-subgroup as required. Hence .R ;: :; (1) and the minimality of X(l) implies that X = Xl is irreducible. Let ~ be a 1f-subgroup of maximum order. If p E: 1f and p II ~ I let P be a p-element in >a # and let '13 be a Sp -subgroup of (\j with P E: '13. Thus ('.1J, 58) £ C(,,(p) (,I. By induction C(i,,(P) contains a S1f-subgroup. Thus the maximality of 1581 implies that ~ is a S-subgroup of (~. Therefore it suffices to show that if q E: 1f then q II ~ ,. Let q E: 1f and let p E: 1f such that p 11581. Let P, Q be elements of order p, q respectively. Since X(Q) < q - 1 and X(Q) is a sum of X(1) q-th roots of unity not all 1 it follows that ~q is the smallest cyclotomic field containing X(Q). Similarly ~ p is the smallest cyclotomic field containing X(P). Hence by (6.13) there exists G E: (\\ of order pq. Let G PI Q1 Q I P l where P l is a p-element and Q l is a q-element. Replacing G by a conjugate it may be assumed that P l E: ~. Thus (Qu 58) £ CN(P I ) N. By induction C('Ij(P l ) contains a S1f-subgroup. Thus the maximality of I >a I implies that q 11>a ,. Since q was an arbitrary prime in 1f the proof is complete.
*'
*'
*'
*'
*'
*'
§25. FROBENIUS GROUPS A group N is a Frobenius group with Frobenius kernel ~if
(i) i> <j
(\1,
(1)
*'
.p
*'
(i,I.
134
C H A RAe T ERS 0 F FIN I T E G R0 U PS
(ii) c~ (H) f: S) for H E: S) #. Frobenius groups arise naturally in many contexts. This section is concerned with investigating some of their properties.
(25.1) Let (}; be a Frobenius group with Frobenius kernel £). Then Sj is a S-subgroup of N. Proof. Let 'l3 0 be a Sp-subgroup of S:1 for some prime p and let 'l3 be a Sp-subgroup of N with 'l30 S;;; 'l3. Assume that 'l3 0 ~ (1). Then Z('l3) S C(" ('l3 0 ) f: S)(J 'l3. Hence Z('l3) f: 'l3 0 • Thus 'l3 f: C('j(Z('l3» f: S. Therefore 'l3 = 'l3 0 and the result follows. (25.2)
The following statements are equivalent.
(i) N is a Frobenius group with Frobenius kernel £"f,
I ~ I = hand I N I =: he. (ii) I N I = he with (h, e) ::: 1. If G E: N then either h G = 1 or Ge 1. (> =: {G I ah ::: 1}. (iii) lris a in N with N(" «%) = (% ~ Nand I <.! I = e.
Proof. (i) (ii). This is immediate by (25.1). (ii) (iii). By the Schur-Zassenhaus theorem there exists a complement C! of ~ in (". Suppose that Ge ::: 1 for some G E: N. Let {E} = (! (\ G.\'>. Then (E).\'> = (G).\'> and by the Schur-Zassenhaus theorem G is conjugate to an element of (E) f: (%. Hence every element G with G 4= 1 and Ge ::: 1 is conjugate to an element of (%. The number of such elements is 16~ I - \.\'> \ (e 1)h. Thus lUG (,!#G \ ::: (e - l)h. The number of distinct conjugates of (,! # is 1N: Nbj «,!) 1 :::; h and each conjugate of (t# contains e 1 elements. Thus I (~j: NN (~) I ::: h and no element is counted twice. This proves (iii). (iii) :9 (i). By (23.2) l! has a normal complement.\) in (~'. Let I.\'> I =: h ~ 1. The assumptions imply that lUG <.! #G I (e - l)h. Hence every element in (,\ is conjugate to an element of ~ or of (i#. Let H E: S)# and assume that
FROBENIUS GROUPS C(~) (H)
1;
135
~. Then replacing H by a conjugate it may be
n
assumed that there exists E C ~# C(\i (H). Hence H c ~ (,!H and so Ci :=: (,!H contrary to assumption. Hence C(~ (H) .£ ~ as required. Let (~ be a Frobenius group with Frobenius kernel.\). Let (,! be defined as in (25.2). Then ('1\ ::: ~(,! and.\) n (! = (1). Furthermore (,! acts as a group of automorphisms of ~ under conjugation such that no element of (,! # fixes any element of ~#. Conversely if (! is a nonidentity group of automorphisms of ~ such that no element of (,!# fixes any element of ~# then the semidirect product \!.\) is a Frobenius group with Frobenius kernel.\). The group (l is sometimes called a Frobenius complement.
n
(25.3) Let N be a Frobenius group with Frobenius kernel ~. Let 9[ be a subgroup of (\l. Then one of the followlowing must occur: (i) ~{ .£ ~. (ii) ~ 5) = (1). (iii) ~ is a Frobenius roup with Frobenius kernel ~ (>. If furthermore ~( < ~ then either ~ c?( or ~ c ~ and ('l)j~ is a Frobenius group with Frobentus kernel ~/~L-
n
n
Proof. By definition C~ (A) ~ ft (1 .\) for A c (~n ~)#. The first statement follows. Suppose that ft n ~ it follows that either 5)/~0 ~ ::: (1) or W/~ n ~ ::: (1)0 Thus either ~ ~ ~ or ~ .£ .\), If ~ ~ .\) then ~/11
136
C H A RAe T ERS 0 F FIN I T E G R0 UPS
Proof. The last statement is an immediate consequence of the first two. To prove the first two statements it suffices by (9.11) to show that if ~ is a nonprincipal character of (> and E E: I,j# then ~E ~. Let -~t be a conjugate class of (>, J! =I- {1} and let K E: -~!. Then UI,j Jt E is the conjugate class of 6\ containing K. Since I®: C6;(K)1 = !I,jll(>; C(>(K)\ it follows that IUI,j_\iEI = \I,j 1 1-\I I. Thus -~! E .Il for E E: I,j #. Let k + 1 be the number of conjugate classes of (>. The group I,j permutes these conjugate classes and there are exactly 1 + k/I and so by (12.0 there are exactly 1 + k/I I.! I orbits under the action of (t. Thus ~E ~ for E E:
*
*
*
(25.5) Let (.!"* (1) be a group. There exists a Frobenius group ® with Frobenius kernel (> and 6;/(> isomorphic to . By the Sylow theorems 1('1\: (>1 I IN(~ ($)1. Hence by (25.3) N6; (~) contains a complement of .\) in 61 which is necessarily isomorphic to ~. Thus it may be assumed that ~ f: N(~\ (~). Hence (!~ is a group. Let ~l be a minimal normal subgroup of (!~. By (25.3) I.!" is a Frobenius group. Furthermore ~ has exponent p. Since ~ is minimal the action of I.! on ,. gives rise to a faithful ~ -irreducible if -representation IT of (l where if is the field of p elements. Since ,. = 1 then by (4.4) there exists
FROBENIUS GROUPS
137
a representation 3 of (i with underlying vector space 'll over a finite field of characteristic p such that the restriction of ~ to any subgroup l!o : /:- (1) of (! does not contain the trivial representation of l!o as a constituent. Let H :::: {Hv I v E: 'll}, where Hv Hw :; ;: HV.f.w' Let N be the semidirect product of l! and .p where Hv E ::: ERv~ (E) . It is easily seen that (\; is a Frobenius group with Frobenius kernel .p as required. (25.6) (Burnside) Let (» be a Frobenius group with Frobenius kernel .p. Let l! be a complement of .p in 0; and let q, r E: 1T ( l!) (q, r not necessarily distinct). Then every subgroup of 6) of order qr is cyclic. Proof. It may be assumed that I (j I = qr. If q = r then is abelian. By (25.5) l! has a faithful irreducible character. Thus (.! is cyclic by (1.8). If q : /:- rand l! is not cyclic then l! is a Frobenius group. This contradicts (25.4) and (25.5), Thus (! is cyclic in all cases. (j
(25.7) Let (~l be a Frobenius group with Frobenius kernel .p. Let IZ be a complement of .p in (~l. Then a Sylow group of l! is either cyclic or a quaternion group. Every abelian subgroup of (.! is cyclic. Proof. By (25.6) a Sp-subgroup of (.! has a unique subgroup of order p for p E: 1T ( i.f). This is known to imply the first statement. The second statement is an immediate consequence of the first. By using (25.5) it is possible to give a complete classification of groups l! which can occur as the complement of a Frobenius kernel .p in a Frobenius group (\l. The conditions in (25.6) are in general not sufficient. However by using (20.16) it is easily shown that they are necessary and sufficient in case all Sylow subgroups of (.! are cyclic. Much less is known about which groups .p can be the Frobenius kernel of a Frobenius group. Before looking at this question we prove another consequence of (25.5).
138
CHARACTERS OF FINITE GROUPS
(25.8)
The Frobenius kernel of a Frobenius group is
unique. Proof. Let .i) and .i)l be Frobenius kernels of the Frobenius group (~L It may be assumed that .i)l 4: .i). Thus there exists p E: 1f(.i)1)' p Ef 1f(.i». Let 13 be a Sp-subgroup of .i)l' By (25.3) .i) ~ .i)l' By the Sylow theorems Iill: .i)l IIIN(~ ('V) I. Thus by (25.3) there exists a complement lj of .i)l with G ~ N(,,; (~). Hence ~ 'V is a Frobenius group with Frobenius kernel 'V. Thus if X is an irreducible character of (% 'V then by (25.4) either 1~ :; XI' or 1 ~ !; Xlj' This contradicts (25.5) since ljlJ.i) is a Frobenius group with Frobenius kernel .i). (Burnside) Let N be a Frobenius group with Frobenius kernel .i). If I N: ~ I is even then .i) is abelian. (25.9)
Ht : :
Proof. Let J be an involution in ()j. Suppose that Hil Hil for HI, H2 E: .i). Then H2 Hil :::: (H 2 Hil )J. Therefore HI = H2 . Thus there are I.i) 1 distinct elements of the form H-I HJ in .i). Consequently if G E: .i) then G H-l HJ for some H E: .i). Thus GJ ::: H- J H :::: G-I. Therefore if Gu G2 E: ~ then
Hi
-I)J -_ G -J G -J_ GI G2 -_ (G-1 2 GI 2 I - G2 GI as required. It is possible to give an elementary proof of (23.2) and thus of (25.2) in case Ilj I is even by elaborating on the above argument. See Burnside p. 172. The following result is deeper.
