Lecture Notes in Mathematics Editors: J.–M. Morel, Cachan F. Takens, Groningen B. Teissier, Paris
1773
3 Berlin Heidelberg New York Barcelona Hong Kong London Milan Paris Tokyo
Zvi Arad Mikhail Muzychuk (Eds.)
Standard Integral Table Algebras Generated by a Non-real Element of Small Degree
13
Editors Zvi Arad Netanya Academic College Kibbutz Galuyot, St. 16 Netanya 42365, Israel and Dept. of Mathematics and Computer Science Bar-Ilan University Ramat-Gan 52900, Israel e-mail: aradtzvi.biu.ac Mikhail Muzychuk Netanya Academic College Kibbutz Galuyot, St. 16 Netanya 42365, Israel e-mail:
[email protected]
Cataloging-in-Publication Data applied for Die Deutsche Bibliothek - CIP-Einheitsaufnahme Standard integral table algebras generated by a non-real element of small degree / Zvi Arad ; Mikhail Muzychuk (ed.). - Berlin ; Heidelberg ; New York ; Barcelona ; Hong Kong ; London ; Milan ; Paris ; Tokyo : Springer, 2002 (Lecture notes in mathematics ; 1773) ISBN 3-540-42851-8 Mathematics Subject Classification (2000): 13A99, 20C05, 20C99, 05E30, 16P10 ISSN 0075-8434 ISBN 3-540-42851-8 Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specif ically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microf ilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. Springer-Verlag Berlin Heidelberg New York a member of BertelsmannSpringer Science + Business Media GmbH © Springer-Verlag Berlin Heidelberg 2002 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specif ic statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Camera-ready TEX output by the editors SPIN: 10856649
41/3143/du - 543210 - Printed on acid-free paper
Preface
of
Properties
products
of
conjugacy classes of finite
groups
4r6radoldbrancf,
of finite group theory. This topic was intensively studied in the 1980's. The book [22] "Products of Conjugacy Classes in Groups," edited by Z. Arad and M.
Herzog, gives period. It
was
comprehensive picture of the results obtained during this
a
realized
by several authors that
this research could be extended to
of irreducible characters. We refer the reader to the papers
products
[1, 2, 11,
13-16,21,23,35,40,51,52,651. In several of these papers the authors found an analogy between prodconjugacy classes and products of irreducible characters which led to
ucts of
the notion of table
algebra,
introduced
H.L.Blau and Z. Arad in
by
[7],
in
uniform way the decomposition of products of conjugacy classes and irreducible characters of finite groups. Since then, the theory order to
study
in
a
algebras was extensively developed in papers of Z. Arad, H. Arisha, H. Blau, F. B-dnger, D. Chillag, M.R. Darafsheh, J. Erez, E. Fisman, V. Miloslavsky, M. Muzychuk, A. Rahnamai, C. Scopolla and B. Xu [3-5,7of table
10,12,17-20,25,29-33,35,41]. algbras, as defined, may be c onsidered a special class of C-algbras by Y. Kawada [49] and G. Hoheisel [48].:More precisely, a table algebra is a C-algebra where the structure constants are nonnegative. Each finite group yields two natural table algebras: the table algebra of conjugacy classes and the table algebra of generalized characters. Table
introduced
Both table
algebras arriving
from group
erty: their structure constants and degrees
theory
have
an
additional prop-
nonnegative integers (we refer the reader to the Introduction where these notions axe defined). Such algebras were defined in [30] as integral table algebras (briefly, ITA). Generalized table algebras (briefly, GT-algebras were introduced in [20]. They generalize properties of such well-known objects, e.g., homogeneous coherent algebras, Iwahori-Hecke algebras, etc. are
integral table algebra may be rescaled to a homogeneous one [32], algebra whose non-trivial degrees are equal. This common degree is a natural parameter which may be used for a classification of integral table algebras. The first result in this direction was obtained by Z. Arad and H. Blau in [7] where homogeneous table algebras of degree 1 were classified. The classification of homogeneous integral table algebras of degree 2 with a faithful element was obtained by H. Blau in [31]. This research was continued in [10] where a complete classification of homogeneous integral table algebras of degree 3 with a faithful element was obtained provided that the algebra Each
i.e.,
an
does not contain linear elements.
VI
important class of ITA is comprised of so-called standard integral algebras (briefly, SITA) which axiomatize tile properties of Bose-Mesner algebras of commutative association schemes. The standard algebras are also involved in the study of homogeneous ITA. Each element of a table algebra is contained in a unique table subalgebra which may be considered as a table subalgebra generated by this element. So it is natural to start the study of integral table algebras from those which are generated by a single element. Table algebras generated by an element of degree 2 were completely classified by H. Blau in [291 under the assumption that either a generating element is real or the algebra does not contain linear elements of degree a power of 2. If a table algebra is generated by an element of degree 3 or greater, then its structure is more complicated. If a generating element is real, then we are'faced with a classifleation. of P-polynomial table algberas which would imply powerful consequences for a classifleation of distance-regular graphs. In contrast, if a generating element is non-real and of small degree, then either a complete classification or important structure information may be obtained. For example, standard integral table algebras generated by a non-real element of degree 3 were classified in (5], [33] under the additional assumption that there is no nontrivial element of degree 1. In this volume we continue the investigation of integral standard table algebras generated by a non-real element of small valency. More precisely, we collect here the recent results about integral standard table algberas generated by a non-real element of degree 4 or 5. In all the examples known to us of SITA generated by a non-real element of degree k, the degrees of all basis elements are bounded by some function f (k). This gives evidence of the following Another
table
I
Conjecture generated by bounded
are
a
There exists
are
function f : N -+ N such that if a SITA is of degree k, then all degrees of the algebra
by f (k).
The results of
and
a
non-real element
[29]
show that this
conjecture
is true if k
=
2. If k
=
3
SITA does not contain nontrivial elements of degree 1, then all degrees bounded by 6 and the conjecture is valid. The paxtial classification of a
standard ITA
generated by
an
element of
degrees 4,5 obtained
in this volume
also supports this conjecture. It is not difficult to show that the conjecture holds for the table algebras of generalized characters of a finite group even without the assumption of
being
non-real.
[22] and the paper [7] countries to work on table algebras, The book
attracted
many researchers from various
products of conjugacy classes
and related
topics. At Bar-Ilan
Z. Arad and his students H.
Arisha, V. Miloslavsky, Fisman, jointly with his colleague M. Muzychuk, performed extensive research on table algebras. In the academic year 1998/99, H. Blau from Northern Illinois University (deKalb) and two postdoctoral University,
and his former student E.
Vii
F. Bfinger from Germany and M. Hirasaka from Japan, joined the University group in order to further advance the theory of table algebras. This volume, together with [5] and [331, collect most of the results obtained in this period at Bar-Ilan University. This volume contains 5 chapters. The first chapter is an Introduction, which contains all necessary definitions and facts about table algebras. The second chapter, Integral Table Algebras with a Faithful Nonreal Element of Degree 4, deals with standard integral table algebras generated by a non-real element of degree 4. The contribution of one of its co-authors, H. Arisha, is a part of his Ph.D thesis. Another co-author, E. Fisman, was supported by the Emmy Noether Research Institute at Bar-Ilan University. The third chapter, Standard Integral Table Algebras with a Faithful Nonreal Element of Degree 5, and the fourth chapter, Standard Integral Table Algebras with a Faithful Real Element of Degree 5 and Width 3, are devoted to standard integral algebras generated by an element of degree 5. F. BiAnger, one of the co-authors of these chapters, was supported by the Minerva Foundation in Germany through the Emmy Noether Research Institute at Bar-Ilan University. The last chapter, The Enumeration of Primitive Commutative Association Schemes with a Non-symmetric Relation of Valency at Most 4, classifies primitive commuta, tive ass'ociation schemes which contain a connected non-symmetric relation of valency 3 or 4. Its author, Mitsugu Hirasaka, was supported by the Japan Society for Promotion of Science, and worked in both the Graduate School of Mathematics at Kyushu University and the Emmy Noether Research Institute at Bar-Ilan University.
students, Bar-Ilan
We also would like to thank Mrs. Miriam Beller who corrected the ous
misprints
in the text and
Ramat-Gan and
July 1999
Netanya,
prepared
the final version of the
numer-
manuscript.
Israel
Z. Arad
M.
Muzychuk
Contents
Introduction
I
Z.
Arad,
..............................................
1. 1 Main Definitions
1.2 Basic 1.3 Basic
examples properties
.............................................
........................................
...........................
1.4 Basic constructions
.........
Arad,
a Faithful Nonreal Element of Degree 4 Muzychuk, H. Arisha, E. Fisman examples .............................................. .........
2.2 Proof of the main results
Z.
SITA with
Arad,
F.
a
..............
E.
Element of Degree 5 Fisman, M. Muzychuk
5
........
.................................................
3.2 General facts and known results
Degree
.........................
Faithful Nonreal
Biinger,
3.1 Introduction
3.3
........
...............
................
....................................................
3.4 Case 3
......................................................
3.5 Case 5
......................................................
4
F.
6
7 13
SITA with
a
Faithful Real Element of
Degree
14 19 43
43 44 61 62 66
5 and Width 3 83'
Bilnger
4.1 Introduction
.................................................
4.2 Case 1
......................................................
4.3 Case 2
......................................................
5
1
4
M.
2.1 Known
3
......
...........................................
2. SITAwith Z.
I
Muzychuk
M.
83
83 87
The Enumeration of Primitive Commutative Association
Schemes
Mitsugu
......................................................
5.1 Introduction 5.2 The
case
5.3 The
case
5.4 The
case
References Index
105
Hirasaka .................................................
of
valency of valency of valency
2
1
or
3
...................
4
.........................................
.....................................
......................
....................................................
105 109 110
117 121
125 ...............
Integral
2
Nonreal
Arad',',
Z.
Algebras of Degree
Table
Element
Muzychukl,2,
M.
Department of Mathematics Bar-Ilan University Ramat-Gan 52900, Israel Department of Mathematics Netanya Academic College Netanya, Israel
2
Arishal,
H.
with
a
Faithful
4
Fisman'
and E.
and
Computer
and
Computer Science
Science
Throughout this chapter we assume that (A, B) is a algebras generated by a non-real element of degree from the class A strongly of the algebras 3. The structure b_ where b is a non-real faithful of the multiset structure
table
degree 4. A direct for [6'b-]: possibilities
shows that
verification
11 4,621, 11 4,43],
It
not
difficult
3).'If
[6-b]
is
tion
=
algebra Bb is exactly 11, cl is uniquely defined we have the following element
basis
Let
1. 2.
If If
[6-b] [6-b]
=
=
b
isomorphic to 1, cl. by bTb
an
(A, B)
of degree
4], [14
4
min(B)
depends
on
element
multisets
! the
of B of
exhibit
all
41],
2
integral
a
[Cb]
an
Assume that
EE
table
standard
be 4.
Z,,, ? 11, cl where [ 14 4', 4', 4'1, [14 81,411,
product
wreath
For
min(B)
algebra
> 3.
with
non-real
a
Then
F,. [14 41,41,41], then Bb F',. [14 41,81], then Bb
(The algebras F, The remaining that
3
4 and
[14 41,41,41], [14 81,41], [14, 121]. to show that [Cb] V j[14 6 2], [14,43], [14 42 41]1 (Proposito this book, [14,34], then by Theorem 5 of the Introduction
the
Theorem 1.
[14
following
the
integral
standard
element
and
F,,n
case
x E
B of
are
[6-b]
=
defined
in the
[14,121]
degree
4
is
if
is'strong
section).
next
the x
most
difficult
is nonreal
one.
and
[xyx]
Let =
us
say
[14, 1211.
c Supp (X2) Of always 2. If x" then by Theorem 5 [Xa-X7a is real, [14 3 41. If Xa is not real, then either [14 3 4] or [Xayal [14, 1211. Thus, in any case, either x" is strong [Xa7Xc, or [Xa-X7a [14 34]. In the second case we have the following:
element For each strong degree 4 (Theorem 3). It
x
there
turns
exists
out
a
that
unique
the
element
width
of x'
x" is
=
=
=
=
Theorem 2. element
Let
b which
(A,B) satisfies
be
the
integral following
an
standard conditions:
Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 13 - 41, 2002 © Springer-Verlag Berlin Heidelberg 2002
table
algebra
with
a
strong
Z. Arad et al.
14
min(B)
1.
[6-b]
2. 3
=
b2
.
Then
=d+2f,JfJ=6 sgn
exactly
(see
we
algebra isomorphic
to the
:
in the
1, -11
Z2n
(x)
1 and sgn
-
next
=
is
sgn
sign function
a
(-x).
section).
if and only if n 2. In this An (sgn) is group-like group-like algebra over the group Z4 (1) Z4
case
Z4
(9
2.1.1).
b and b'
the
have
< m< n
is defined
Subsection If both
1, 0
3 4].
[14
where sgn
> 2
n =
The
Remark 1.
[cFdj
An(sgn)
(The algebra is
and
An(sgn), (2m + 1)
Bb _9',;
such that
it
3;
>
[14, 121];
same
then
strong,
are
alternative.
we
Thus there
may consider
b ,2
two
possibilities:
N such that
[F_77
are
(b)'
=
for
which
either
bl'-
is
for
strong
every
natural
(minimal)
m (=-
m
or
there If B is first
case
a
exists
a
group-like If
occurs.
B
algebra over is a group-like
an
abelian
algebra
Z2- 6) Z271 then [F-7177] [14 3 4] for each case Bb is always group-like.
group over
an
[14, 34].
of odd
order,
abelian
then
the
Z271
group
-
=
mG
N. We conjecture
that
in
the first
examples
Known
2.1
algebras
Group-like
2.1.1
Z4
H be an arbitrary subgroup of the symmetZ4 coordinates. The subgroup G by permuting group I (x, y, z, w) I x + y + z + w 01 is an 11-invariant subgroup. So we can define table algebra O(G, H). Let now a standard an S-ring, and, therefore, integral H be an arbitrary subgroup of S4 that contains a. full cycle (1, 2, 3, 4). Then the set 0 J(1, 1, 1 3)) (1,1, -3,1), (1, -3,1, 1), (-3, 1, 1, 1)1 is an H-orbit
Consider
the
group
ric
S4.
H acts
.
Let
on
=
=
=
-
is easy to see that 0 is a non-real faithful element of O(G, H) of denon-trivial elements gree 4. If H G JA4) S41 , then 0 (G, H) does not contain of degree less than 4. The constructed algebra is of dimension RO. In order to on
G. It
build
dimensional
finite
examples one has to take an H-invariant subgroup of example, kG, k E Z) and factor out with respect to the is not difficult to see that such a quotient with respect > 5 gives us a finite example of the above algebra.
(for finite It chosen subgroup. index
G of
to
a
subgroup
2.1.2
Weshall
The build
following multiplication
the
kG,
k
algebras a
series
F.,,, of
6-dimensional table
F'
M
algebras of dimension 4m+ 2, m > algebra (H, E) where E 1, h,
of which looks
as
follows:
I
-
Westart
C1, C2) V) U)
with
1,
the
ITA with
2
h
Faithful
a
C2
Cl
h +
Element
Nonreal
of
Degree 4
U
V
V
3cl + 3C2
h
41+h+c,+C2
h+v
C1.
h +
v
4C2
41 + 4u
3c,
3h + 3v
C21
h +
v
41 + 4u
4c,
3C2
3h + 3v
31 + 2u
3h + 2v
3C2
3h + 3v 3h + 3v 3h + 2v 121 + 9cl
v
+4u +
3c,
+
+4u +
hold
gebras U
=
jh,
v,
By direct
I
I
Define
case.
21 + 16 2
=
I
all
shows that
this
in
s
v
check
A routine
3C2
3c,
V
U V
(CI
C2)7
+
computations
the a
+
h
v
2s + 4t +
S,
3h+v
tj
2v
41+cl
,
3c,
v
A+
+
tables:
2v
v
1 2s + 4tl
4v + 6h
4u +
(2.2)
4s + A
t
S
3h +
3C2+
al-
follows:
4s + A 2s + 4t
5v + 3h
I
+C2 + h
table as
t
S
V -
H
multiplication
following
v
.
10s + 8t + 3h + 2v 5v + 3h 4v + 6h
v
h h
+
V
2s + 4t +
h + 2s
h
U
the
obtain
(2 1)
standard
integral
HQ := Q (9z " + C2)1(Cl 2
U C
subset
t
we
v
+ 9C2 +3h + 8u + 2v
,
of
axioms
15
2v
v
v
3h+v
5v + 3h
2v
4v + 6h
4v + 6h
5v + 3h
3c, + 3C2+ 121 + 9cl + 9C2 +8u + 3h + 2v 4u+v
(2.3) 4u + 3cl
61 + 3c,
+ 3C2
+3C2 + 2u
I
h
let
S
t
2s+4t+v
3h+v
2v
h +
v
3h + 3v
2s + 2t
2s + 2t
C2.
h +
v
3h + 3v
2s + 2t
2s + 2t
v
2v + 3h
V v
+ 4t + 2s
'
13h + 2v
I +8t
2t +
s
I
I
3h + 5v 4v + 6h
+ 10s
(2.4)
t + 2s
j
that QUis an ideal of QE. multiplicatively cyclic group of order m written (a) For each a E C, table algebra. Cm) be the corresponding
from the above formulas
follows Let
61 + 3c, +3C2 + 2u
3C2
Cl
U
It
+
V
h + 2s
h
4u + 3c,
C,,,
(CC,,,,
=
be
and
a
Ua
E,a U,a:7
=
1; 1.
we
set:
Z. Arad et al.
16
Therefore
C,,, a
0 U is
0 h is
A direct of
U
basis
JUj,,cc_
=
4 of the
check shows that fusion
a
The
2.1.3
subalgebra
is
Introduction
of
of
set
a
this
to
C,"-graded
book the
fibres.
F,
set
:=
the
by H(m, n),
denote
following n E N,
H(m, n) C'
e,
(F,,,
\ JC1) C21)
Fn. Its dimension
equal
is
U
fC1
+
C21
is
a
to 4m + 1.
co,
two-parametric
R>O.
mE
I
cn-1
...,
U
of table
series
is defined
It
by
foo, ..-On-11
the
f
U
eo,
algebras following
we
I
en-1
...,
which data:
oa
eg ec,+o
(m (M
is
F' M :=
the set of
,
ec,+,3
where the additions
n)
f
=
cp C,,+O
0,3. Ocl+,8.
an
An(sgn)
algebra
with
Westart
H(m,
that
by Proposition a distinguished
standard table algebra. Since integral non-real a faithful element, the algebra F.. is an example of algebras looking for. The dimension of this algebra is 4m + 2.
we are
basis
(2.2)-(2.4)
from
follows
It
00,+0
1)c,,+j3 1) 06,+0
-
-
(M
is done modulo
algebra
table
2)e,,+,3
-
'
of indepes
standard
(m
+
which
1)oa+,a
-
+ Me",+'3+1
Mcc,+'8+1
A direct
n.
check shows that
if 'M G N. integral Let now C2n be a cyclic by g (E C2n- In a group algebra group generated basis xi (CC2n we choose the following sgn(i)g' where sgn : Z2n -+ 1-1, 11 function which satisfies is an arbitrary We 1, sgn (-x) sgn (0) sgn (x). a
is
=
=
have x,,xp
=
S(a,,3)x,,+,3
S(a, -a)
symmetric,
sgn is
the
Now we consider
and commutative
elements
form
basis
0+
=
4eO, +
Ca
:=
f
a
=
4c,,,
An(sgn).
through the Clearly, lAn(sgn)l
structure
constants
+
+
Oa 00
in the
0-
54 0;
Ce
:34
CO + XO,
a runs
set
+
0).
Since
10, =
0
=
-
5n.
-
'
(CC2,rb
Clearly,
-
check shows that
I
-
this
is
the
following
an
as-
ca+0+1 + (1 + S)e a+o+l 4c+0 + e+, 0 a+ 0 + 1 0 =
2X2ce+l)
-
2ea
0,
f
0, 3co
2X2.,
-
a
=
a
0 0;
0
0; -
xO,
a
=
0
11 C Z2n We denote this show following equalities -
The
above basis
3o,
:=
C
n
-,
=
e,
0; a
(1)
CC2n:
(1)
0
a
=
sgn(a)sgn(,8)/sgn(a
A direct
2X2ce+l,
2e, + 2X2co
ea
where
+
o,,,
H(4, n)
sum
algebra.
H(4, n)
of
=
1.
=
direct
sociative
a
S(a,,3)
where
are
+
nonnegative
s)e,+,,+,,
basis
that
integers.
a
+
0
+
10 0;
by the
ITA with
2
where
s
=
S(2a
1, 2,3
+
a
Faithful
Nonreal
Element
of
Degree
417
1).
+
2.
where
(3 s)e+ 4co3c +,+,+3eo,a+,3+1=0 +
0+0-
=
S(2a
s
-
1, 2,3
+
3+3so+ 2
Oe0 s
=
S(2a
where
s
=
S(2a
+
+ c,
+,,
10
a
+
0
a
+
)3 +0
+
0
0
0-
a+ 'a,
2
3oa,,O,
+
0
0
1).
+
4. 0
3-3s
+
a+18
3o
1, 2)3
+
s)e-
+
1).
+
3.
where
(3
+
+,
a+
+
3-3so+ 2
a+
e
)3+3+so
-
a+
2
)3,0: O
0
1, 2,3).
5. +
6.
o+c
Oa+)3+Oa+,3,0: O
c+-
0 ,+
+ "3 Oa
,q
=
0.
o-
=
ce
7.
9c,,,+,,+,
0-0
where
s
(9
+
s) ece+,6+1
S(2a
+
1, 2)3
9-3so+ 2 a+
+
Oa e,8 s
S(2a
+
+
0-6a
3ol
0
=
3o,,,
9+sO 2
+
43
)3
-
0,3
3o
a
C+ ,q 6
+
cl
+
Q+0
2
0
a
+
9+3so+ )3
10.
oc
s) e,+,,+,,
a+
0"3
0
0
1, 2,0).
9.
11.
(9
1).
+
8.
where
+
12c++8c-+9e,+,a+0+1=0 0 0 0
=
9-so2
a+,6
10 0
0
0.
3o+
+
3o
2o;;,)3
a+,6 +
3o,+,,,,O =
0
0
0.,
2o,-.
12.
6ct
+
3co+, +0 6c+0
4(et +
(2
+ +
+ 2c-0 +
cc-,),,3
=
s)e+a+O
2e+, 0
a
0
+
+0
(2 =
-
0
s)e-a+ 'e,
a
= A 0,)3 =A 0,
a
+0
Z. Arad et al.
18
where
s
=
S(2a, 2ß).
13. 0 0"3 6c+ + 4(e+ + 6-), -p ß =
3c+",+ß
where
s
=
0
C,
(2 s)e++ + 4c, +2e.,a+0=0 +
-
ß
a
+
(2
+
s)e-
+
S(2a, 2ß).