(25.10) (Thompson) The Frobenius kernel of group is nilpotent.
Frobenius
Proof. Let (~; be a Frobenius group with Frobenius kernel ~. Induction on \.i) I. If Z(~) "* (1) then Z(.i» <1 6J. Hence by (25.3) and induction ,p/Z(.i» is nilpotent and so .i) is nilpotent. Assume that Z(.i» (1). Let p € 1f(.i» such
AN
E X CUR S ION
I N TON U M B E R THE 0 RY
139
*
that .t> does not have a normal p-complement. Choose p 2 if possible. Let ~ be a Sp-subgroup of (\I;. If p :::: 2 then ~ <1 .t> and so T(~) <1 .t>. Suppose that p 2. By the Sylow theorems I(\1;: ~. liNN (~)!. Hence there exists a complement (! of .t> in (\\ with CZ ~ N~ (~). CS = C.t>(Z(~» .t> since Z(~) :::: (I). By (25.3) CZCS is a Frobenius group and by induction (£ has a normal p-complement. By (22.2)' N'.t>(T(~» does not have a normal p-complement. Hence T( 13) by induction. Thus in any case T(~) <1 ®. Let ~ be the subgroup of Z(T( ~» generated by elements of order p. Then ~ <1 ("'. By (25.3) and induction .t>lft is nilpotent. Let 5B be the inverse image in .t> of the normal p-complement in .t>/~l. Thus ft ~ ~ and
*
*
*
§ 26. AN EXCURSION INTO NUMBER THEORY
Let 5= be a finite field of characteristic p and let = q. For any integer n > 0 define
I 5= I
I
(~n = «an, b) a E:: 5=*, b E:: 5=, (an, b)(a 1 n, b I
»
(26.1 )
It is easily seen that l~n :::: (!n ~n' (!n n.t>n :::: ({1, 0» (I), I.t>n I : : q and I C!n I : : (q - U/(q - 1, n). Furthermore if I IS- n I :I 1 then (\;n is a Frobenius group with Frobenius kernel ~n' Clearly ®n = C;)m,q-ll' .t>n .t> (fi,q-l) and CZ n :::: (! (ll,q-u' A simple computation shows that (\;n is isomorphic to a group of permutations on the
C H A RAe TE RS 0 F FIN I T E G R0 UPS
140
elements of W where x(a n, b) = an x + b for all x E: W. Questions concerning €F can often be translated into questions conc.erning groups related to the groups ®n for various n. We will illustrate this by proving a result closely related to a theorem of Hua-Vandiver [1] and Wei! [1]. The proof is in the spirit of the proof of Feit-Thompson [21 Lemma 38.9. (26.2) Let W be a finite field with I U' I = q, Let n 1 , . ' " ns be integers and let di = (I ni I, q - mar i = 1, ... , s. Let Cu " . , Cs E: W with nT=1ci if:: 0 and let k be the number of solutions in W of ~r=1 Ci xri :::: 0 with nT=1xi *- O. Then Ik-(q-1)s/ql
1
l::::
1
d 1 ••• d s m where m is the number of solutions of s d· for some Xi E: w, * ~i=1ciYi = 0 and Yi = xiI i=1, ... ,s. For i = 1, .. '..1 S let ®i = ®n.,.\li .\In. and ~i 1 1 ~n.' where ®~, .t'n. and ~n. are defined by (26.1). Let 6) ; @ x .. · x @ 14) = .t' x 4)S and (! = 1 S' 1 (!l X ••• x l!s' Define the subgroup ~ of 4) by s ft = {«1, b l ), " ' , (1, bsnl 'E b. O} i=1 1 Let If = «1, c 1 ), ••• , (1, c s » E: .\l. Let .\1 be the conjugate class of @ with H E: .\1. Since.p C® (H) it follows that l·Q I = I (! I and .\1 = {«1, Cly 1 ), . . . , (1, csYsnlYi x~i,
i ...
Xi
E:W*, i
= 1, ... , s}
Therefore m = I .\1 n " I. Let lW denote the character of .p induced by 1". Then for K E: ~, 1! (K) = q or 0 according to whether K E: ft or K tf: W. Thus
AN
EXCURSION
INTO
NUMBER THEORY
mq
141
kq
Let 1~ = I; aj Xj' where Xj ranges over the irreducible characters of @. @/,p has I (! I : : (q - l)s / d1 ••• d s linear characters and for each of these aj Xj (H) 1. Thus if S is the set of all j' such that Xj does not have ,p in its kernel and aj "* 0 then (26.3)
d
kq 1 • 0:.
d
S
_ -
~
L.J
a1-Xj (H)
:::: (q -Os + :6 a- X- (H) d 1 • •• d s jf S J J
Suppose that j E: S. By (6.3) Xj :::: n~=1 Xji where Xji is an irreducible character of @i' Suppose that ,pi is in the kernel of Xji for some i. Then Xj is a character of @/.pi' Since ,pi n ~ : : (I) it follows that ,p :::: ,pi". As (Xj!'" 1,.) :::: aj "* 0 we get that 1,p 5; Xj!,p' Hence by (9.10) ,p is in the kernel of Xj contrary to the fact j E: S. If (!i = (I) then Xji is linear. Thus by (25.4) Xji is monomial in any case. Hence by (8.2) Xj is monomial and so Xj for some linear character ~-j of ,p. By (9.11)
Aj
this implies that I;j£ S Xj l,p is a sum of pairwise distinct linear characters. Thus for any linear character A of ,p ( ;\,
~
Xj l,p),p
:5 (;\,
p,p),p :::: 1
JfS
and so
(1~{, _:6 li: S
Consequently
Xj!,p),p
:5
q
142
CHARACTERS OF FINITE GROUPS
Hence (26.3) implies that if I Xj (H) I with j E: S then
Ik
- (q
~
1)s
I :: d
1
I X(H) I ...
is maximal among
d s \ X(H)
I
The result now follows from the fact that
I X(H) \2 < 6 I Xj (H) 12 = I C@(H) I = l.p I = qS §27. CN GROUPS A group @ is a eN group if C@ (G) is nilpotent for every G E: @#. Suzuki [2] first showed that if @ is a group of odd order such that C@(G) is abelian for every G E: @# then @ is solvable. The purpose of this section and the next is to prove the more general result that eN groups of odd order are solvable. The proof given here follows the original one in Feit, M. Hall, and Thompson [1] except insofar as (22.2) can be used to simplify parts of the argument. In the process of proving this result it is first necessary to give a description of solvable eN groups. In proving that groups of odd order are solvable the first part of the proof is subsumed in a much more general argument. The second part of the proof is however still an essential ingredient of the general proof. (27.1) Let @ be a eN group and let p, q E: 1T(@), p:# q. Let_ 'l3 be a Sp-subgroup of @ and let Q be a Sq-subgroup of ~. If PQ = QP for some P E: 'l3# and Q E: Q# then 'l3Q = 'l3 x O. Proof. Since (P, Z(Q) ~ C®(Q) it follows that (P, Q) ~ C@(Z(Q)). Thus P E: C@(,o) and so (Z('l3), Q) ~ C@(P). Hence ('l3, Q) ~ C@ (Z('l3)) which implies the result. (27.2) Let ~ be a nonnilpotent eN group which contains a normal Sylow subgroup p ~ ( 1). Then @ is a Frobenius group with Frobenius kernel .p and @/,p is nilpotent.
eN
GROUPS
143
Proof. Let ~ be a maximal normal nilpotent S-subgroup of @. Thus (1) "* ~"* @. If @ is not a Frobenius group with Frobenius kernel ~ then there exist q €: 1T (8,)), and r €: 1T(~) 1T({» such that QR = RQ for some elements Q, R of order q, r respectively. Let 0 be a Sq-subgroup of .t'> and ffi a Sr-subgroup of ® with Q €: 0 and R €: ffi. By (27.1) ill ~ C~(O). Thus ffi.t'> is a S-subgroup of @ and ffi.t'> f: CC@ (0)
n
(27.3) If @ is a CN group, ~ then @/.t'> is a CN group.
@ and ~ is solvable
Proof. Induction on 1@ I. Let G.\) "* ~. Suppose that C@/ ~ (GS)) is not nilpotent, Let <S be the inverse image of C@/.n (G~) in 0;. By induction it may be assumed that ~ = -<S. By taking a suitable power of G it may be assumed that G~ has prime order q. Suppose first that 8,) is an elementary abelian p-group for some prime p. If q "* p then it may be assumed that G has order q. Since G~ 5; Z(@/~) it follows that for K €: @, G-1K.pG=K.p. Since 1 K.\) 1 is a power of p it follows that G-l KHG KH for some H €: ~. Hence ~ ;;;: C@(G).p and so ®/~ is nilpotent as C@ (G) is nilpotent.