14.
e "ß +
ea
=
4(ep
+=
Co
e++", c,+ 2e0
,
0
eß ),
+
a
=:
0
7 01,6:A: 02 C+ 7 0 0. 7 0, ß:7 0, a + ß
a a
=
15.
e
+c-0 ea
ce:7
+ ei, +
3e0
,
a
=
0.
16.
0,a e
==
3e,
e
6c+0 where
s
+ß
0
0
or
+(2+s)e
+ 2c-0 +
a+
2e+, a:7 0
0,
0,
a
+
j3
0
S(2a, 2,6).
17. 0
0,a
+
e-c a
e+"+O,
=:
0
2e0 18.
ei cJ0
=
2e+
+
ej,
a
0
=
:7
* 0, ß * 0, ce 7 01 01 a
ce
,a
+
ß :A 0 0.
0.
19. +
ca
3
=
4c",+" a + ß 7 0; a + ß 4(c+0 + c-), 0
20.
4 cj 21.
c0 ej
=
3co+
+
2co-
3c+,
a
=
Co
,
a
:7 0;
=
0.
=
0.
7
0
ITA with
2
Nonreal
Element
Degree
19
4
Preclassification
2.2.1
Proposition either Aabc
B# be
a, b (=-
Let
1. E
simple
following
from the
Westart
10, 11 for
each
E B
c
elements
two or
the
of
one
of degree 4, a =7 T. cases holds: following
=lhl=4 2f +g+h,lfi=lgl 2f +g,lf l=4,l gl=8 6, I gi4 2f + g, if I
24
8 2f,I fI 2f + 2g, i f I= IgI 3f + g, if I= IgI4 4 f,IfI + 3, IgI 4f g, if I
32
24 28
=
=
32
4
40
64
=
1 Proof.
Then
(ab, ab)
ab
=
=
1
4
5
Routine.
As
a
consequence
all
b E
be
following
the
obtain
we
(A, B)
Let
Theorem 3.
for
of
main results
of the
Proof
2.2
Faithful
a
exists
JbI
such that
table algebra integral b E B* with b 0 b
standard
a
BO. Assume that there
and
JbI
=
4.
3
Then
either bb
41 +
=
4c, b2= 4d, Icl
=
3, Idl
=
(2.5)
4,
or
b2= Proof. only
Since 7
IxI
d + >
2e, Idi
3,
possibilities
G
x
for
=
4, jej
=
6 and
B# and 4 divides
Ab;,ICI
for
f
+ g +
4 +
f
h, If I
(bb, bb) =
IgI
=
8, IgI if I 4 + f, If I 12, IgI 2f + g, if I 6 4 + 2f, I f 1 4 4 + 3f, i f 1 3 4 + 4f, i f 1 + g,
=
IhI
=
=
4,
=
4
4 +
now
the
of the
decomposition b
2 =
product
1: beB#
4,
28
28
28
=
Consider
all
bb:
bL 4 +
(2.6)
V, ESUPP(b-b)#Ab c
/\bbeC.
36 40 52 64
c
G
B#,
there
are
Z. Arad et al.
20
If all
coefficients
nonzero
Therefore,
there
(bb, bb) 28, (b2, b2)
that
=
c
(b2, b2)
E
(b2, b2)
then
ones,
B# with
C-
Abbc :
f 28, 40, 52, 64}.
(2.6)
then
=
axe
exists
holds.
Thus
16
=
(b2, b2)
If
have to
we
contrary
(bb, bb)
to
follows'from
It
2.
64, then (2.5)
=
deny
28. I
holds.
following
the
>
Proposition
If
possibil-
lities.
bb
=
41 +
2 3c, b =4d+e,ldl
bb
=
41 +
2 2c, b =3d+e,lcl
Assume first
equality
(2.7)
of
sides
of this
lcl
4, I\jdc
=
b satisfies
that
db
equality =
Consider be real
that
in bc is not
b
we
0. Therefore
now
in this
by
the
than
greater
It
9,
a
lcl)
=4.
(2-7)'
=4.
(2.8)
from the directly A 36. Multiplying
follows
second
both
=
e)
+
lei
=lei
Idl
Since
121 + 9c.
=
=
3 and
contradiction.
(2.8).
when b satisfies
case
gcd(lbl,
=6,ldl
d(4d
obtain
remaining
=
equivalently,
or,
Aj,
Since
case.
(2-7).
3b,
=
3,lcl
=
Note that
2, the number of
=
c
should
summands
nonzero
Moreover,
2.
(bb, c)
(bc, b)
12 =: -
/\bcb
12 = -
=
=
3.
Therefore, bc The second g
0 b, lgl
=
Af, f =/= b, Alf I
3b +
equality
(2.8)
in
bd
implies
12.
=
3b + g for
==
some
g
E
B#,
4. Wehave
(b2)b
(3d
=
b(bb)
b(4
=
+ +
e)b
2c)
=
=
3(3b
+
g)
2(3b
4b +
eb,
+ +
Af).
Consequently, eb Therefore
f
=
g, A
immediately
b +
=
implies
and for
JR(e some
Therefore
(C2 6b)
to
+
=
Proposition
d)l
<
St(e
+
d)
Idl =
3f, bc
3e +
3g.
-
=
3b +
3f, bd
St(e =
d,
6(e
+
d)
=
+
d)
=
1. The latter
Supp(6-b).
case
But
C
Supp(6-b)
84.
On the other
in
is
this
St(e+d). St(e
contradiction.
=
(bc, bc)
=
Since +
d)
impossible, case
hand,
(c2, b-b) a
f.
4f.
or
6-
=
3b +
4b +
=
[20], Supp(bb)
4.4 of
8, either
d E B with
2Af
that
bf
According
b +
3 and
=
e6 This
=
72,
C2
=
=
lSupp(6-b)l
Supp(Cb)
7 Uf
dj
since
min(B)
61 +
5c, implying
! 3.
ITA with
2
(2.5),
b satisfies
If
Z,,, ?
11, c}
Proposition
2.
B
(2.6).
.
gcd(IbIjej)
Since 3
=
J I from (be, be)
f
exists
be
==>
sides
both
pf
for
equality,
latter
of the
f
we
satisfies
=
Now Abbe
2.
:5
suitable
a
book
one.
of degree 4 that be 3b + f.
element
21
4
this
to
interesting
12 such that
=
ISupp(be)l
2,
=
3b +
=
Introduction
5 of the
non-real
a
BO, if I
E
Degree
of
Element
is the most
case
BO be
Let b E
Nonreal
Faithful
by Theorem
then
So the second
Then there
Proof. Abeb
a
obtain
1-tif I
12/,t
> 24.
2
=
BO and
G
=
implies
/.t
E N.
12.
The
that
Taking identity
(bb, ei ) implies
=
E
A,e,lcl
12 +
=
r_GSUPP(bb)\j1j The left-hand
Therefore
it
In what rem
3 and
side
Let
if bx
Properties
Proof.
a
2f, R
=
tt
=
b E BO of
element
divisible
by
12.
0
1.
degree
4.
By Theo-
Jbi
3b + g,
=
Idl
=
6, Igi
4, jfj
=
(2.9)
12.
say that
[bb]
that
is
element
an
one
following: -of degree
of the
Supp(bb)
x
G
a
basis
[14,41,41,41], 4 is
a
starting
Ed.
=
of the
starting
If
Supp( b_)
3.
starting
For the
element. contains
of degree 4,
element
then
Supp(66b)
one.
Eb
product
there
possibilites:
two
are
juj=jvj=jwj=4, U:
b
Icl us
30 and is
always'have
Eb=41+u+v+w,
Let
161
-
implies
in turn,
non-real
a
we assume
us
Proposition contains
which,
we
JeJ2
than
of Theorem I
*subsection
[14, 81, 41]. element
d +
=
Proof
In this
greater
24
to
we fix follows, Proposition 2,
b2
2.2.2
is not
equal
is
=
8, jhj
compute the product
(bb)E
=
(d
+
2f)E
=
b _b =
=
4,
V,
V
7-L
W,
-
c
+
h,
41 =
h,7!
=
:7
W
u
(2.11)
c
in two ways:
dE + 2Cb
=
dE + 6b
(2.10)
+
2g.
Z. Arad et al.
22
b(blb-) After
=
reduction
of
d
4b
common
+ 2b +
Assume first
t
+bu
bc + bh if
terms,
bc + bh holds.
Abwg 11 Abug ! 17 Abvg < 1. So, we have the following equalities: '5
=
=
is
w
element
starting holds,
a
(2.11)
If
in this
Abcg 'A
Clearly,
Abcg (2.13)
2.
=
hand, Together
implies
bh
Proposition 1.
Let
4.
x
for
a
=
Further
1.
Supp(bE),
(2.12)
d.
=
6
=
Abcglgl 2A -bc
+ 6b +
(2.13)
2g. is
2 this
=
gives
bc
us
least
at
Abcg
=
=
=
2b +
On
2.
< 3.
Thus
2g. Now El
Supp(bE)
E
=
0(mod 8). Therefore Abcg 32 implies :5 Jbilcl
=-
inequality with Abcb Ed..
=
y + 2f
dT
Abcglgl
But
0.
the
other
=
case.
4b + bc + bh
the
holds
b + g; b + g;
bw Thus
holds
Abub Abvb Abwb renaming of the elements of
up to
bv
holds
(2.10) (2. 11)
if
Then
bu
holds
obtain
we
bu + bv + bw if
2g
(2.10)
that
(2.10) (2.11)
+ bv + bw if
suitable
be
a
element.
starting
B#, jyj
y E
=
Then
4;
2.
Cbd, Ed) Proof.
(f, dY) < Idl
AdTd for
a
2.
(bx, bx)
0, then (d, dY) -
1
suitable
( d, Ed)
Proposition Proof.
=
=
(bx,
if
0
y
d
otherwise
=
Cbd, bx) A
5.
that
[bq
of the elements
(6-b, uUU)
=
(6-b, v-v)
(d T, b2)
clear
in
the
[14 41,41,411, u, =
v
=
element
starting
is =
=
=
(b2, dY)
=
(d
+
2f, dT)
=
B# \ ff I.
y E
d)
the fact that 4AdYd >' 16. But that contradicts 2. Thus dY AdTf 0 0 implying AdTf y + 2f
=
Therefore,
3.
Our claim
assume one
(bx, bx) =24 28
1.
16 <
If
=
should
(bv, bv),
is
case
i.e., be real.
(d, y)
+ 24.
always of
=
[bE]
b satisfies
W.l.o.g.
0
real. =
[14 81,41].
(2.12). u
=
If
w
U. Since
Thus is
16
we
nonreal, =
may
then
(bu, bu)
ITA with
2
A,-,-, A,,7u-, A,uu Av-,, Av-vw
Nonreal
Faithful
a
=
0 ==>-
=
0
A... AvUU Awuu Auvv Avv, AWvv
0
=
=
0
=
0
0
=
== '
Element
Degree
of
4
23
0
=
0
07
(2.14)
0
0 0.
=
Further (2.12)
(bu, bv)
Auvu
(2.14),
with
Together
+
implies
this
uv)
16
=
16
Auvv
+
uv
4w. Romhere it follows
=
Auvw
=
=
4.
that
uw
=
4v.
Thus
(6-b, uw)
(bu, bw) Since
b + g and /\bwb 16 contrary bw)
bu
=
(bw,
and
1. dY
d
2.
to
=
Proposition
If
6.
=
d +
=
x-x
x
1, (bu, bw) Proposition
=
Supp(bE)
E
is
=
12Abwg.
Hence bw
=
b + g
4.
element,
starting
a
16.
4 +
=
then
2f;
=
b.
Proof. 1. Assume the contrary, i.e., dY y + 24. By Proposition 4, (bx, bx) Proposition
2f and y 0 d. Then, according 1, bx is either of type [4 2 8] since Abxb 1. Therefore occur
=
to
or
bx are
=
[4 2 4,4].
of type =
b + 2u + v,
distinct.
pairwise
b(bx) (bb)x
=
The first
where u,
Now we
b 2 + 2bu + bv
(d
+
type
2f)x
=
=
cannot
basis
are
v
can
=
elements
of
degree
=
u, v, b
write
d + 2f + 2bu + bv
dx + 2fx
4 such that
d + 2bu + bv
y + 2f + 2fx
=.
y + 2fx.
(2.15) It
follows
from Afff
=
Afxd
2 that
=
3.
Together
with
gcd(If 1, IxI)
=
2, this
implies
fx=3d+[tz, for
a
suitable
z
E
BO \ Idl.
After
2bu + bv
b2X On the other b 2X
bEd
=
=
(2.16)
[tGN,zGB of
substitution =
y + 5d +
y + 2f + 6d +
(2.16)
in
(2-15)
we
obtain
2ttz
(2.17)
2/,tz.
(2.18)
hand, 4d + xd +
cd
cESupp( b)\11,xI
=
4d + Y + 2f +
cd.
CESupp(bb)\{I,xj
Z. Arad et al.
24
Comparing this
with
(2.18),
we
(
T
obtain
cESupp(0b)\f1
'.
I
C)
d
=
+ 2d.
2/-tz
in the left-hand side do not exceed 4, /t < 2. By (2.16), /-tlzl Hence, jzj E f6,121. According to (2.17), bv =_ y + d(mo 'd2). Therefore 0. Hence AbvzlZl :5 8. Together with AbvziZl =_ O(mod 12), this implies Abvz d + p z. But now b(u + v + b) 3d + y, bu bv 5d, which is impossible distinct. since u, v, d are pairwise d as required. Thus y
Since coefficients 12.
=
=
=
=
2. Since
y
=
d, (bb,xTY)
(bx, bx)
=
E
Therefore
28.
=
Ax-,,,Icl
12.
=
CESupp(1;b_)\{1j
Together
with
1cl
=
12
ESupp(6_b)\{1j and Theorem 3 this xx
=
that
implies
Ax-x,
=
I for
each
as
before,
c
G
b.
Since
bx
Td, Cbd, Ed)
=
=
Arguing
28.
Supp(bU)
we
obtain
111. that
6-b
Hence
=
du. 0
In
bE
41
-
sum
what -
of two basis
So
we
c
Proof.
for
some v
E
(fd, Td)
=
dh
=
B# \ jhj.
,
set
c
:=
=
=
d,
dh
exists
v
d + 2f.
=
c-
=
Now we
(2.19) =
ff
d +
2f, dU
=
h
2
=
B# such that jvj =
Therefore
Td
(Ff dU)
h. We also
=
There
By (2.19) gives us
as
of
Td
2, this
element
starting
then c is a basis element, otherwise c is a degree 4. Moreover, it follows from the proof of to u + U, u E BO. In both cases x-x o c is proportional x-T o c Ad-x, by the equality A, yxcc. bh
7.
the
(2.11),
satisfies
elements
may define
Proposition
denote
we
5 that
Proposition c.
h.
follows, If 6-b
=
12 and
(2.20)
3h +v.
ATdh
3h + pv, can
(2.19)
bb.
3.
Together
with
ged(If 1, Idl)
p
write
jd, Td) (bf,Ef ,6b)
36 + 12 p (2=9) 48
p
=
1, jvj
=
12.
ITA with
2
Proposition
a
Faithful
Nonreal
Element
of
Degree
4
25
8.
hc
2h +
=
2v V
(2.21)
v.
=
Proof. (2.9)
E2 d
=
(2.20)
d3 + 6h ( + 2?)d E2 d (2.19) Ebh=4h+hc =
=
Since
and h
c
Proposition
are
real,
9.
There
is real
v
exists
B* \ jcj
hc
such that
Jul
Cf
61 + 3c +
Arf,.u
C2
81 + 4c +
4Ai!f-uu
Arfu Proof. Weuse the following form: following
U)
=
2h + 2v.
+ h2
well.
as
u E
+ 2v
E
f3,61
and
(2.22)
{1, 21.
(6-b)2
identity:
E
b
=
2(E)2
.
The left-hand
side
has
the
(41+c+h )2 =161+8c+8h+h 2+C2 +2ch The
right-hand
side
is
equal
d3
=
h2,
=
161+c
2
+h2+12h+8c+4v.
to
=O)d3+2(6h+2v)+4Cf.
(2.2
(d+2f)(d+2f)=dd+2(df+f'd_)+4Cf By (2.19)
(2.21)
hence
4Cf
=
161 + 8c +
C2.
Comparing
of c in both sides, the coefficients 8 + Ac,,. we obtain 4Arfc I 7. Thus Acc, E Accc is divisible by 4. On the other hand, Accc < Icl 2 =#0, then Aqc 10741. If Acc, Arflcl 0 0(mod Ifl), a contradiction. Therefore 3. Accc 4, Arfc =
Hence
-
=
=
=
Let
4Arfu 24),
which
On the and
E Supp(c 2) \ 11, cl be an arbitrary basis element. Then Accu Since Arfulul =Ac,ulul 0(mod 6), A,culul =- 0(mod 4Afrfulul. implies 24 < Acculul. other hand, Acculul : IC12 24. Therefore 24 JCJ ACCCICI A,,ulul
now u
=#-
Supp(c 2)
equality
A,,,,
=
-
contains =
4Arf,,
-
=
from cj. exactly one element, say u, distinct the proof. implies that Aff,, E 117 21 finishing
=
The 0
Z. Arad et al.
26
g (xy, xy) (x-x, y-y) 4, Isl unique r, s G B0, Irl
x, y c-
grading G, we
there
exist
The
Proposition
Proof. :
we
x
shall
Thus z-z
41 +
=
Now let
+ h
c
x
x, y,
:7 T, T 0
z,
(y
x *
If
(y
z)
*
If
definition. z
The
h
If
x
Xyzv
=
follows
of
=
zj
that
xy
in
the
gh
Proposition
Proof. other
At the a
a
h otherwise.
=
z
h
*
as
G.
E
x, y
Denote
of
So
*.
z(mod 2), and, consequently, -=
(mod 2).
+ h
c
E G.
G. If
g
gh(mod 2).
=-
unit.
a
from the definition
each g (=-
triple.
x(yz)
=-
h*z
=
*
Denote
h,
=
there
then
\g-g-h
But
1
=
v
*
*
y
:=
from the
y)
*
y,
x *
:=
=
I.
h E
If
z.
following:
z(mod 2).
otherhand, x*(y*z) z by v y 7 v, then y other hand, A-g,z I implies
Onthe
z.
But
(x
=-
On the
h.
u
law follows
(xy)z
1, A-g,z v
Supp(c). (y * z)
=
If
=
Supp(z"T)
Supp(Cc).
h
while
z.
=
*
=
it
=
analogously.
== -
So
Furthermore,
[xyl (x
x
y =7
x *
may
we
assume
now
v
*
y
=
implies
that
* z
[4', 621
=
y)
* z
=
from which
it
h. The
case
of
Y
h,
way.
G,
g
to
=7 h,
that
mention
to
the
equality
x *
=
0
x.
there
exists
an
element
sg
(=-
134 of degree
2sg,
The mapping g
same
contradiction.
h.
=
1. Since
enough
is
g E
g +
x * v =
invertible
7 is
(glh,92h)
10, 41,
g
may be treated
proof
11.
[4,62].
y
similar
a
Assume that
hand,
=
z) \xyr,=-'T(mod 2)
For each element 6 such that
then
=-
xy
g for
=
U
(y
x *
G shows that
E
g
=-
1. Hence
=
To finish
y
-
7 is treated
=
directly
h)2 (mod 2)
+
Ay,v
x * z
U, then
Xyjx-
z)
then
and
(c
h
G whenever
C-
y
case
arbitrary
*
f 1, hj z,
x *
that
Define
+ 28.
r
each
For
implies
I
=
and
with
then the associative
Since v,
=
*
(x*y)*z v.
y
case
'77 % jx,
that
x
*
x(y
=-
Supp(y-y)
c
Therefore
u
'g
=
6-bj.
=
p(mod 2).
an
0 7,
v
z)
*
then
=g,
x
7,
u
*
x
=-
0 h,
-=
G be
z
h
g
h
*
=
xy
x
group
b& Consequently,
=
If
nothing to prove. 1, whence g I\ghg
if
r
follows
77. In this
that
is
Let
that claim
our
x
check
us
y
gT
Proposition
28, 6, such that
abelian
an
x7V-y(mod 2)
=-
zz
is
check
that
assume
may.
h)
*,
F, then
=
4 &
=
and
(2-19).
(G,
we
y. If
x *
I IgI
E B
=
G by setting
on
10.
First x *
=
*
h E G by
Note that
z
G
=
=
operation
binary
Let
group.
have
36
gi
=
time,
-+
92 and sg,.
(h 2, 92711).
[h2l
E
sg, g =
Since
G G
h}
S92. Then
gj,T2
j[j4,4,4,4],[j4,41,81]j.
is
injective.
(g, h,
E G
92
h)
=
24.
On the
and 91 7 92; [92TI] Hence (h2, 9 2 T-)
E
0
ITA with
2
Proposition
Jwl
Faithful
a
arbitrary
Let x, y E G be
12.
hy
If
0 T,
x
then
there
exists
=
q Cz
Y(x
Proof.
Denote
z
x *
z
then
x,
=
Supp(c)
left-hand
the
U jhj.
A,yylyl
8
together
with
(C2, y-y)
A,y,,Iul
==>
Jul
64. Thus
=
+ pq
==>
+ q, cy
=
2y
n
C=
Let
m,
(2.23),
m=
TZ
cy +
2y
=
lcllyl
+
First
and Sh
mn
sh
n
0
there
h
-1c) 2
we
we
=
may
==*
exists
obtain
we
(2.25)-
+ cy.
1, from which elements
follows
it
that
G. Therefore
from
6. We claim E BO, Jul =_ (mod 24). But A,,.,,Iul 24. Therefore 24 which Acyujul (cy, cy) 2y + 4u. But this contradicts ==>. Tz. ==> u (E Supp(hy) hy y + 2u
suitable
a
u
=
then
=
=
=
=
=
1 that
we
obtain
4y
Let
27xw.
=
=
E
now q
2ATwqlql. Together with implies Acyqiql 2,\-Zwllql Supp(cy) f y, qj and Yw
this
=
=
that
(cy, cy)
But
+ cy
Acyqlql
2ATwq =* O(mod 6),
elements
Y
41 + h +
=
=
64
=
==>
1, i.e.,
p
Tw
m
0
Set y
TF.
=
m
n,
x
By
M.
=
2 of the
2
q G
=
as
that
so we
B#, JqJ
=
for
statement, =
2y
(2.26)
to
the
case
3(m
*
n)
each y C G there
Wenote that
2t,.
+
of
th
exists
V.
+ trrt*n-
special case of n Then, by the definition
assume
y,
we'may extend
a
3m + tm,
=4
latter
12 and cy
=
consider
.1c.
So
c.
(2.26)
+ 28m*n-
1c. 2
21 +
=
Vm,nEG Tnsn
13.
21 +
=
+
hy
+
contains
and part
arbitrary
G be
71, then by setting
Proposition Proof.
4y
Supp(cy), Acyylyl
-
2/Lq.
(2.24)
2q.