144
CHARACTERS OF FINITE GROUPS
p let ~ ::: (G, ~) so that ~ is a p-group. Let L be an element of prime order distinct from p in @. Since L -1 G ~L G~ and I G~ I is a power of p it follows that L centralizes an element of G~ and thus of ~#. By (27.1) L E: C@ (~). Thus by (27.1) @ s;: C@(~()'.l3, where '.l3 is a Sp-subgroup of @. Let ~1 = ~ n Z(13). Then "1 :/: (I) since ~
If q
(27.4) Let @ be a solvable CN group. Assume that @ is not and @ is not a Frobemus group. Then there exists a normal series (I) C ~ C \' C @ where ~ and @/I!. are p-groups for some prime p, \'/~ is a cyclic {2, p}' -group and @/ \' is either cyclic or a quaternion group. Furthermore \' and @/ ~ are Frobemus groups with Frobenius kerne~~ and \.' I ~ respectively. Proof. Let ~ be a maximal normal nilpotent subgroup of @. Let p E: 1T{~), let '.l3 be a Sp-subgroup of .~ and let ~ ;:: 'P x ~o. Since ;3C@ (;3) is nilpotent the maximality of ~ implies that ~ : :.; 'PC@ (;3). Thus ~o is a S-subgroup of @ by (27.1). Hence ~o::: (I) by (27.2). Consequently C@ (~) C ~ 'P, By (18.10) @ contains a Sp' -subgroup !( . Hence by (27.2) ~t == t' is a Frobenius group with Frobenius kernel ~ and 1( is nilpotent. By (27.3) @/ ~ is a CN group and by (25.7) every Sylow subgroup of ~ is either cyclic or a quaternion group. Since ~ is a maximal normal nilpotent subgroup of ® it follows that ®/~ contains no normal p-subgroup, Thus @/.p is not nilpotent. Hence
145
eN GROUPS
there exists a subgroup ~ of ~ with ~~/~ <1 ®/~o Since W ~ C®(~) and ~ is a S-subgroup of ® it follows that W~/~
'*
(27.5) Let ® be a solvable group in which ever element has prime power order. Then u(®) :s 2. Proof. Clearly ® is a CN group. If ® is nilpotent the result is clear. If ® is a Frobenius group with Frobenius kernel 5) then IU(5)) 1 = 1 by (25.10) and u(®/~) = 1 by (27.2) If ® is neither nilpotent nor a Frobenius group than lu(\,' /.p) I 1 where \' and .p are defined in (27.4) and the result follows from (27.4). A group ® is a minimal simple group if 6J is a noncyclic simple group and every proper subgroup of @ is solvable. (27.6) Let ® be a CN group of odd order which is a minimal simple group. Then every Sylow subgroup of ® is a T.I. set in ®. Proof. Let p E: 1T (®) and suppose that a Sp -subgroup 't3 of ® is not a T.I. set in ®. Choose G E: ® such that (1) 't3 n 't3 G 't3 Let IDl be a maximal subgroup containing N® ('t3 't3 G ). Then IDl is a solvable CN group and a Sp-subgroup of IDl is not a T.1. set in IDl. Thus IDl is not nilpotent. By (27.2) IDl is not a Frobenius group. Let!" be a maximal normal p-subgroup of IDl and let 't3 0 be a Sp-subgroup of IDt It may be assumed that 't3 0 ~ 't3. By (27.4) (1) '!" 't3 0 and 'l)l ~ 't3 0 j( where " is cyclic, ~ 't3 0 (1), '!"~ is a Frobenius group with Frobenius kernel !" and ID1/!" is a Frobenius group with Frobenius kernel !'~/tl. Thus CIDl (Z('t3 0 )) ~ 't3 0 and so C~Dl(Z('l3o)) has a normal p-complement. Since 9)1 does not have a normal p-complement it follows by (22.2) that N9): (T('30)) does not have a normal p-complement. Suppose that
n
n
'*
'*
'* '*
146
CHARACTERS OF FINITE GROLIPS
T(t\o) 1; 'r'. Since iUlj'l) is a Frobenius group this implies that 'I'N:I)l(T('Uo)/t' = X'ilJo/!'and so N~Dl(T(iSo» =: iS o contrary to the fact that N~)l (T(~o» does not have a normal p-complement. Thus T(~o) S 1'. Hence 9)1 S N® (T(iS o The maximality of 9)1 implies that = N® (T(iSo »' Since NiS ('l3Q ) N® (T('l3 o it follows that ~o = ~ is a Sp-subgroup of (~i. As N® (~) N~ (T(ll» we also see that N® (~) = N~m (~) = ~. Thus we have shown that if G €: (lb - ~ then ~ n ~G ~ N® (T(~» IDl and so ~ n ~G £ l' since the Sp-subgroups-of IDl!'!' are T.1. sets in IDl/!". By (20.1) ® has a normal complement of is over l' contrary to the simplicity of ®. The proof is complete. It should be observed that (27.6) is false if the assumption that I ® I is odd is dropped. The proof breaks down because (22.2) is no longer valid if p = 2. It is easily checked that the simple group PSL2 (7) of order 168 is a minimal simple eN group in which the S2 -subgroup is not a T.1. set.
»
s
rol
»'
s
(27.7) Let ® be a eN group of odd order which is a minimal.simple group. Let 'Ul 1 , ••• , IDlk be a complete system of representatives of the conjugate classes of maximal subgroups of ®. Then each IDli is a Frobenius group. Furthermore if i is the Frobenius kernel of IDli then ~i is ~ S-subgroup of ~ which is a T.!. set in (lb, (I ~i I, I ~ j I) = 1 for i :;t! j, n~ I ~. I = 101 I and every element of ® is con1=1
1
jugate to an element of
~
i for some i.
Proof. Let IDl be a maximal subgroup of ®. By (27.6) IDl is a solvable eN group in which every Sylow group is a T .1. set. By (22.8) £lJl is not nilpotent. Hence by (27.4) IDl is a Frobenius group. Let p E: 'IT ($"i ) and let 'l3 be a Sp-subgroup of ~i' Since (ji is nilpotent and IDl is maximal it follows that IDl ;:: N® ('l3). Thus "'i is a S-subgroup of (lb, If p E: 'IT(~j) then choosing a suitable conjugate it may be assumed that is £ ~j and so IDli N~ (is) = IDl j • Hence (I ~i \, I ~j \) =: 1 for i :/: j since p was arbitrary in 'IT«(j). If p €: 'IT«(ji n (jiG) let l' be a
NO N SIMPLI C I TV OF CERTA I N G RO UPS
147
Sp -subgroup of ~ i (1 f)i G. Thus (!',!,G) ~ II and so G ~ N® (~) = 9)li = N® (f)i)' Since p was arbitrary in l1(~i) this implies that ~i is a T.1. set in ®. Let q E: 11(®) and let 0 be a Sq-subgroup of ®. Let IDl be a maximal subgroup of ® with N® (0) ~ IDl. It may be assumed that IDl = IDl i for some i. If 0 sf;: ~i then by (25.7) and (27.2) N® (0) is cyclic contrary to the simplicity of ® by (20.13). Hence 0 ~ f)i and so n~:::11 ~i I = 1® I. If G E: ® then some power Q of G has prime order. Replacing G by a conjugate it may be assumed that Q E: ~r for some i. Thus G E: C® (Q) ~ ~i as required.
§28. NONSIMPLICITY OF CERTAIN GROUPS
OF ODD ORDER Let IDl 1 , • • • , 9)lk be a complete set of representatives of the conjugate classes of maximal subgroups of ®. Groups satisfying the following conditions will be studied in this section. (28.1)
(i) For i ::: 1, ... , k, 9)1 i is a Frobenius group
with Frobenius kernel .t>i' 1~i 1 = hi and 1IDli I = hi e i . (ii) For i = 1, ... , k, .t>i is a S-subgroup of ® which is a T.!. set in ®. k i "* j, 16)1 = n'h .. Every element (iii) (hI" h J·) ::: 1 for 1- l 1 of ® is conjugate to an element of ~i for some i = 1, ... , k. (iv) 1 ® 1 is odd. Observe that if ® satisfies (28.1) then each f)i is nilpotent. If furthermore ® is simple then ® is a minimal simple group. The purpose of this section is to prove (28.2)
If ® satisfies (28.1) then ® is not simple.
Before proving (28.2) we will deduce the following consequence from it. (28.3)
A eN group of odd order is solvable.
148
CHARACTERS OF FINITE GROUPS
Proof. Induction on I @ I . By (27-3) and induction every subgroup and every quotient group of @ is solvable. Thus it may be assumed that @ is simple. Hence @ is a minimal simple group. By (27.7) @ satisfies (28.1). Thus the result follows from (28.2). Throughout the rest of this section @ is a simple group which satisfies (28.1). A contradiction will be derived from the assumed existence of ili proving (28.2). (28.4)
1
1
= IlUl '21
+
k h· - 1 ----=h--
",1 L.J
i=1
i ei
Proof. Since each .pi is a T.1. set in @ it follows that the number of elements in ili which are conjugate to an element of is I® I/e i hi (hi - 1). By assumption every
.pt
element in @# is conjugate to an element of unique i. Hence
.pt for some
k (hi - 1)
6
i=1
h.e.
I@I
1 1
as required. For each i = 1, ... , k let Si be the set of irreducible characters of IDl i which do not have .pi in their kernel. If ~ E: Si then ei I ~ (1) by (25.4). Let {e i dis} be the set of integers which are degrees of characters in Si' Choose the notation so that dh < db < ... . Let Sis = {~ I ~ E: Si' ~(l) = ei dis}· Let wis = I Sis I. By (12.4) wis is even since I @ I is odd. Thus wis ~ 2 for all i, s. Since ~i is nilpotent (25.4) implies that db = 1 and wit = (I.pi: ~i 1- 1)/ei for i = 1, .," k. Choose the notation so that (28,5) Let Sis = {~ist I t = 1, ... , wis }, Let
~isu) and let 0istu = aistu' By (23.1)
_for 0' istu
e·1 = e 1
= (~ist -
NON S IMP LI CIT Y 0 FeE RTAl N G R0 UPS
149
Since wis 2: 2 (23.3) implies the existence of a sign €is = ±1 and a set ::l is = {Xist} of irreducible characters of ~ such that
I ::lis I
(28.6)
= 'wis
If (]
E:
and 8istu ::; fis (Xist - Xisu)'
g~I~I/~ then ::lrs
=
::lis' If i"* j then
Xist I -Pj is rational valued. Proof. Since sfs ::; Sis and xrst -
~st
=
€ is (~rst -
~ist)*
the first result follows. As (~fst - ~ist)* vanishes on ~j we also get that xrst (H) = ~st (H) for H E: ~j which implies the second statement. (28.7) Xist:: Xi's't' if and only if i i', s = s' and t = t'. Furthermore Xist "* 1~ for all i, s, t and {1~} U i ,s ::l is is the set of all irreducible characters of ~. Proof. Clearly (l~, 8istu ) :: 0 and so Xist "* 1~ for all i, s, t. Suppose that Xist = Xi's't'. If i "* i' then Xist is rational valued by (28.6) contrary to (12.4). Thus i = i' and
«~ist - ~ist)' (~is't' - {is't'))IDl.1 «Xist - Xist)' (Xis't' - Xis't'
»G)
=
2
Hence s = s' and t = t'. Since ~s I ::lis I = ~s wis is the number of conjugate classes of IDC i which lie in ~t it fol-
lows that ~i ,s I ::lis I is the number of conjugate classes of
150
C H A RAe T ERS 0 F FIN IT E G R0 UPS
in @#. The last statement now follows from the first two. Let .\)1 = ~, hI = h, e 1 e and Wu ;:: w. Let ~t ~Ut and Xt = XUt for t = 1, •.. , w. Let .6. = E~l Xt . By (15.4) there exist nonpri. ~.Lpal linear characters CfJl' ••• , CfJ w of ~ such that ~s = CPs is the character of ID11 induced by CfJ s ' Let f3 (1S) - CfJl)*' Then (f3, XistXist) ;:: 0 for i 1 since f3 valushes outside U® S)#G and Xist Xist vanishes on U~S)#G. By (23.1) 11f311 a = e + 1 and
(I,;
'*
Therefore there exists an integer a and a real valued character r with (r, 1@) 0 (r, Xt) for t :: 1, ... , w such that (28.8) Let a ;:: {j I j > 1, (Xjsb (3) = 0 for all s, t}. Let
'"'" (h j LJ
jE:
hj e j
1)
e - 1 2ea
Proof. H j E:
'"'" (h. - 1) 1 LJ J -< h·e· - w jE:a J J
'*
Proof. Suppose that j E: a. Let H E: ~j' H (X, (3) 0 for an irreducible character X of @ then the definition of a and (28.6) imply that X(H) is rational and so X(H) = X(H). Thus .6.(H) and r(H) are even. Since f3(H) = 0 (28.8) implies that
NONSIMPLICITY OF CERTAIN
o == f3 (H)
==
1 -
£ 11
GROUPS
151
Xl (H) (mod 2)
'*
Hence Xl (H) O. Thus I Xl (H) I ~ 1 since Xl (H) is rational. As Xl Xs vanishes on flj for all s = 1, ..• , w we see that 1l:!.(H) I = w I Xl (H) I ~ w. Let @Q be the set of all elemen~s in @ which are conjugate to an element of ~j for some J E: a. Then I @o I = I@ I ~jE:a(hj l}/h j ej . Thus
w
=
hj - 1
h·e·
1 1
This implies the result. If we now combine (28.4), (28.9), and (28.10) we get that (28.11 )
1 =
1
I@\ 1
<-
- I@I
2: k
+
i=l
(hi - 1) h. e· 1 1
h-1 he
+--+
1 e-1 +-w 2e 2
Suppose that w > 2 so that w : : :- 4 since w is even. Since e z ~ e 2: 3, (28.11) implies that 1 ::::;
~
1
f(il 1
111
1
+-+-+--e 4 2 2e
3
I@ I + 4"
1 + 2e
1
"S
3
I@ I + 4"
1
+
"6
Therefore 1/12 ~ 1/1 @ I or I @ I :::; 12 which is not the case, Thus w 2. Hence by (28.5) e z > e and so e 2 2: e + 2 since e and e 2 are odd, Now (28.11) becomes
152
C H A RAe T ER5 0 F FIN I TE G R0 UPS
1.
<:
2 -
1
+
I~
1
h-1 e-1 ~- + -7--=-'he 2(e + 2) 1
1
1
3
---+-+----:--I® I he e 2 2(e + 2)
Thus (28.12)
o <: -
_1__ ~ + (-e + 4) I~)
I
he
2e(e + 2)
5 this yields that 0 < 1/1 ~ I - 1/he or I~ I < he which is impossible. If e < 5 then e = 3 since e is odd and e :f. 1. Since w = 2 this implies that IS): .p'1 = we + 1 7. Consequently .p' =:: (1) and l.pl = 7. Now (28.12) becomes
If e
Therefore I ~ I ~ 70 which is not the case. The proof of (28.2) is complete.
§29. PROPERTIES OF INVOLUTIONS The results in this section are due to Brauer and Fowler [1}. While these results are quite elementary they are of fundamental importance for almost everything that is known concerning groups of even order. (29.1) If I, J are involutions then (IJ)1 = (IJ)J (IJ)-l. If furthermore IJ :f. JI then (I, J) is a dihedral group, Proof. (IJ)1
=::
(IJ)J
= JI
This implies the result.
=::
(IJ)-l
PROPERTIES OF INVOLUTIONS
153
(29.2) If I, J are involutions which are not conjugate in @ then there exists an involution K E:: @ such that (I, J) s;; C@ Proof. Let 1(1, J) I :: 2a. If a is odd then any two involutions in (1, J) are conjugate by Sylow's theorem contrary to assumption. If a is even then by (29.1) (I, J) s;; C@(K) where K ;:; (IJ)a/2 and K is an involution. Let 'l3 be a Sa -subgroup of ®. Assume that ~ is (29.3) a T.1. set in @. Then either ~
«
154
CHARACTERS OF FINITE GROUPS
(29.4) If Gi '" 1 then ai is the number of involutions I with GI :::: Gil. If Gi E: 9)1 then ai :::: b i - 1. If Gi :::: 1 then ai :::: m. Proof. If Gil:::: Gi- l for I E: 9)1 let J :::: IGi . Then IJ:::: Gi and ~ :::: GlGi :::: 1. Conversely if IJ = Gi with I, J E: roc then by (29.1) Gil = Gfl. If Gi 2 '" 1 then J '" 1 and the result follows in this case. If Gi :::; 1 the result is clear. If Gi E: 9)1 then J E: ID1 unless I :::: Gi and so ai :::; b i - 1 as required. (29.5) If Gi is not strongly real and Gi 2 '" 1 then ai :::: O. If Gi E: 9)1 then a i 5 Ci - 2. In any case a 5 c i . i Proof. The first statement follows by definition. If Gi E: 9)1 then b i 5 Ci - 1 and so ai 5 Ci - 2 by (29.4). If Gi :::: 1 then ai 5 g :::: c i . If Gi'" 1 then a i 5 IC@(Gi)1 1C@ (Gi)1 5 ci proving the last statement. s (29.6) m 2 5 m + 6 (b. - 1) ~ + (t - S - l)g . 1 1 c·1 1= 5
cg + (b - c) m - c
Proof. The number of ordered pairs of involutions in @ is m 2 • Thus m2 :::: L; ~ g/ ci' The first inequality follows from (29.4) and (29.5). Since m :::: L;:=1 g/c i we get that m2
5
mb + (t - s - 1)g
However the number of strongly real elements is at most g. Thus t-l l + m + £...J ' " c. 5g i=s+1 1
PROPERTIES OF INVOLUTIONS
155
Consequently (t - s - l)g :s; c (g - m - 1) which implies the second inequality. (29.7)
t - 1
2':
m(m + 1) g
Proof. Since b i :s; ci - 1, (29.6) yields that s 2 m :s; m + sg - 2 Jl + (t - S - l)g c· i=l 1
I:
As m !;~:::::1 g/c i this implies that m2 :s; -m + (t - l)g as required. (29.8)
.p
;#
> 2 there exists a subgroup l.p 13 > g.
If g
~ and
Proof. Induction on inequality in (29.6)
I ~ I.
Let n
.p of
1~ II m.
~
with
By the second
g :s; cn2 + (b - c)n ::: cn (n - 1) + bn For some j with 1 :s; j :s; s, b j ::::: b :s; s !;i=l 1/ ci we get that n :s; Cj' Thus g :s; CCj (Cj - 1) + Cj (Cj - 1)
Cj
= (c
1. Since lin :::
+ 1 )Cj (Cj - 1)
If Cj:S; C then g:s; c 3• As c::::: IC~(Gi)1 for some strongly
real element with Gi ;# 1 it follows that C~ (Gi ) ;# ® as required. If C < c j then I @ I < cj and we are done unless Cj g. In that case G :; : Gj is an involution in the center of ~. If g 4 the result is trivial. If g ;# 4 and g/2 is even then by induction there exists .p such that .pI (G) ;#
156
CHARACTERS OF FINITE GROUPS
[Sj/ (G) [3 > I @/(G) 1 3. This implies that I Sj 1 > I @ I as required. If finally g/2 is odd then by (20.14) @ has a normal 2-complement ~ and I Sj I = g/2 > ~ as required.
<»/ (G) and 3
§30, GROUPS WITH QUATERNION S2 -SUBGROUPS In this section the following result of Brauer and Suzuki [1] will be proved. The result is still true without the hypothesis that I ~ I 2' 16; however all known proofs for the case I ~ I = 8 depend on the theory of modular characters. (30.1) Suppose that the S2 -subgroup ~ of ® is a quaternion group and I ~ I 2' 16. Let IDe be a maximal normal subgroup of ® of odd order. _Then ®/IDl has a nor mal subgroup of order 2. Thus in particular ® is not simple. Proof. It may clearly be assumed that <» has no normal subgroups of odd order. Let 'l3 = (A, B IA2 n = 1, B2 = A2 n-\ B-1 AB = A-I) Then I ~ I 2n+1 and n 2: 3. Let U = C® (A2) and 91 = N6\ «A2 ) ). Then (A) is a S2 - subgroup of u since n 2' 3 and the center of a quaternion group has order 2. Thus 1m: u I 2 since ~ S; 91 and 9l/U is an automorphism group of the cyclic 2-group (A2), By (20.14) u has a normal 2-complement Sj. Thus Sj .p
r
r
QUATERNION
5 -SUBGROUPS 2
157
vanishes on m - ?f. It is easily seen that ~ is an irreducible character of m. Thus II 0 1191 :;: 3. By (23.1) and (30.2) 110* II@ = 3. Since 0 * (1) ;:: 0 this imples that a * :;: 1~ + Xl - XII where Xl and X2 are nonprincipal irreducible characters of ®. As 0* vanishes on involutions we get that if J is an involution then (30.3) 1 + Xl (1) - XII (1) = 0 (30.4) (~i contains a unique class of involutions jf since a S2 -subgroup of ~ contains only one involution. For G E: (\) let a(G) be the number of ordered pairs of involutions 11 , 12 with 11 ~ = G. Suppose that G has even order and a(G) '" O. G is not an involution since (\) contains no two distinct involutions which commute. Thus by (29.1) a S2 -subgroup of (11' ~) has at least two distinct involutions where G = 11 12 , This is impossible as a Sl -subgroup of O} has only one involution. Hence if a(G) '" 0 then G has odd order. Therefore a(G) a * (G) :;: 0 for all G E: (~. Consequently (a, a *)N = O. Let 1(\;, Xl' XII' ..• be all the irreducible characters of 6). By (2.15)
_ L!f '"
a(G) -
IN I
"t
Xt (J)2 Xt (1) X(G)
Hence (30.5)
Substituting (30.3) and (30.4) in (30.5) yields that if X ;:: Xl then X(J)1I _ 1 + X(1) -
or equivalently
{1
+ x(J)}2 1 + X(l)
158
C H A RAe T E RS 0 F FIN IT E G R0 UPS
x(l)
+
X(1)2
+
X(J)2
+
x(1)X(J)2
x(l) + 2x(1)x(J) + x(l )X(J)2 Hence {x(1) - x(J)}2 = 0 or xU) = x(J). Thus also X2 (1) Xa (J) and so J is the kernel of Xl and Xa' By (30.3) either Xl or X2 is nonlinear. Hence J is in the kernel of some nonlinear irreducible character of N. Let ~ be the normal subgroup of ru generated by the involutions in (\J. Let t\o be a Sa -subgroup of (~~ with t\o s;;: t\. Thus t\o
COHERENCE
For any set S of characters of a group let s(s) denote the set of all integral linear combinations of characters in S. Let So (s) be the set of all a E: S(S) with a (1) == O. Observe that the character ring of a group has a natural metric whi ch is defined by the inner product. Suppose that 91 is a subgroup of (\'. Let S be a set of characters of ~l such that So (s) O. Assume further that there exists a linear isometry T from So (s) into the character ring of N such that aT (1) = 0 for a E: So (S). The pair (s, T) is coherent if it is possible to extend T to a linear isometry of S(S) into the character ring of (\;. In case T is determined by the context we will also say that
'*
COHERENCE
159
the set S is coherent. In case S is coherent an extension of T to s(s) will also be denoted by T. If S is a coherent set and ~ is an irreducible character of 91 in S then II~TII~~ = 11~llm = 1. Thus ±~T isanirreducible character of6~. In this way it is possible to construct irreducible characters of (~. This method proceeds in two stages. First it is necessary to construct a linear isometry T on So (s) with a T (1) = 0 for a ~ So (S). Then it must be shown that S is coherent. In one case such a T has already been constructed on So (S). Namely if ~ is a T.I. set in N with 91 = N{'I) (~) and if S is the set of characters of 91 which vanish on 91# - ~ then (23.1) asserts that T is an isometry on So (s) where aT = a * for a ~ So (s). Clearly T is linear. In this section it will generally be assumed that T has been defined on So (S) and So (S) oF O. We will be concerned with finding criteria which ensure that S is coherent. Suppose that S = {~i} consists of irreducible characters with ~i (1) = ~(1) for all i and I S I ~ 2. Let 8 ij = (ti - ~j )T. Then (23.3) implies the coherence of S where ~! = EXi' It was precisely this fact which was exploited in S~ction 28. The first result in this section is a generalization of that situation. See Feit [3] and Feit-Thompson [2, Theorem 10.1]. The following situation will be studied. (31.1) (0 91 is a subgroup of N. S is a set of irreducible characters of 91. There exists a linear isometry T from So (s) into the character ring of 6; with aT (1) = 0 for a ~ 9.0 (S). (ii) S = Uf=l Si and Si = {~is I s = 1, ... , ~}. For each i either Si is coherent or all the characters in Si have the same degree. (iii) There exist integers lis such that ~is (1) = lis~ll (1) for 1 ~ i ~ k, 1 ~ s ~ llt and In1lis' (iv) nl ~ 2. For any integer m with 1 < m ~ k m-l ni