According to part ty G BO such that Ityl
So
2w,
y +
x *
have
we
m(21
=
27
12,
by Y,
sides
E
u
implies cy 0 Supp(cy)
u
=
mn = m* n
m=
xy
4
(2.23)
does not contain
if
:
6
=
Txy
=
y + 2u, for
=
from (2.25) 2. It follows Supp(Yxw), q: y. Then Aqyq Acyqlql =_ O(mod 8), 1\"Ewqlql (mod 24). Arguing as before,
If
Write
Degree
y),
*
both
side
Supp(c)
But
y=h and hy=Yz. So, z 0 x, whence zY that u Supp(cy). Indeed,
3y 3y
of
Yw=3y+q.
Multiplying
y.
Yz + 2Yw
y E
elements.
BO such that JqJ
cy=2y+2q,
If
Element
Then
6.
2.
Nonreal
h. Wehave
=
of t,,,,
mc
m, * n =
2m +
m*
h
=
m
2t,,,
desired. n
0
may
h. Set
x
=
M,
y
=
apply Proposition
m* n.
12, part
12 such that
Ysn
=
3y
+ q; cy
=
2y
+
2q.
Then xy 2, which
=
n
+
28n,
says that
Z. Arad et al.
28
But
cy
3(m
*
2y
=
n)
+
2ty
==: -
q
Proposition
3Y
=
ty
+
Tnsn
14.
hg
Wehave
ggh
+
13, sgT
(41 Together
g +
=
(g
=
By Proposition
+
+
2-3g)
=
=
3h +
3h + th
h)2
+
c
Aq.u.
61 + 3c +
=
(2.27)
Therefore
2sg.
2sg)(T =
ff
=
g-gh
=
=
g-g +
2(sg
+W-gg)
-9-gg
==*
v
+
4sgg-g.
3h + th
=
+ 12h + 4v +
g g-
A+
v.
Thus
4sg 7gg
with 161 + 8q + 8h + 4h + 4v +
we
Ts,,
0
VgEG sg--g Proof.
Therefore
t..
=
+ tm*n-
c
2+
h
2
(41
=
+
c
+
h)2,
obtain
4sg"-g By (2.22)
right-hand
the
side
of Bh.
The structure
161 + 8c +
=
4fT
exactly
is
The purpose
C2.
of this
section
following
the
is to prove
result:
11, h,
Theorem 4. Our first
v,
step is
to
ul
U Supp(c)
< B.
Jul
show that
=
3, where
u
is the
defined
element
by
(2.22). order
In
show that
to
the
of
case
Arfu
=
1 is
impossible,
we
need the
following Proposition
Let
15.
k, 1,
elements
tains
m,
(A,B)
be
k2
=
n m-
where x
Proof. k
1,
mE
BO,
c
First
ki
==
+
B and either
then 1
we
T1,
ki
m and
=
k (E B
11, 11
that
note
G B.
standard
a
table
algebra
which
81 + 4 k + 41
=
61 + 3 k +
or
k
=
U Supp(k)
AWy-k
is
follows
directly
ISupp(mk)l
2.
It
integral
con-
such that
(2.28)
1,
kj + 71, ki is
a
correctly from
table
E B.
subset
defined
(2.28)
that
for
Iml
If JxI of B.
> 3
each
Ill
y
for
each
E
B if
8. 6, Jkl 4, this implies =
-
Together with Amkm BO\ Iml. Using the identity 4Ta + /-tn, for a suitable Tnk n (mk, mk) 4m + 3n 8, i.e., mk (k 2, rrOM) and formulae (2.28), we obtain M 3, Inj 8. with n C B, Inj If
k c
B, then
=
=
=
=
=
=
ITA with
2
If
2,
k
=
ki
Tnki
:!
(2.28),
mulae
=
n k-j
holds
it
cases
(Trz7k-j,Tr:k-j),
+ tt% obtain
Using
-
M
nk
E B
n
or
n
for
every
According
C-
=
12.
Therefore In2j. (k 2, m7mg) and for
=
-
4 and nj
=
Milnil
n2.
Thus in
-
both
=-
Ini
3n,
(2.29)
8,
=
G B.
0(mod 4). There-n(mod 4). By (2.28), V 0(mod 4), which, in turn, implies that A,,7n--- E
(k2, rrn)
32/\n:n-k
64 +
JnJ2
An-nk
+
Inj
-
7,
=
Jkl
24An7n-1.
10, 41.
A,,7n-1
If
29
E B.
x
(2.28),
to
=-
4m +
=
Since
/\rrn-k
Ind
=
n2 E B and
A2,1nil (mk, mk)
=
identity
+ n2, ni
ni
=
mk
write
can
0(mod 4),Ti-n
=-
10, 4, 81
4
that
where either fore
/-tl
the
3, Inil
=
mk
Now we
Degree
=
where nj,
2m + /-12n2,
=
=
we
of
=
(mki,mki) 4m'+ /inj
=
Element
Nonreal
4 which implies that ki E B, then 1kil jSupp(rakj)j Together with Anikim A,,i-k 'M 2, we obtain
2.
2,rn + pini,
=
Since mk
Tj-,
+
jSupp(mTkj)j
Faithful
a
=
8, then for each
A,,7nxlxl
=-
0(mod 8).
implies
jxj
<
2,
Thus An7nk
a
3
x
:5 8 Supp(nWn) \ 11, 11 we have. A,,7n-,;Ixl 8, which together with A,,7nx Ar,-xjxj
Cz
Therefore
=
and
!
4
contradiction.
1\7M-1
E
10, 41,
and
we
following
have the
(0, n7n)
possibilities:
ArMl An7nk
64
0
64+96
4
0
64+128
0
4
64+96+128
1
0
741
1
4
constants Of Ankx) product (nk, nk). Since all structure is 41nkj 464. It 4, the minimal value of (nk,nk) 4 and (nk, nk) 64 + 96 + from the above table that Arni follows ArMnk 4 and Ankn Furthermore,
Consider
x
(=-
now
Supp(nk)
are
the
least
at
==
=
=
=
=
1:
AnkxjXj
=
(2-30)
32.
xESupp(nk)\jnj Denote X
=
f x,
e
Supp(nk)
\ f nj I /\nkx
equivalent (2.30),is 4 28+4 21XI (nk, nk)
Then
=
41,
Y
=
41XI + 81YI 821yj. Combining
to
+
=
=
fx
G
32.
this
In
Supp(nk) \ f nj I Ankx addition, with
the
=
81.
64 + 96 + 128
previous
equality,
128.
Z. Arad et al.
30
JYJ
obtain
we
namely: Together
U
=
with
Proof of
=
1, JXJ
111. Iml
6. But there
=
Therefore
n
Theorem 4. Let g E G be
sg3-g If
Arf.
=
=
u
b, b,
But
Jul
2 =#-
Ff
=
1. Now the
arbitrary
an
14
Arf,,u.
finish
to
Supp(d-c)
that
implies
u
2
?
?
IV!
3c
?
?
proof,
table:
V
81+4c+8u3c?
we
? ?
1?
need to settle
St(c)
subset
coefficient
? all
question-marked
consists
of basis
8. Therefore
St(c)
elements
cases
from
11, ul which,
31 + 2u.
applied
law
hv
=
product
to the
3c + 4u +
h 2V in two ways
Computing
u
2h+2v
in c7c with
=
The associative
C
2h + 2v
u
The closed
appear
c
c
the
above table.
8u.
=
following
have the
In order
turn,
becomes evident.
By Proposition
81 + 4c + implies that C2 incomplete multiplication
3. Now (2-22)
=
we
h
in
1,
1,
h 41 + h +
the
=
Supp(V).
then c and sg satisfy the conditions of Proposition 15. Hence each g E G. Now it follows from Proposition 11 that I GI :! 2. distinct h are pairwise elements of G, a contradiction. Hence Arf.
Thus
in
element.
61 + 3c +
=
claim
m c=
for
*
sg
m=
JUI
U c B with 4 ==>
=
implies
6, this
=
is a unique subset By (2.29), Akkm
k.
=
gives
us
v
cv
hu
==: -
=
h2C yields =
v.
6h + 6v.
Furthermore UV
V2 Now we may
=
complete
41 + h +
c
c
2h+2v
v3c+4u+v
Remark I
If
table
subset
f 1, h,
table
of 11, h,
way.
This
=
3h + 2v,
C
U V
V
3c + 4u +
81+4c+8u
3c
6h + 6v
3c
31 + 2u
3h + 2v
r -6h+6v
13h+2vl
B, then Table 2.27 gives If c % B, then c v, u, cl.
v, u, cl,
=
Tl-}
to.Table
may be. reconstructed
2.3.
v
(2-31)
l2l+9c+3h
G
c
leads
u2h
121 + 9c + 3h + 8u + 2v.
2h + 2v
V
U
=
=
the above table:
h
hl
u2 h2
the cl
multiplication + 71-
from
of the multiplication Table 2.27 in a unique and the
table
ITA with
2
Bb.
of
The structure
gh Therefore
sg
According
(89,g- g,C2)
=
exists 4
=
=
is
sg,
gP(h)
easy to
closed
imply
we
that
to the
a
E
Proof. an
Since
tg
(G,
*,
mapping
+
4rg.
=
b.
gBh
from
polynomial
=
h with
in
(g*f)Bh.
Thus
Moreover,
the
sg 7 rational
tg, rg I
U
UgEGgBh is
Ig,
:=
may be
coefficients.
quotient
algebra
where y
c
U/BK
a
is
h).
W:
written
as
U -+
proof
yP(h)
G & Bh
h & P(h) y &
the
(2.32)
2sg
+
if
hP(h)
as
Y
if
=
11
U G\
jhj.
follows: 1
Y rz G
(2.33)
111.
of Theorem 1.
16.
and injective; well-defined is an homomorphism. algebra W g & v,r o 9 gw g 0 h,s o 9 g & s,ff 9 =
=
=
=
g ot
1, then our claim is evident. (i) If in (2.33) y We have to show, that gP(h) element. arbitrary is positively definite, =
=
gP(h) '
=
W is
gP(h) But g7j
3g + tg
a
=
2sg
g +
U may be
below finishes
Proposition
be
:5 2. Therefore, =
=
contains
(yP(h)) '
(i) (ii) (iii)
(sgc, sgc)
Therefore,
jSupp(sgc)j
4 and
each element
that
group x
may define
The claim
tg
61 + 3c + 2u.
=
=
P(h) is (gBh)(fBh)
of B that
Each element So
By the
=
where
see
subset
isomorphic
+
=
These formulas It
3g
=
h 1.
G
(=-
g
31
4
directly
hsg htg
as
have
Degree
-
hg
written
element we
of
=
other
Ig,
=
2sg; sgh
g +
Element
4sg + Arg, A E N. rg 0 sg and sgc 192: computed from the equality (s g T_g' c 2) 6. Thus Ig,sg,rg,,tg} C gBh, implying IgBhl > 28. Irgi 28. Finally hand, Supp(gg-) C Bh, therefore IgBhl jBhj have the and we equalities: following 1, tg, rg
==>-
On the
gB h
E
A is
The coefficient A
=
to
'
there
an arbitrary Proposition 13,
gBh 14, sg"g-g Proposition 192. Furthermore, \,.,c,_, a unique rg E B# such that
t9
7
Nonreal
Take
of sg, g E G and
definitions
Faithful
a
=
0 <=
(gP(h),
gP(h))
=
0
<=
Let =
now
(gT, P(h)P(h))
=
jhj
g E G
hP(h)
0
=
0.
hT. Therefore =
Here s, v, t
0 * are
(hT, P(h)P(h)) the elements
=
defined
0,#
(hP(h),
hP(h))
in
Subsection
2.2.
=
0 <=>
hP(h)
=
0.
0.
Z. Arad et al.
32
(ii)
Let
nothing Let x
to
xy
If
=
7 F,
x
y
gBh) Y E fBh, f Q(h) for suitable
E
now x
gP(h),
=
arbitrary
x, y E B be two
=
gf P(h)Q(h)
=
(g
f
*
the
other
h &
(h2)
If
-
=
(41 + c + h). Using (2-32) we obtain
two
Bh then there
x, y E
,
(g
=
=
(41
+
h)W
+
c
implying
7
h
g E
G,
(g
*
is
elements.
=
Then
Q[x] implying
cz
f )hP(h)Q(h).
h2p(h)Q(h)
f
*
arbitrary
P(x), Q(x)
*,)P(h)Q(h)
+ sf
* f Bh and (xy)w then (xy) ' xPyw. If x x 0 h, T'P hand, xP
=
(iii)
Bh be f polynomials
9,
then g
hQ(h))
elements.
prove.
(g (2) hP(h)) (f (41 + c + h). On (x (9 h) (Y o h)
=
h (9
xPyw
h &
sg
tg
for 1
=
rg
(h
2
02C 1(v 4
=
=
-
-
=/=
g
that
I
1)g; 1)g; h
-
.2 2
-
21)g.
+
Therefore, SW = g (g (h-I)h 2 9
tgw rw 9
Proof
2.2.3
In this
g 0
=
=
Theorem 5.
g (9 v;
'go(vh-h2
-h-2
4
2
+2h)
is based
If
on
=
(2.34)
following
the
b12
41 +
following
the
satisfies
b
b satisfies
that
we assume
(2.34)
'
result
There
17.
df
Cf
61 +
=
=
by (2.9)
2b12 + 41 +
2
121 +
Eg =
121 +
f
C B such
AUfklkl
b12 + AUfkk,
df b 12
Cg
jk, 1, ml
exist
Arf.Tn7
AdUll, 7
2(Uf
=
+
b12
?d
-
=
E:
Td)
+
2Arfnm
+
+
real,
=
6;
f 3,4,6,121;
4Arf,,,Tn +
then
[d-jj
(2.35)
12;
0 b121 Arfniml
+
b12
=
is
that
m
1, Ill
-[25].
of H. Arisha
defined
and d
equation:
-
[14,34]. Proposition
=got.
of Theorem 2
subsection
approach
1))
-
I;bOur
S;
g
(h(-2 2
+
(2.37)
Ad-d1l.
f.
/\rfm (mb12)
(2.36)
(2-38) (2.39)
+
2AdUl
(2.40)
ITA with
2
Proof
1, this
implies
12A'jfbl2 k
case
bl2)-
=
Proof
Arfb12
2. Now
PE2
(d
=
Comparing
b12
=
Ef
AUfkk
+
for
Degree
Together
3b + g.
some
12Arfb12
3b + g, 24 + follows.
12
33
AEdb
with
(-df b'G)
<
7
allow
48
=
=
=
,
(we
k GB
(bb, ff)
=
4
the
implying
(2.38). (2.36) that
+
2f) (
this
2?)
+
=
dU + 2d?
+
2f d-
+ 24 +
8b12
+
4Af7f
..
m.
with
( -b)2
2
8b12 + b127
161 +
obtain
we
2
b 12
Consider 1 in the left
an
d
=
arbitrary
hand side
2dT
+
+ 81 +
Supp(d3) \ 111.
1 (E
(2.41).
of
2fd-
+
Then
A111
=
Ad-djIlI
implies
turn,
proof,
it
is sufficient
Proof
of
Multiplying
R
equality -
(d Now (2.36)
-
coefficient
0(
mod
12), Arf,111
by
Ad-d1l. (2.41).
Clearly
1. To finish
E,
obtain
41 +
1
of
hand,
12
which,
in
the
(2.40).
-
the
=
dU into
substitute
to
(2.39)
Thus dd
12.
AdU1111
=:
the
On the other
is divisible
number
=
=
A denote
Let
0(mod 12).
=
(2.41)
4Arf,,,m.
Since AJ1111
All IlI(Ad31 +2A3fj+2Ad-fj)+4Arf,,,ImIJ,,,j24, the 0(mod 12) and 4Arf,,jmI
by
3b + g
=
-2
-
+
2f)f
=
b f
+
2Af -f-,,,m
we
3bb +
=
bg.
implies
Eg as
of
consequently,
and,
=
easily
=
Element
-
from
follows
R
Since
(2.36)
12A- dg
3f
and
Nonreal
(2.9),
to
12 +
Agfbi, .
(2.37)
of
Proof It
(2.36)
of
According
(Ed, R)
Thus 1 <
.
=
(2.35).
of
Proof.
Faithful
a
=
Uf
+
b12)
desired.
Equating
(bf)(bf)
9bb +
=
3bg
+
3Eb
+
Cg
with
(6-b)(Ff) and using
Proposition
the
If
(41
(2.38)
equalities 18.
=
1
=,A b12
+
b12)(61
and and
(2.39),
Ill 0 3,
+
2b12 +
ArfmTn)
obtain
(2.40).
we
then f is real.
0
Z. Arad et al.
34
Proof. According by Theorem 5, Ill
to =
( .37),
Ill
3 which
13,4,6,121.
G
contradicts
Ill
If
14,61,
E
then d is real
Thus Ill 12. assumption. implies (bf, bf) Cbf,R) If _7 [bf] [43, 121],
the
R to (2.9), 3b + g which G [bf] j[122]'[62,62]'[43'121]j. b2f bf =- 0(niod 2) which implies df 0(mod 2). Thus [df] E 1 [1221, [62, 62], [34, 621, [34, 34] 1. It follows from (2.35)
According
=
Therefore
=
(df, df) First
show that
we
48,
k
if
12 (1 +
:7 b12. Indeed,
k
(dU, Cf)
hand,
=
=
48.
thein
=
(df, Uf)'=
=
and,
=
if
k
'\Ufk))
if
k
=
=k
b12;
b 12
-
(dU,Cf)
b12 then
=
that
On the
48.
=
m and 7 b12) 12Arfl. Aff, Ill 3, contrary to the assumption. Iml Thus k 0 b12 and (df, df) 12(1 + AUfk) with 1 < AUfk :! 4. This implies b12 + 3k. Thus (-df, if) 3, IkI 4, i.e., Uf [df] C- f [122], [62,62]1 and AUfk 2. Since 10 b12) 1 48 which implies 2, Ill 3, contrary to Arf, m, Arf', the assumption. Thus [bf] [43,121], i.e.,
other
Arf,,,
=
2. But in this
=
24 +
Hence
case
=
Since
2.
1
1
=
=
=
=
=
=
=
=
=
=
bf If 'rn4 that
then f appears M4
:7 -
Consider
V4 (=- B with
some
twice
the
IV41
=
3Tn4
=
t2
in
and the claim
M4-
product
4. Thus
we
each
For
Therefore
equality M4 c
may
of
(2.43)
(41
+
T,
by
b12)b
(bb, dd) Supp(bd) follows
=
that from
Multiplying
implies
both
fE, FI-21 V4
0
which
=
2f + V4 for
(2.44)
=
obtain
we
Ed
+
-bf.
implies
M4
E,
=
Ca
B12-
0(mod 12).
-=
ISupp(Ed)l
b + e12, e12
=
2AEfxlxl
=-
=
If
V4
Now the
2. =
contradiction.
U,
then
Hence
we
d.
(2.44) sides
Ed
may assume
we
"Em4
V4 + 2f.
Supp(-bd) \ JbI, we have /\b3.2bxlXl 0(mod 12) and, consequently, A-gdxlxl 16
So
2,
(2.43)
E
x
assume
It
sides
=
2f7
d +
EM4 both
follows.
A-g,,,,,f
Since
have
b2
Multiplying
(2.42)
+ M12-
( b-,
that
of
Tn4W4--)
by E
(2.44)
-jM4
=
=
and
28. Ther efore
comparing
Tn47ff4-
modulo
2,
=
we
41 + b 12
obtain
bV4(mod 2).
in dM4 are ones. (dU, M4774) 16, all non-zero coefficient bV4 and (dV4, bTn4) (UM4) EV4) 16. Since (bTn4, bTn4) 28, 16 bm4 X4 + 2X6 -===>4AdV4X4 + 2AdX4x66 The latter equation has solution 1 which contradicts a unique the fact that 11 XdX4X6 AdV4X4
(UM4, UM4)
Since
Hence
UM4
=
=
=
=
=
=
=
=
.
=
6AdX4X6
7-
Starting
=
0(mod 4) from
now we
El
shall
assume
that
Ill
=
3.
ITA with
2
Proposition
If Ill
19.
Proof.
Since
ISt(d)l
:5
3, d 4, 11, 11
Ill
=
Nonreal
41 + 41. Therefore
=
Degree
of
Element
4
35
12 =31+21 andm=l.
3 then
=
Faithful
a
dl
3d and
=
11, 11
C
St(d).
Since
31 + 21. B, i.e. 12 df b12 + k12 Assume now that m0 1. Then (d , Cf) ='24, and, therefore, both sides of the of 12. element basis for some Multiplying k12 =A b12 degree b121 + kl2l. Thus jb12, k121 is an 11, 1}latter by 1, we obtain Udf equality that from (2.41) of b12, k12 It follows combination coset and b121 is a linear I and b121 4 and, therefore, b12 + 2k12- Since b12 and \Ib,2b.2 /412bl2l ?d 1 are real, b12 + k12. Hence k12 is real as well and Uf
Idl
=
St(d)
=
is
table
a
of
subset
=
=
=
-
=
=
=
J)
(d 2, f2) =(df, If
d is not
then
real,
d2
Hence d is real
diction.
=
g
=
Since
(d 2, f2)
=
\ffv,
16,
a
contra-
41 + 41.
=
(2.39),(2.40)
follows
as
2b12 + k12 + 2Arfm, 121 + b12 + 2k12 + 81 +
(2.45)
Arfm(mb12)-
8 and Therefore 1 V Supp(mb12). Argi fbl2,kl2l7 Supp(lb12) 16. that which 12 24 Together with implies Igj1, 111 1) B of degree suitable for E a obtain we W4 19,W41 gJ1,11 gJ1,11, =
1 +
g EE
4.
=
V4 E B and 24 =.
and d 2
Now we may rewrite
gg
4V4)
24.
=
=
=
=
Clearly,
9
W41. Now we
=
99
Since
W474 either
=
can
write
A W47-12 4
=
W474 --(31+21).
b12 appears in g-g, either b12 EE W4T4_ or b12 E lW4U4- In the first case, 0. Thus, in any case, 41 + b12- In the second one, b121 nW4T4 W4U4 41 + b12 or W4T4 41 + k12=
=
=
In the first
case,
Cg
=
(41
b12)(31
+
+
21)
5b12 + 4k12 + 81
121 +
=
and, therefore,
,\q-,,,(Tnb12) In the
=
4b12
+
2k12-
second case,
Cg
=
4b12
121 +
+
5k12 + 81
and
,\rfm(mb12) It
follows
sible.
from
In the
first
(2.38) case
that we
/412mb12 obtain
=
=
that
mb12 W474
=
=
2.
3b12 + 3k12-
Therefore,
/\rfm
=
2, Iml
2b12 + k12) 41 + b12-
the second =
3 and
case
is
impos-
Z. Arad et al.
36
(2.39)
Now
implies
Eg
2b12
=
k12
+
+ 4M-
hand, the equality Eg EW41 implies b12 E EW41. Thus EW4n equivalently, bW4njb12k12j : 0. This implies that all coefficients in EW4 are ones. Therefore, CbW4 EW4) 16, contrary to (6-b, W4T4) 28. 0 Since '( f, 48 and f f) (dj, f?) k b12 + A3fkk, -either b12 or 4. In the latter 2 and kd jkj Akdf for some AUf k case, '3, V4 + 2f, On the other
b121 0 0,
=
or,
=
7
=
=
=
=
X4 E B.