6 6
i=l s=l
l~ IS
>
U ml
160
CHARACTERS OF FINITE GROUPS
(31.2) Suppose that (31.1) is satisfied. Then 90 (S) "* 0 and S is coherent. Furthermore l' is uniquely determined unless k 1, n1 ::: 2 and ill = i 12 • Proof. Since n 1 ~ 2, 9 0 (s) "* O. Induction On k. If k = 1 then S is coherent either by assumption or by (23.3). If iu = lIS for s ::: 1, ... , n 1 then the uniqueness of l' follows from (23.3) in case n 1 ~ 3. If ils > ill for some s then ~;1 is uniquely determined by I(~Z., ilS~;;' - ~is)1 > 1. Thus ~'[s = liS ~'[1 - (lIs ~ll - ~lS)1' and so l' is uniquely determined in this case also. Assume that k > 1. Let ::I = U~;l Si' Then ::I is coherent by induction. Thus there exist irreducible characters Xit of N such that ~it ± X it defines a linear isometry l' on 9(::1) which extends l' defined on 90 (::1). Since iit~~(l) ~tt (1) ::: 0 for ~it E: ::I it follows that ~lt EXit for ~it E: ::I where E = ±1. Since ~~;1 iis ~ 3 the first part of the result implies that l' is uniquely determined on ::I. For ~it E: S let 0it = (iit ~ll - ~it )1'. Thus 0it ::: E(iit Xu - Xit) for ~it E: :J. Define the integer a by (Xu, ilkl ) = £ (l.kl a). If ~it E: :J, ~it '" ~ll then
::
(Xit ' ilkl )
:::
(iit Xu, ilkl ) - (iit Xu - Xit' ilkl )
= lit (Xu, il ) kl
::: EiOt (£ 1
kl
de it ,
- a) - €loti 1
il
kl
)
kl
- Edit Furthermore II ilkl 112::: l.lu + 1. As .g(:J 1') .go (s 1') = .go (:J 1') it follows that ilkl 4.g0 (:J 1') since lk \11 - ~kl 4 90 ( :J). Hence there exists an irreducible char~cter X of (\) such that X ~ :J l' and (Okl ' X) "* O. Therefore
n
COHERENCE
161
or equivalently k-1 ni
U kl a + a 2
:E :E
~it
::; 0
i=1 t=1
If a*-O then, (31.1) (iv) yields that
Hence a 2 < a. This is not the case as a is an integer. Thus a O. Consequently ek1 ::: e.(lkl Xu - Xkl) where Xk 1 i~ a generalized character of N which is orthogonal to ~ • Hence II Xkl 112 = 1. Since 8k1 (1) = 0 this yields that Xk1 is an irreducible character of N. If nk =:; 1 setting 'T
~ kl
= e. Xkl
completes the proof. Suppose that nk > 1. Then in any case Sk is coherent by (23.3). Thus there exist irreducible characters xict of
(\j
and E.k ::: :1:1 such that (lkt ~kl - lkl ~kt>7 = E.k(i'·kt Xk1 1 k~ Xkt). If Xu ::: Xict for some t then choosing s -:I t we get that
o :::
(e 12 , (f kt ~ks - fks ~kt
f)
::: -EEk{l12fks + (X 12' Xks)lkt} which is impossible as (X12 , Xks) ~ o. If nk = 2 and tkl lk2 choose E. k =:; c.. Thus Xk1 = Xkl (6 kl , (~kl ~kt)7) -1. Hence also Xk2 = X'k2 and the proof is complete in this case. If n l ~ 3 or l k2 > I kl it follows that E.Xkl E k X'kl since (b k1 , (l kt ~kl - I kl ~kt)7) = -£ kt for all t. Thus E E.k and 7 defined by ~kt ::: EXkt is the required isometry.
162
C H A RAe T E RS 0 F FIN I T E G R0 UPS
In case S contains some reducible characters the situation is a good deal more complicated though there is a similar criterion available. See Feit-Thompson [2, Theorem 10.1]. The remaining results in this section are concerned with conditions under which (31.2) may be applied. See Feit-Thompson [2, Section 11] and Feit [2, Section 2]. If S is a set of characters of a group 9l and ~ 4(d2 - l)/a + 2 and if S(~l) is coherent and contains a character of degree d then S is coherent. Proof. Let ~2
*'
(31.4)
COHERENCE
163
Since tJ/f:>a is nilpotent it follows that tJZ/~3 n Z(tJ/tJ a ) :# (1). Since tJ Z/tJ 3 is a chief factor of 91 it follows that ~2/tJ3 ~ Z(tJ/~3)' Thus if cp is an irreducible character of S)/~a then cp(1)21 tJz: f:>al :"'0 1tJ: -Pa i . Since 9) is nilpotent this implies that cp(l )211 tJ: -P21. Let b be the square free part of 1-p 1 : ~21 and let c :::: (a, b). Then the square free part of I-P: -P21 is ab/c z. Thus cp(1)z II-P: -P2 i c2/ abo Every character in Si is a constituent of some character of 9l induced by an irreducible character of -P. Hence if ~ E: Si ~(1) :"'0 19l: -P Ivll-P: -P21 c2/ab. Thus (31.4) implies that I
~1:
-P21 - I
~1:
-P I
:"'0
21'i d2
2d 191: -P I
:"'0
VI
-P: !2 I c
Therefore I-P: -P212 - 21-P: -P21 + 1 :::: (I-p: -Pl)1 - 1)2
:"'0
4d2 I-P: !J I c
2
Since ab/c2 is a common denominator this yields that
Thus
•. I.'0.
c:;. 'VI
1_ 2
<: -
(4d2 - 1) c2 a b2
contrary to assumption. Hence -P2 coherent.
<:
-
(4d2 - 1) a
= (1)
and S
=
S (-Pz ) is
(31,5) Let -P be a subgroup of @ such that ~1 =. N@(.p) is a Frobenius group with Frobenius kernel ~. Let S be
2
164
CHARACTERS OF FINITE GROUPS
the set of irreducible characters of 9l which do not have .p in their kernel. Suppose there exists a linear isometry 'T from So (S) into the character ring of @ such that O''T(1) ;; 0 for a E: So (S). Let e ;; hn: ~ I. Then one of the following possibilities must occur. (i) e + 1;; I-PI andSo(S) O. (11) .p is a nonabelian p-group for some prime p and IS;;: -P/I ~ 4e 2 + 1. (iii) S is coherent. Proof. If So (S) :::; 0 then I S I : :; 1. Thus I S;; I : :; e + 1 as required. Assume that So (S) "" O. Let ~ :::; ~I X ••• x ~n where ~i is a Spi -group of -P for Pi E: 1T(~), i 1, ... , n. Then I~i: ~i I ;; 1 (mod e) for each 1. Thus
1S (.,6') 1 : :; I-P: ~ 'I -
(31.6)
e
1
~ -1 nn
e i=1
(I~·
1
: ~~ I -
1)
1
Hence if I S (~') I : :; 1 then n :::; 1 and (ii) holds. Assume now that I S(~') I "" 1. Thus S(~") is coherent by (31.2). If .p' : :; (1) then S :::; S(~') is coherent. Thus it may be assumed that ~ is nonabelian. Hence by (25.9) e is odd. By (25.4) e divides the degree of every character in Sand S(~/) contains a character of degree e. If I~: ~'I > 4e2 + 1 then S is coherent by (31.3). Assume that I~: ~' I ~ 4e2 + 1. If n ;; 1 then (11) holds. Suppose that n ~ 2. If Pi is odd then I~i: ~i I : :; 1 (mod 2e). Hence if 1T(~) contains at least two odd primes (31.6) implies that I~:~' I ;: : (2e + 1)2 - 1 > 4e2 + 1 contrary to assumption. Thus it may be assumed that n 2, PI :::; 2 and P2 = P is an odd prime. Let I~1: ~~ I : :; 2s and I ~2 : ~~ I : :; pt. Let 2s - 1 :::; mi e, pt - 1 :::; 2m2 e. Thus 4ell + 1 ~ I~: ~' I Hence m 1 pt
:::;
>
2ml mil ell + 1
mil = 1. Consequently
2e + 1 :: 2S +1
-
1 (mod 4)
COHERENCE
165
Thus t is odd. Hence p! a where a square free part of 1.\1: ~' I. If (p - 2)e2 < p then e 2 < p/(p - 2):=; 3 which is not the case. Thus (p 2)e2 ?,: p and so pe 2 ?': 2e2 + p. Thus
Hence
1.\1: .\1'1
>
2
4e 2
4e + 2 p
a
+ 2
and so S is coherent by (31.3) as required. For the next result it is necessary to know more about the original mapping 7" defined on do (S). Generalizations of this result may be found in Feit-Thompson [2] Section 10. Let ~ be a T.I. set in @ with 91 = N@ (.\1). As(31. 7) sume that 91 is a Frobenius group with Frobenius kernel .\1 and e = 91:.\1 I. Let S be the set of irreducible characters of 91 which do not have .\1 in their kernel. For Ci E: do (S) let = Ci • Assume that (S, 7") is coherent. Then there exists a rational integer c such that if t E: S then
(7" (H)
=
{(H) + (1) c
e
for H E: .\1*. If furthermore X is an irreducible character of @ such that ±X ¢ S7" then X is constant on .\1*. Proof. Let S {(i} where ~l (1) = e. Let (~191 = ~j asj ~j + ~ s where ;\ s is a character of 91/.\1. By the Frobenius reciprocity theorem, (23.1) and the coherence of $ we get that
166
CHARACTERS OF FINITE GROUPS
~j (1)
-e- aS1 -a·= sJ "
= ( ~s'
* *)
~j (1) ~1 -e-
-
~j ®
Thus
Since As (H) ;::: As (1) for H exists an integer c s with
"
~s 1.\1#
€.:
,p this implies that there
~s l,p# + C s
By (23.1)
~s (1) " - e - ~11,p
, , _ (s (1)
- ~s l,p - - e - (11,p - ~s I~
and so C s : : : (~s (1)/e) c 1 proving the first statement. Let Xl9l = ~bj (j + A where A is a character of 9lj,p. The Frobenius reciprocity theorem, (23.1) and the coherence of S yield that
~j(1) b e
1
b. J
~j
-_ ( X, -(1) -~
e
* J*)_ ® - (.
1
- 0
c LAS S
0 F DO U B L Y T RAN S I TI V E G R0 UPS
167
Therefore
A(1) - bl completing the proof.
§32. A CLASS OF DOUBLY TRANSITIVE GROUPS A group ® is a Zassenhaus group if it has a faithful permutation representation which is doubly transitive and in which no nOnidentity permutation leaves 3 or more letters fixed. If ® is a Zassenhaus group the follOwing notation will be used: h + 1 is the degree of the defining permutation representation. 9l is the subgroup of @ consisting of all elements leaving a given letter fixed in the defining permutation representation. I.! is the subgroup of 9l consisting of all elements leaving a second given letter fixed in the defining permutation representation.
I IS! : :;
e
I~ll : :;
eh
I @ I : :;
eh (h + 1)
If e :::; 1 it is easily seen that @ is a Frobenius group whose Frobenius kernel has order h + 1. If e :f 1 then 91 is a Frobenius group whose Frobenius kernel has order h. In this case Sj will denote the Frobenius kernel of m, Thus e I (h 1). It follows directly from the definition that ~ is aT.!. set in @ and '9l :::; N® (.f)). Zassenhaus groups have been completely classified. See Zassenhaus [1], [2], Feit [2], Ito [3] and Suzuki [4]. In this section we will only give an intermediate step in this classification to illustrate how the results of Section 31 can be applied, The following will be proved. (3~.1)
Let ® be a simple Zassenhaus group, Assume
168
C H A RAe T E RS 0 F FIN I T E G R 0 UPS
that e is odd. Then either .p is a nonabelian p-group for some prime p with 9,):.p' I ::s 4e2 + 1 or e (h - 1)/2. Thus in any case .p is a p-group for some prime p. The proof will be given in a series of short steps. Until further notice it is assumed that @ satisfies the hypotheses of (32.1) and e < (h -1)/2. Observe that the simplicity of 6; implies that e > 1. (32.2) ~ is cyclic and a T.1. set in ®. N® (\!) is a Frobenius group with Frobenius kernel ~ and IN@(~): ~I
= 2.
Proof. If E E: ~# then E leaves exactly two letters fixed in the defining permutation representation. Thus if ~ '" ~G for some G E: ® and E E: ~# It ~G then E leaves more than two letters fixed and E '" 1 contrary to the definition. Thus l5' is a T.I. set in ®. Hence N@ «(.! )/(.! is a permutation group on the two letters fixed by all the elements in (.! # and so I N@ «(.!) ! ::s 2e. By (25.7) every Sylow group of N@ (\!) is cyclic. By (20.15) N@ «(I)' is cyclic and N® (\!)/N® «!)' is cyclic. Suppose q is an odd prime and q II N@ (<.!): N('9 (\!)' I. Let (1 be a Sq - subgroup of N@(~). Then Cis a Sq-subgroup of @ and N6;(Q ~ N® (~) since \! is aT.!. set in @. Thus by (20.10) ® is not simple contrary to assumption. Hence ! N~j(l5'): N® (<.!)' I = 2 and so (.! N~; «(.!)' is cyclic. If E E: (.!# and C® (E) g; <.! then replacing E by a suitable power it may be assumed that E has prime order. Since <.! is cyclic E E: Z(N('9 «(.!» and by (20.12) @ is not simple contrary to assumption. Thus C@ (E) ~ G for E E: G# and so N6; (i) is a Frobenius group with Frobenius kernel
C LAS S 0 F DO U B L Y T RAN SIT I V E G R0 UPS
169
Proof. Assume first that h is even. By (25.10) ~ is nilpotent. Thus a S2 -subgroup of @ is a T.I. set in ~ and so by (29.3) @ contains exactly one class of involutions. There exists an involution J E: Z(~) and so I~: C~ (J) I ;;; I @: S) I ;;; e(h + 1). Thus @ contains exactly e(h + 1) involutions. Thus every involution in ~ is in Z(~). Suppose that H is strongly real and H E: ~#. Then H IJ where I, J are involutions and (I, J) s; N@ «H») s; 91. Thus (I, J) s; S) as 1m: S) I is odd and so (I, J) s; Z($)). Thus IJ = JI and IF = 1 as required. Suppose next that h is odd. If H E: g,# and H is strongly real then H = IJ where I, J are involutions. Thus (I, J) s; N(~ «H») ~ 91 contrary to the fact that I 91 I is odd. Thus f)# contains no strongly real elements and f)1 f)J for I, J involutions implies that I :::: J. Thus no right coset of f) contains more than one involution and 91 contains no involutions. Therefore the number of involutions in ~ is at most e(h + 1) - e ;;; eh. Let J be an involution in (~L By (32.2) (eh, C® (J)) == 1. Thus I C@ (J)/ j{h + 1). Hence the conjugate class of ~ containing J contains I@: C@ (J)! ~ eh involutions. Thus 6} contains exactly eh involutions and they are all conjugate. @ contains at least e conjugate classes which (32.4) have no elements in common with f).
Proof. Let t - 1 be the number of strongly real conjugate classes of @ which do not contain involutions. By (32.3) (~ contains at least t - 2 conjugate classes which have no elements in common with f) and are not involutions. Suppose that h is even then (29.7) and (32.3) imply that t _ 2 > e(h + 1) {e(h + 1) + 1} 1
-
I@I
=
eh + e + 1 - 1 h
=
e +
e + 1 -1>e-1 h
170
CHARACTERS OF FINITE GROUPS
Thus t - 2 .~ e as required. If h is odd then (29.7) and (32.3) imply that
2
t
>-
-
eh(eh + 1) - 1
I@I
eh + 1 _ 1 h + 1 (e - 1)
-
1
- e - h+1
>
e - 2
Thus t - 2 ~ e - 1. Since .p contains no involutions the result follows in this case also. Let S be the set of irreducible characters of 9l which do not have .p in their kernel. For a E: So (S) let aT a *. By (23.1) T is a linear isometry. If S is not coherent the result follows from (31.5). Thus it may be assumed that S is coherent. Let S = {~) and choose the notation so that ~l (1) = e. Let ~ = ~l'
II ~* II@ = e
(32.5)
+ 1 and ~* (H)
= ~(H)
for H
E:
.p#.
Proof. If H E: .p# then ~* (H) ~(H) by (23.1). Since (1) = (h + 1)e and ~* (G) = 0 if G is not conjugate to an element of .p this implies that ~*
II ~* jI2
=
l'e
1 @
-
1
- I@
{e I
e(h + h
2
(h + 1)2 +
2
(h
~ .E I ~(H) 12} he .p#
+ 1)2 _ e2 1@1+ I ® I 1; I ~(H) 12} he
1) _ e
h
+
II ~ W 9l
he .p
e + 1
c LA SS
0 F DO UB L Y T RAN SIT I V E G R0 UPS
171
q:::; q
(32.6) + ~i (l)/e r where r is a sum of irreducible characters none of which are in ± Proof. If X is a nonprincipal irreducible character of @ which is not in ± S'T then ~ is not in the kernel of X and so by (31. 7) (~*, X) :f; O. By (32.4) there exist at least e such characters. By (31. 7) and (32.5) there exists e E: ± S'T such that (~*, e) :f; O. Thus by (32.5)~* e + r where r is a sum of irreducible characters of @ which are not in ± S'T. If I s I : :; 2 then 'e + 1 :::; I.\): ~I I and ~ is a p-group for some prime p since (h -l)/e > 2. Thus it may be assumed that I S I 2: 3. There exists E :::; ±l such that (32.7)
*
~i (1) *
~i - -e- ~
=E
Il 'T
\~i
~i (1) - -e~ J 'T\
If E -1 then (~[, ~*) :f; 0 for i > 1. Hence e : :; ~::::; ~; which is impossible. Thus E :::; 1 and so e = ~'T. Now (32.6) follows from (32.7). Let 1}1' ... , 1}e-l be all the irreducible nonprincipal characters of 91/(! where 1}j :::; 1}j + {e 1)/2 for j 1, ... , {e - 1)/2. Then 1}j (l) :::; 1 for all j. It follows easily from (32.2) that
(32.8)
1}:' (E) :::;
(E) + 1}. (E) for E J ~ (H) :::; 1 for H
J
II
1}.