=
Now we
write
can
Cbd, J) All
coefficients
in
Ed
are
Cbd, Ed)
since
ones,
E
24 <
b 2) > 24.
(dk,
=
A kgxjxj
Therefore
16.
=
<
16,
mESupp(_bd which
is
impossible.
Thus there
the formulas
may rewrite
exists
a
(2.35)-(2.40) dJ
unique possibility
_Gf
2b12 and
we
follows:
as
41; +,21; Uf 2b12 Cf 61 + 2b12 + 21; 121 + 8b12 + 121; b12b12 Eg 3b12 + 41; 121 + 9b12 + 81. Cg 41 +
12
31
(2.46)
=
=
=
It
closed
follows subsets
from the above equalities of B.
Proposition
tegral)
where B
Proof. finite,
is
20.
which
m
a
E R and a,
H(m, n) Consider c
is
(A, B)
Let
contains
where the
of finite
b, n
be
a
faithful
standard
element
mi + ma,
m(c
C of B: C
M
-
M,
(not necessarily
in-
d),
Since
n.
algebra
distinct
with
(c)
U a (c)
Icl
=
1,
Then
c.
1,
1XI
+
pairwise
of
subset say
table
are
b such that
6_5
are
1, 11, N := 117 1, b121
M
b2 c, d E B is the order
following order,
that
1,
(c), a(c), G b(c),
X
E
x
E
x
U b (c).
Since
B
ITA with
2
(c),
the sets
Indeed,
St(b)
a
xb, =
b (c)
(c),
xa
distinct,
pairwise
c-'d
c-'b,
and
i.e.,
ac
77
c":: 7 a&:
7
I b (c)
n.
that
i
=
I
-
are
=
-
following
(m
-
1)
1 + (m
-
2)
a
holds:
table b&
(m (m
b&+f'
1)c0'+0
-
1)
-
(m
+
-)-a-&--+
7
2
-
C is
(m
-
I) b&+#
m&+f'+'
bc+#
shows that
table
of B. Since
subset
table
a
+
mac"+O+
b is
faithful,
C
B
follows.
and the claim Consider
(m 1) b, a2
ab
ac'+O
[FC-T. b&:1-7, This
a.
=
37
assumption 0,..., n
the
ac , bc ,
elements
that
the
I a (c) I
that
from
4
n.
=
Therfore
follows
ac"
C
7
lb(c)l
=
assumption
=
If
Degree
of
Element
Weclaim
oint.
(c).
E
x
Nonreal
Hence the
=
from the
It follows
each
111. ja(c)l
11, al, St(a)
disj
pairwise
are
for
E B
Faithful
a
algebra
factor
the
According
by attaching-.
B
B/M.
=
We denote
the
images of
x
E B
to the definition 'L
(bM)+
IMI and
lbl It
follows
from
(2.46)
(2.47)
and
b12
that
1 -Zbl2) Jb121
=
Idl 1d, 4 I
(b
-Z
(2.46),
to
we
n
structure
of
as
algebra
=
41 +
4b12;
4d +
47f.
the order
&
B from
generated
(a)
o"
follows
from
following
20 that
Proposition
by
is
isomorphic
:=
the
CC,+,6
table
multiplication
o,3. o.,
oa
0"+'6
e,,,+,a
e,q e,,,+,3 3
3c,,+,q 3o,,,+,3
+
the
c".
l ea
ca
H(4, n)
a
form:
c'3
to
C';
12C
ea
has the
3.
of d in B. Our purpose is to reconstruct First some notation: we introduce c"
It
4, lfl
=
subset
20, the table
is defined
bl), lbl
+
obtain
2
where
4;
1;
=
bb
By Proposition
(2.47)
-
-
-
Applying-
JbJ2 ( b, M+)'
=
2e,,,+,3 .
3e,,,+,3 4c,+,3+1
+
4e,,+,3+1
of these
elements
Z. Arad et al.
38
where the addition The next
of indeces
result
Theorem 6.
is done modulo
establishes
Let
G
a
the structure
Zn be
arbitrary
an
n.
of M-cosets. element.
Then
I.
SUPP(c") where
ct
B and
E
Supp(o,,)
2.
lot,
=
Ictl I
o;
f Ct},
a
0
a
0 0,
4;
=
ot,
where
lotl
E B and
o
=
4, lo I
Ce
=
0
a
=/= 0,
12;
=
3.
SUPP(e.) where
et,
E B and
e-
Je+J
Jb121) f et, e 1,
=
6, le- I
=
6;
=
4-
bo,+
If
a
+
a
2e+ a+ 1,
a:
U12; + 2e+ a+ 1 Ce+ I
=
bc,+
(i).
b122
c+ a+ 1
=
a
3c+
b,ot
+ 0 ;'
+ 4e-a+ 1,
a
=
0
a
=6
0.
0, then the equality
=
0.
41 +
bo-a
Proof.
41 +
Supp(co)
a
=,A
f 1, 11 is Supp(c,), J-; l then IxMI linear, Supp(x-x) =
of cc,. For arbitrary x E quence of the defintion =. 4. If xM contains more than one element, JxMI B has
since
impossible,
linear
no
element.
lxJ
because
=
2.
If
a
(ii)+(iii)+(iv). a
:7
0. Weuse induction,on
Clearly
12.
lot
Since oj7; 41
or
o ot
=
a
=
C
28.
with
Jul
=
u,
v
E B
to
is
;
0, then Supp(oc,)
Supp(oc,)
Let
=
=
lot,
direct
a =
=
I.
conse-
Therefore
2 and
JxJ
=
C M which
2, is
lb, gj. Now assume that 4, lo j o I with lotl =
0-
f 1,
121, oto,+,,
ot
M
=
Therefore =
1, 1, b12j.
C'
b12- In the first
41 +
o,+ 1, contrary ( b_, otot)
=
a.
Since
(2.48)
0.
4, JvJ uu
a
a
and
=
ai
otot
=
41 +
=
=
2
b12
EE
O(mod 2).
=
a
-
Thus
6. Now we may write
bbolot
either
3ot consequently o1M 41 + b 12 and (bot, bot) ot ot u + 2v for suitable [4', 6 2], i.e., bot
case,
I ot, o; 1.
[botj
oll
Therefore, =
=
ITA with
2
Combining
obtain
that
Together
tru
Supp(6-bojo )
9
41 + 41. Therefore
=
Supp(trv)
with
Nonreal
Element
Degree
of
4
39
with
this
Supp(tru) we
Faithful
a
and
u
Supp( -botot)
9
11, 1bl2j)
E-
belong
v
J1, 1, b121,
=
M-cosets.
to distinct we
obtain
uUv
=
2b12-
Thus
(41
+
b12)2
Now (2.46)
b ojot
=
that
implies
(u
=
vU
uM
Since
Jul,
=
Since
ju-j
=
consequently,;U
and 2
12 are
that jvMj jvMj, then there jxj jvMj :! jxjjMj,
v1 less
of
v
=
degree A
four,
=
2b12-
=
+
8b12 + 4'v-v.
31;
(2.49)
4. Since
2 and v1
vM
both v
=
3, ii
=
Jv,
=
x,
A and /-t
are
6 < 12
jxj
yj
Avjv
that
+ 2x +
=
with
E vM
(2.49)
from
Ivi
because x
consequently
exactly
o0o"
=
1, le,,+11
=
=
c,,+I,
=
JvMj.
12.
=
follows
-12ot
product
jvMj < jvMj,
element
an
=
41 + 41 +
1c,,,+11
1
A+ p
p
=
4e,,+,
with
2. Indeed
It
=
in the
appear
exists
3.
+ Ax + py with
than
uv
3 and
=
2-v)
2b12 41;
Therefore
=
elements
41 +
+
e,,+l.
<
=
=
1 and ii
Weclaim If
tru
4c,,+,
o0o,,
61 +
=
+
2b12 + 21, and finally
61 +
=
vV
4 times.
2v)(U
+
Since
3.
:!
where x, y E B 1. Therefore
=
even
2y. Purther,
and
the
strictly
inclusion
2 and 31 + 21. Therefore Axlx 11, 1, b121 implies that x-T obtain vM > Thus contradiction. we a I I 41, yielding consequently (x + v)l vM Jv, wl for some w E B of degree 6. We set 2, i.e., Supp(e,,+I) e++, := v, e-+, := w. In these notations, bo+ u + 2v c++, + 2e++, which in (2.48). is, the first equality + e+ that le,,+, I implies Now A,+ + 2ea+ Together with a+, oo+1, we obtain
Supp(x-x)
9
=
=
=
=
=
=
=
=
=
a
ce
bo-
=
a
which
is
Since
merely bca
c+a+ 11
+
the second
00c"
Supp(o,,)
=
the
jo+,
o-0,
of
the third
equation of (2.48). we set follows,
=
3c+a+ 1
+
2e+a+ 1
+
4e-a+1
(2.48).
of
equation
0",
elements
=
2e+a+ 11
bc is a linear combination 16, we 1. Together with (6b, c ct) product
=
of the
obtain
In what
e+ 0 Since
'5-,
=
;!7
c,,
o-,,-,
implies
:=
=
that
b12 C+' 0 7
c-,,,
01
=
which =
1,
0
=
implies
O+c, -1 1
C-
1,
e-0
that
and oa
=
0-
c+c,. Analogously,
ct =
o-,-,.
It
follows
from
Z. Arad et al.
40
7;
that
e-,,
=
the function
either u :
Z,,
1-4
et 11,
eta
=
e-,, follows
ec,
7
11
-
as
17 follows
directly
Proposition
1,
from the definition For each
2 1.
bel
a
=
be;
3ol
Proof.
follows
It
bc+
=
the first
case,
be+
A
+ o-, a
=
=
et
be+ Cz
we
l(et
equality
Now the
(2.49)
from
ol
a
(Y
If
+
Oz
be,
and
=
2o-,
+
o; if o-(a)
2o,-,,
o-(a)
if
1;
(2.50)
-1.
(2-51)
a
3(e+
=
in the
4 which
implies
=
a
=
OL
a
a
a
this
3o+ + o3ol + o; in
=
a
one.
Together
with
oa
o-"'
=
we
o-(a)
Analogously,
-1.
=
be+
be
that
Combining
24.
or
a
0,
a(a)
that
=
2o;
=
a.
second
4.
=
(bu, bv)
be+
e-) implies
+
0,
2o,-,,
=
a
either
that
a
3(bl).
3b +
(bc+, bel)
that
e-)
=
e
=
obtain
I in
the 0
.
.
orthogonal
idempotents
:=
31-1 ;
4
e:=
-4
;j
associated
with
the closed
+e
jA(DeA. Since jA --- A/M, the structure algebra eA. the algebra eA, we define the following set
of A is
=
if if
a
is
0; odd;
if
a
is
even.
e, X,
.ieo+ 2
=
1
following
formulae
I
2 -'
+-
e6a 2
The
that
M:
Thus A In
holds
it
=
1 +1
the
o-(-a).
=
0,
3ol
=
bb12
=
A 0;;-, ' t
then
3
by
o-(a)
:
be,-,,
obtain
3o+ + o.case a There are two pairwise subset
e-,
Wedefine
0, then
be+ 0 with
a
o,-,,,
+
ea
e+-,,.