J
J
E: ij if E: ~ #
By (32.8) 11} r (G) I ~ 11m (G) I for all G E: @ and 11}91 (E)l < 2 11;1 (E)I for some E E: (!. Hence J I'll jw < 1/191112. As @ is doubly transitive on the cosets of 91, 111;1112 ;: : 2 by (9.9). Thus II q 112 1. Hence (32.8) implies that {IJ; I j 1, ... ,(e 1)/2} is a set of pairwise distinct irreducible characters of @. Furthermore 1@ + ~ where ~ is an irreducible character of 1M with ; (H) 0 for H E: 6)#. Thus ~ $: ±I'T and I/j if: ±Ir by (31.7) and (32.8). Since Ii> = 1(\j + ~ + :6 (e'l)/2 21) itfollows that (~ 4>' loP) :::;
j
191 : :;
I
r
172
C H A RAe T E RS 0 F FIN I T E G R0 UPS
(~T , 1~)
==
O. The Frobenius reciprocity theorem and
(32.6) imply that (~r91' ~i) Hence in particular (32.9)
~
T
==
0 for i
> 1.
Thus ~r9l
==
~.
(1) ::: e
The proof of (32.1) can now be given by applying an argument due to Brauer. By (32.8) e 1 ::: 1, ... ' - 2 Thus ~2(1) e
1"T ':>
*
11 j +
1"T ':>2
=
)"T '>2
*
11 j
+ ~2(1) e
1"T ':>
r
Hence ~T S ~T 1/ and so 11 j S ~T ~T by (6.6). Since l1j(1) = h + 1 and 1(}J C ~T~T(32.9) implies that
Thus e + 1 ;::: (h + 1)/2 or e ;::: (h -1)/2 completing the proof of (32.1).
§SS. ISOMETRIES Let 11 be a subset of (}J with ~ ~ N(}J (~) == 91. Let d ("n) be the set of all generalized characters of 91 which vanish on 9l - ~. If ~ is a T.I. set in @ with ~( = ~# then by (23.1) the map T defined by aT = a* is a linear isometry from d (~) into the character ring of @ with aT (1) == 0 for a C d (,.). For some purposes the assumption that ~ is a T.I. set in (}J is too restrictive. In Feit-Thompson (2) Section 9 such a map T was constructed under weaker hypotheses. Dade [2] has simplified and generalized that construction and his method will be presented in this section.
ISOMETRIES
173
The following notation will be used: is a fixed set of primes. For G E: @, G1T is defined by G = G1T G1T , = G1T , G1T where G1T , G1T , is a 1T-element, 1T' -element respectively. ft is a subset of @ consisting of 1T-elements. 91 is a subgroup of @ such that ft s; m ~ N@(ll). 11(11), e(~) is the set of generalized characters, complex valued class functions respectively, of m which vanish on m- ft. The following assumptions are relevant. 1T
(33.1) (i) If two elements in ft are conjugate in @ then they are conjugate in m. (ii) If A E: ft then C@ (A) = Cm(A),p(A) where Cm(A) n ,p(A) = (1), ,p(A) C@ (A) and ,p(A) is a S1T' -subgroup of C@(A).
Assume that (33.1) holds. If a a T (G)
=a
E:
e(ft) define aT by
(A) if G is conjugate to A 11
E:
I.
= 0 otherwise
By (33.1) (i) aT is well defined. Observe that if 71 is a T.1. set in @ with " = 1# and 91 ::: NO) (ft) then (33.1) is satisfied with 1T 1T(~;). By (23.1) aT a * for a E: e(I). The main purpose of this
section is to prove Assume that (33.1) is satisfied. If a, f3 E: e(l) then (aT, f3 T )@ = (a, (3)m' If furthermore a E: u(l) then aT is a generalized character of @. (33.2)
Since a *(1) = a (1), (33.2) implies that the mapping T satisfies the assumptions needed in Section 31 provided that 1 = 1#. In case (33.1) holds let .11 1 , Sl:i!1 ••• be all the conjugate classes of m which lie in t. Let @i = {G I G E: @, G1T is conjugate to an element of .Il i }.
174
CHARACTERS OF FINITE GROUPS
As an immediate consequence of (33.2) one gets the following analogue of the Frobenius reciprocity theorem for T. See FeU-Thompson [2, Lemma 9.4}. (33.3) Assume that (33.1) is satisfied. Let a E: e(!l). If 6 is a class function on ® such that for all A E: !l, 6 is constant on the coset A.p (A) then
(a T, 6)® = (a, 619l)9l Proof. Let K,. E: ~'i' Then 6(G) ;:: 6(Ki) for G E: ®i' Thus if 60 is the class fUnction on 9l defined by 6 0 (A) = 6(A) for A E: ,. and 6 0 (N) = 0 for N E: 9l !l then 8{G) ;:: 8~(G) for G E: U®i' Hence T _ 1 T-(a , 6)® - ~ ~ a (G)8(G)
since aT vanishes outside U®.. Similarly 1 1 __ (a, 619l)9l = ~ a (G) 6(G)
Til
= (a,
80 )9l
The result follows from (33.2). Suppose that S is a set of irreducible characters of 9l with ~o (s) .s; e(!l). Assume further that (S, T) is coherent. Let S = {~i}' It is useful for many applications to
ISOMETRIES
175
have an analogue of (31.7) available for the mapping I defined in this section. By using the theory of modular characters it can be shown that ~! is constant on A~(A) for all 1 A E: t( then (33.3) can be usee to prove su.ch an analogue of (31. 7). The proof of (33.2) will be given in a series of steps. Assume that (33.1) holds. (33.4)
If a, (}
Proof. Let ~
! @i I = I @: = \ @:
E:
e(t() then (ai, (}I)®
E:
·R'i' Then by (33.1)
C® (~)\! {a 1 a
E: @,
C@ (~ )1\
E:
{a I a
=
(a,
Mm.
an = ~}!
~(Ki)} I
= I@: C@(~) 1\ ~(~) \ \@\
= .--::\c:::-@-;-(~-=-.-:-): :......,~-:-(~-• .,..-,-)I
I®I Therefore
1
=
I'~i\
= TmT Hence
--
IC91(Ki ) I a(~)(}(~) --
a (~)(}(~)
176
CHARACTERS OF FINITE GROUPS
= (a, (3)91
(33.5) If ~ is a 'If' -group and G is a 'If-element with G E: N® (~) then
G~ = U {GC.\'> (G)}H .\'> Proof. Since ~ <1 (G).\'> it follows that (G~)H = G~ for H Thus U ~ {GC.\'> (G)}H ~ G~. Suppose that L E: G~ then L'If is conjugate to G in (G).\'> by the Schur-Zassenhaus theorem. Thus L'If = GH for some H E: .p. Hence L E: L'If C ~ (Ln) = {GC~ (G)}H as required. If SB is a nonempty subset of I let .p(SB) = n SB.\'>(B).
E:T.
(33.6) For any nonempty subset 5B of I N91 (5B) ~ N® (.p(5B» and 14''.» (SB) n .p(SB) = (1).
Proof. By (33.1) ~(1;\) <1 C® (1;\) and ~(1;\) is a S1f' -subgroup, of C® (1;\). Thus ~(1;\) is characteristic in C® (1;\) and so ~(t\) <1 N® (t\). Hence N9l (t\) ~ N® (~(~». For B E: t\ N9l (SB) n ~(SB) ~ 9l
n
~(B) ~ C~l (B)
n
.\'>(B)
(1)
For any class function a of N9l (SB) define (33.7) a SB (NH) = a (N) for N E: N9l (SB), H E: ~(SB). By (33.6) a is a class function of N91(t\)~(t\). Further5B more at\ is a generalized character of N (t\).p (5B) in case
m
a is a generalized character of N91 (5B). (33.8) If Ci E: e(l) then aT -
- -
2: 5B
(-1)1 SBI
I ~l:
Nm (SB)
I
0'*
SB
177
ISOMETRIES
where
~
ranges over the nonempty subsets of !.
Proof. Define
a*
~
where ~ ranges over all the nonempty subsets of ft. Thus y is a class function and for G c ® (-1) I
y(G) ;: -
5l\1 0' (GM)
~ ~ I ~l: Nm (~)IIN:(~)4.1(~) I
where for each ~, M ranges over all elements of ® with G M c N (~)4.1(~). m By (33.6) INm(~)4.1(~)1 ::: IN91(~)II~(t\)I. If N c N~l (5l\) is the unique element such that GM c N4.1(~) then 0' 5l\ (GM ) ;: 0' (N) by (33.7). Thus 0' ~ (GM ) ::: 0' (N) ::: 0 unless N c ft. Hence (33.9)
_ y(G) - -
1
~
Ti1
( -1) 15l\ I0' (N)
I ~(5l\) I
~,M,N
where for each Sl\, (M, N) range over all ordered pairs such that N c _Nm (~) n ft and GM c N4.1(~). Thus by (33.5) N is conjugate to G1T if it occurs in the above summation. Thus y(G) = 0 = O'T (G) if G ¢ U®i. Hence it may be assumed that G c U®i and by changing notation that G c ®l' Since y is a class function it may further be assumed that G1T c "ll' If GM c N 4.1(~) then by (33.5) G1T is conjugate to N in (N) 4.1. Thus 0' (N) = 0' (G1T ) and N c .n 1 • Interchanging the order of summation in (33.9) implies that 0' (G1T )
_
y(G) -
-
I 911
~ ~
Nc.5l 1
M,~
(_1)1~1 I 4.1(~)
I
178
C H A RAe T E RS 0 F FIN I T E G R0 UPS
where for each N in .n l ' (5l.\ M) ranges over all pairs such that N E:: Nm(t\) and GM E:: N .\'>(51.\). The inner sum is independent of Nand l·n 1 1 = 191: Cm(G u ) I. The number of conjugates of Gu in Gu .\'>(t\) is I.\'>(t\): C.f)(t\) (Gu)1 by (33.5). Thus (33.10)
y(G) = _
~
a(Gu )
I C91 (Gu ) I
(_1)1t\1 M,5I.\ I C.\'>(5I.\) (Gu ) I
where (t\, M) ranges over all pairs such that Gu E:: N91 (58), GM E:: Gu .\'>(t\) and M E:: C® (G ). Gu E:: Nm(5l\) if and only if Gu E:: N91 ('8 U {Gu}) and GrJ E:: Gu .\'>(58) if and only if GM E:: Gu ~(5l.\ U {G.}). Thus (33.10) implies that Ci
(33.11)
(Gu )
ICm(Gu ) I ~
y(G) = -
(-1 14)(G ) I
M
+
¥
,
(_1)11:1
u
(_1)!1'U{Gu }1 })
II C .\1(1') (Gu ) I + IC.\1(t)U {GuH(Gu>l
where M ranges over C® (Gu ) and 1: ranges over all nonM empty subsets of I - {Gu } with G E:: Gu .\1(1'), Gu E:: N91 ('1:'). Since '\>('1) is a .' -group it follows that C.\>(~) (G u ) = C® (G.)
= .\1(Gu ) n
n
.\1(1')
.\>(t')
= .\>(t' U {G'/I'}) = C.\>('I'U {G } )(Gu )
u
Thus (33.11) implies that
ISOMETRIES
179
where M ranges over C® (G1T ). Thus
completing the proof of (33.8). (33.12)
where ~ ranges over a complete system of representations of equivalence classes of nonempty subsets of " under the action of m by conjugation. Proof, The number of distinct subsets of" of the form t\N with N E: ~l is I ~l: N91 (t\) I for any nonempty subset t\ of ft, Since 0' ~ == 0' ~ N the result follows from (33.8). Now (33.2) is an immediate consequence of (33.4) and (33.12).
NOTATION All groups are assumed to be finite unless explicitly stated otherwise.
II r is the cardinality of ft. 1# = ! - {l}
= B-IAB [A,B J = A-I AB A-I B-1 AB It' = {A B !A E: ft, B E: ti} AB
(A,B ... ) is the group generated by A,B, ...
[1,tiJ = ([A,B]lA
E:
ft, B
E: ~).
= [@,@J
@'
N @Ul) is the normalizer of ft in @ . C@(7l) is the centralizer of ft in @ .
Z(@) is the center of the group @.
.p l> @ means that .p is a normal subgroup of @: If ft 1> @, ft c t' 1> (\;, then till is a factor of @. If t\1 ~ is a minimal normal subgroup of
chief factor of
@
If 5.A/I is a factor l~, then C® (til!)
all B If
1T
E:
@/I, then it is a
{G I (G,B]
E:
I for
5.A}.
is a set of primes, then
1T'
set. 181
denotes the complementary
NOTATION
182
Generally {p} will be identified with p for p any prime. n11 is the largest integer dividing the integer n all of whose prime factors are in 11. (\j
is a 11-group if r~ t11' = [~r.
G is a 11-element if (G) is a 11-group
.p .p
is a Hall 11-subgroup or a S11-subgroup of ~ if
!.p I
= I~ 111
is a Hall-subgroup or a S-subgroup if S';) is a S11-subgroup
for some s,et of primes 11(~)
11.
is the set of primes dividing I~ I.
An involution is an element of order 2.
SL(2,q) = SLa (q), PSL(2,q) PSLa (q) is the unimodular, projective unimodular group respectively of degree 2 over the field of q elements
fJ is the field of rational numbers. fJ n is the field of nth roots of unity over fJ.
s:. ac is an extension field of s: then Tr 3\Js: is the trace of over s:. If ac is a Galois extension of s: then 9 xiir is the
char If
s:
is the characteristic of the field
ac Galois group of ac over S:.
REFERENCES R. Baer [IJ Math. Z. 71, 454-457 (1959). R. Brauer [IJ Ann. oj Math. 42, 926-935 (1941). R. Brauer [2] J. Math. Soc. Japan 3, 237-251 (1951). R. Brauer (3) Ann. oj Math. 5'7, 357-377 (1953). R. Brauer [4J "Proc. International Congress 1954," Vol. 1., pp. 1-9. R. Brauer [5J Proc. A.M.S. 15, 31-34 (1964). R. Brauer [6J "Lectures on Modern Mathematics," Saaty, Vol. 1., pp. 133-175, New York (1963). R. Brauer [7] Math. Z. 83, 72-84 (1964). R. Brauer and K. A. Fowler [IJ Ann. oj Math. 62, 565-583 (1955). R. Brauer and M. Suzuki [1] P.N.A.S. 45, 1757-1759 (1959). R. Brauer and J. Tate [11 Ann. oj Math. 62, 1-7 (1955). E. C. Dade [1] J. oj Algebra 1, 1-4 (1964). E. C. Dade [2] Ann. oj Math. '79, 590-596 (1964). W. Feit [IJ Proc. A.M.S. '7, 177-187 (1956). W. FeU [2J Rl. J. oj Math. 4,170-186 (1960). W. FeU [3J "Symposia in Pure Mathematics," Vol. 6 pp. 6770, (1962). W. Feit [4J Trans. A.M.S. 112, 287-303 (1964). W. Feit, M. Hall, Jr., and J. G. Thompson [1J Math. Z. '74, 1-17 (1960). W. FeU and J. G. Thompson [1 J Pac. J. Math. 11, 1257-1262 (1961). W. Feit and J. G. Thompson [2J Pac. J. Math. 13, 775-1029 (1963). P. Fong [1] Rl. J. Math. '7, 515-520 (1963).
183
184
REFERENCES
P. Fong and W. GaschUtz (1] J. Reine Agnew. Math. 208, 73-78 (1961). P. X. Gallagher [1] J. London Math. Soc. 39, 720-722 (1964). L. K. Hua and H. S. Vandiver (1] P.N.A.S. 35, 94-99 (1949). N. Ito (1] Nagoya Math. J. 3, 5-6 (1951). N. Ito [2] Nagoya Math. J. 5, 75-78 (1963). N. Ito [3] Rl. J. Math. 6, 341-352.(1962). P. Roquette [IJ J. Reine Agnew. Math. 190, 148-168 (1952). P. Roquette [2] Arch. Math. 9, 241-250 (1958). L. Solomon [1] J. Math. Soc. Japan 13, 144-164 (1961). L. Solomon [2] Proc. A.M.S. 12, 962-3 (1961). L. Solomon [3] Math. Z. 78,122-125 (1962). M. Suzuki [1] Amer. J. Math. 77, 657-691 (1955). M. Suzuki [2] Proc. A.M.S. 8, 686-695 (1957). M. Suzuki [3] J. Math. Soc. Japan 15, 387-391 (1963). M. Suzuki [41 Ann. of Math. 75, 105-145 (1962). J. G. Thompson [1] P.N.A.S. 45, 578-581 (1959). J. G. Thompson [2] Math. Z. 72, 332-354 (1960). J. G. Thompson [3] J. oj Algebra 1, 43-46 (1964). A. Weil [1] Bull. A.M.S. 55, 497-508 (1949). H. Wielandt [1] Math. Nachrichten 18, 274-280 (1958). H. Zassenhaus [IJ Abh. Math. Sem. Hamburg 11, 17-40 (1936). H. Zassenhaus [2J Abh. Math. Sem. Hamburg 11, 187-220 (1936).
Books W. Burnside, "Theory of Groups of Finite Order," Cambridge University Press, England, 1911. C. W. Curtis and 1. Reiner, "Representation Theory of Finite Groups and Associative Algebras," Interscience, New York, 1962. M. Hall, Jr., "The Theory of Groups," Macmillan, New York, 1959. W. R. Scott, "Group Theory," Prentice Hall, New York, 1964. H. Zassenhaus, "The Theory of Groups," 2nd ed., Chelsea, New York, 1949.
INDEX Absolutely irreducible, 6 Algebraically conjugate characters, 14 Associated permutation representation, 44
Frobenius group, 133 Frobenius kernel, 133 Frobenius reciprocity theorem, 47 Generalized character, 78 Group algebra, 1
Character, 10 Character ring, 78 Character table, 41 CN group, 142 Coherence, 158 Complement, 56 Completely reducible, 3 Constituents of ff', 3 Contragredient representation, 3
Index of ramification, 53 Induced ff'-representation, 44 Induced function, 46 Inertial group, 53 Integral representations, 23 Irreducible character, 10 Kernel of a character, 13
Degree, 1 Dihedral group, 63
M-group. 58 Minimal simple group, 145 Monomial character, 44 Monomial ff'-representation, 44 Multiplicity, 13 Multiplicity free, 13
ff'-conjugate, 67 ff'-conjugate classes, 67 ff'-elementary, 71 ff' -elementary with respect to the prime p, 71 ff'-normalizer, 81 ff'-reducible, 3 ff'-representation, 1 faithful, 1 similar, 1 Frobenius complement, 135
Normal complement, 98 Normal complement of ~ over .po in~, 98 Normal 7T-complement, 98 Orthogonality relations, 16
185
186 1I"-factor, 98 11"1 -factor, 98 1I"-section, 105 Permutation character, 44 Principal character, 11 Quaternion group, 63 Real conjugate class, 68 Real element, 68 Regular ff-representation, 2 Schur index, 61
I ND EX Splitting field, 9 Splitting field of a character, 13 Thompson subgroup, 118 Transitive permutation ff-representation, 44 Trivial intersection set, 123 Unit ff-representation, 2 Z-group, 111 Zassenhaus group, 167
BENJAMIN BOOKS Of RELATED INTEREST
LECTCRES ON AI.CEBRAIC TOPOLOCY MARVIN J. GREENBERG / Nod/leas/ern l/Ilil'('rsilv This text is int('nded for a ~raduat(' courst' ill al~('hruic topology. Th e four main parts of the book tn'at elementury homotopy theory, singuiur homology theory, orientation and dualitv on manifolds, and producls and iI,e l.e fsclwtz fix f'd point theorem. SET THEORY A[\J) THE CONllNUU;\1 HYPOTHESIS PAUl. J. COHEN / SICln{ord Uni.versily Thi s graduate level text-sllpplement contains a d etail ed expositi on of the author's research on the inderH'ndence of iI" , continuum hypoth esis from the axio ms of set th eo ry. RAPPORT SUR I.A COHOMOLOGIE DES GROUPES SERGE I.ANG / Columbia University Thesf' advanced notes. in French, provide a ddailed introduclion to the Artin-Tate notes on class field tht'ory, for graduate student s interested in homolo gical algebra and co homology of groups. ALGEBRES DE LIE SEMI·SIl\1PLES COl'vlPLEXES JEAN ·PIERRE SERRE / ('ollefje de France This set of notes, in French, is a compact pn'sentation of Ihe str ucture of semi-simple complex l.ie alg('bras and their representation th eo ry. LIE ALGEBRAS AND LIE GROUPS JEAN·PIERRE SERRE / College de France Covers the main general theorems on lie algebras. witl, some results on fr('e lie algebras. plus an introduction to groups and their cnrrespondence to lie theory.
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