6--a) e,-,
=
et
that
Z,,,
E
a
2o;,
bel ff
if
e,,,
et
if
-
It
or
=
7
show that
the
a=
elements
defined
completely
of elements:
a
x,,,
E
Z2n)
a
E
Z2n form.
a
basis
of
ek
b12e a
:7
=
following
=
-e;
eel=
0
Y2a+l
Wehave the
0; el
=
eo+ a
=
"
Y2a, eea
2
3o, -o;; 4
7
e0a
=
-Y2a+l-
equalities: o+ a
=
o+j a
+
o+e, a
=
o,,,
+
2X2ce+1;
-Y2a;
(2-52)
ITA with
2
o.
j
o.
=
+ oce
ee
+
eaj
+ eae +
+
cO
eA is
that
To show
odd a,
isomorphic
this,
we
then
-1,
2
2e,
=
2ec,
e
-
If
a
If
a
72)'
4
(60+)(eo+ 0
by Theorem
=
then
ot-,
=
9
I
In
both
y1yo
eA
=
=
cases
yle
=
a
0;
a
0;
group
we
2
2
ylyc,
and yly-1
=
=
4o-(a/2)
f 3eo:/2
for
yc,+l
-1e6b 4
by Theorem
-
that
obtain
Thus
ebot-j.
1
=
e
=
ce
yo.
6 +
if
eo,/2,
t 2eoa/21
In y1y,, y,,+,. have that y1y,, =
always
if the
case
yc,+l,
of
u(al2) a(a/2) a
i.e.,
y,,,
=
=
=
0, =
1; -1.
we
have
(yi)0'
CC2nDefine
the function
sgn
sgn
Now it
An(sgn ).
is
easy
to
of
even.
2
then
=
cv/2
we
yl.
o;
+
1
ebe+
a(a/2)
0;
c.+,
=
2
Yly.
7
xO.
-
a is
Finally,
E
and non-zero,
even
=
a
6
4y,,,+,.
2
-1, is
0;
obtain
2eet+j
=
41
is odd
a
x,
2
2
3co
4-
CC2n where C2n is the cyclic
xC"
+
ebot-,
=
4
0;
=
a
2X2a7
-
Degree
1;
C1
CO + X0,
=
to
bo.-,. and
2X2cxi
+
of
set
we
MY.
If
=
e
13j
=
YU
Now for
2-12a+
-
0,
COL
2n.
3o,
=
a
ej
ea
order
e
Element
4eo,
+
ea
We claim
Nonreal
Faithful
a
see
:
(a)
that
1, 11
Z2n
xc,
as
follows:
1
a
is odd
o-(a12),
a
is
sgn(ce)(yj)c,
even.
which
implies
that
Bb
and
Integral
Standard
3
Arad',2,
Z.
Element
Nonreal
Faithful
Biingerl,
F.
Fisman',
E.
Department of Mathematics Bar-Ilan University Ramat-Gan 52900, Israel Department of Mathematics Netanya Academic College Netanya, Israel
2
Table of
with
Algebras Degree
a
5
MUZyChUkl,2
and M.
and
Computer Science
and
Computer Science
Introduction
3.1
chapter
This
(A, B)
with
faithful
basis
deals
L(B)
with =
the
11}
and
b of
element
Jbi
for
> 4
degree
all
b E
Starting
5.
GT-algebras integral BO which contain a nonreal
of standard
classification
from this
using the basic
point
identity
A,,y,,Izl one
can
of basis
list
all
=
possible
elements
(cf.
(xy, z)
(x, z-y)
=
of
representations Tables
=
and III
II
\-,-gxlxl,
x, y,
6-b and b2as
of Subsection
z
E
B, combinations
linear
3.3).
Assuming
that
T yields the identity (Cb, Cb) (b 2, b2) which reduces the number of these representations (cf. Table III of Subsection 3.3). Then, using of the associakinds of techniques various (for example repeated application In order treated be will the of cases each separately. remaining tivity law), table base of a specific the following we introduce the main result, to state b commutes with
=
algebra: Example 1. The homogeneous table 11, b, b, b2, b2l whose multiplication
Cb
is
algebra of degree given by
2(b2 b2=3b+2b2 =
51 +
+
72)
72) + b2 Tb2=2(b+T)+b2 bb2
=
2(b
+
b272 =51+b+T+b2+T2
b2=2Cb+72)+b2 2 is
simple
and all
elements
of B#
are
nonreal.
Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 43 - 81, 2002 © Springer-Verlag Berlin Heidelberg 2002
5 with
basis
Y5
Z. Arad et al.
44
The status
of recent
Theorem 1.
If (A, B)
and
JbI
> 4 for
of degree
(i) (ii) (iii)
b E
=
51 +
V
=
11, al
bb
=
51 +
bb
=
51 +
Y5
or
is
given
is
standard
a
B# which
(6-b, 6-b)
5 such that
bb
to
all
research
integral
contains
65, then
4a, where jai 4; and B and a2 41 + 3a. =
following
in the
table
a
one
algebra faithful of the following
L(B)
with
nonreal
exactly
is
main result:
basis cases
isomorphic
to
=
III
element
b
holds:
Z,? V, where
=
2(b + b), and B is exactly isomorphic to Tr,,(5). 2(c+Z) for a nonreal c E B\jb, TI, and B is exactly Tn (5).
isomorphic
Under the hypothesis of this chapter, Table III in Subection 3.3 displays 16 of 6-b and b2. Theorem I treats possible cases for the decomposition comCases 1-4; Subsection 3.4 resolves the particular pletely instance of Case 5 where jb3, b4l 1b, TI. The remaining 13 cases are still open. Three of these realized in the algebras cases are of the form 0 (G, H). To show that consider the following where m > 1. The symmetric 'group group F := Z5 on the elements of F by permuting S,5 acts naturally the coordinates. Thus S5 < Aut (F). The subgroup D : I (f, f, f, f, f) I f E-= ZmI is S5-invariant. Therefore H := FID _-- (Zm)4 and the table algeS5 acts on the factor-group bra 0 (S5, H) is correctly defined. Let G < S5 be a subgroup which contains a cycle (1, 2, 3, 4, 5). Then G is one of the following groups: =
M,
=
algebra B degree 5, where
The
Go
=
G,
=
((1, 2,3,4,5)), ((1, 2,3,4,5),
=
=
A5
G4
=
S5
O(G, H)
T ,is
a
=
5
(5,2) (4,3)),
IGI 1
contains
a
faithful
=
General
facts
Since many make the
partial following
Hypothesis and there
Jai
> n
-
1
is I
for
all
E a
b
=
T of
of
table
=
G1,
algebra
then
we
of
have
and known results
results
(A, B)
an n
element
non-real
(1, 0, 0, 0, 0) + D E FID. standard homogeneous integral
G-orbit
If G Go, then B is a degree 5, which corresponds to Case 13 in Table III. If G Case 11. The groups G2, G3, G4 correspond to Case 14.
3.2
10
((1, 2, 3, 4, 5), (1, 2, 4, 3)), 1 G, 120
G2
G3
:=
IGI I
is
hold
a
in
a more
standard
N>3 such that B E B#.
general
integral contains
setting,
GT-algebra an
element
in what follows
with b
L(B)
of degree
=
n
we
Ill and
SITA with
3
with
Westart
If
Lemma1.
NB\101,
E
u
(c, UACI
then there
=
is
is in fact
yields
immediately
the
Degree
of
a
special
45
5
of the
case
(u'u) (c, u) : -TUT-
such that
*
Corollary
If
1.
only if
if and
holds u,
v
E
(v, v). In particular, (w-W, W-W).
E(c,
U) Id
cEB
0
I (u, V) I' equality
U' U) lul
=
assertion.
inequality).
(Cauchy-Schwarz-Buniakowski
Remark 1
Supp(u)
(=-
c
(u, U) lul lul
(u, U)
cEB
Proof.
(which
observation
Elemeiat
identity
The
and
Nonreal
inequality).
Minkowski
Proof.
simple
a
Faithful
a
and
R>oB\101,
linearly
are
v
then
R>OB
r=
w
u,
v
G
A
(u, U) (V, V),
<-
u
For all
u
=
if
v
(u, u) only if (u, v) only if (w 2, W2)
and
if
U7
commutes
dependent. and
that
Suppose
(u, V)
(u, U)
=
=
(3-1)
(V, V).
Then
(U, V) some
A2
R\f0j.
A G
Since
and
u
(U, U) (V, V), are
v
linearly
R>oB\fOj7
E
u,v
=
A E
dependent, say R>O and (3-1) Since
1, which proves the first part of the assertion. from the first with follows w 0 0, the second part
=
that
[Note
(u, u)
that
Let
Lemma2. E
=
(v, v)
lemma
The next
x
that
yields
and Remark I
2
-
C X
(u, v)
generalizes
X C B
X\f 11 Suppose
Supp(bbb)
and
be
(w 2,W 2)
=
Proposition table
a
subset
u :
in this
3.3 of
[5].
such
that
we
Av for
=
shows that may
and
tv-T
=
u
assume,.
Uw
v
case.] Jxj for
divides
n
all
that
Icl
and
is
by
divisible
n
for
all
c
E
Supp(bx),
x
E
X\fTI. (3.2)
Then b E X.
Proof.
a(bX)+. that
If the
zX+
a
=
we
get b
albX+l
=
JbI
E X since
support
of
(ii) z
of
[20]
=
n
Ce2j(bX)+j
and therefore
otherwise
G X.
=
(3.2)
there
exists
NB is contained
c
IzIX+ ( [20], Proposition
=
ajbjjX+j i.e.,
4.8
By Proposition
4.8
=
yield
=
C- N
X, then
Romthis
(bX+, bX+)
l(bX)+l would
(i)).
an a
in
=
it
such that it
follows
CbbX+, X+)
bX+
is easy to
=
that
Jb121X+j,
JX+j -= 1(mod n). But thisimplies 1(bX)+1 -= O(mod n). Since X
see
Z. Arad et al.
46
following
The
2
=
1).
+
_n
Supp(T2b2), Idl
n
=
1 for
that
d E S
some
=
to
=
n
(d,T-2b2)ldl Supp(T-2b2)\j1j.
:=
3
for
+
all
d E
[Otherwise
n
1)
1 then
By
Lemma I
-
((d,T2b2)d,(d,T2b2)d)
contradiction.]
a
n
-
d E S such that
exists
n(n -1)2 n(n 1)
>
gcd(n,
and
n2 +
<
1),
+
n
_
(d, T-2b2)
A
d E S
Hence
(n2
n
divides
n
=
all
by the assumption
(d,72b2).
divides
n
Ib2l
for
> n
2 (b2b3,b2b3) n(n there U:= EdES (d, b2F2)d,
(b2b2,b3b3) applied
Idl
11
=
Since
contrary. that
we see
-
implies
[5, Proposition
of
=
Assume the
Proof.
isa generalization
If b2 7 T3 and TA b272 b3T3, then (b2b3, b2b3) In particular, Tb, then (Tb, Tb) -- 4 if b 0 T and 6b
Lemma3.
n(n
observation
n
=
1.
-
-
Comparing degrees d and T Set b4 : e
(E T
Idl
that
we see
S u pp (b2
=
b3).
A
n,
n
=
I and
b2b2
applied
to
-
Then LemmaI
n1 +
=
(n
-
I)d-
b2 b3 supplies
u
with
[t
i.e.,
>
+
n
-
1)
<
n
3
+
n
n2
(b2b3, b2b3)
Now
n.
2
n(n
(e, U)
:=
n
jej
yields
n
=
-
1.
Set a2
2
_n+1)
e-
:=
It
follows
that
jT\jejj(n
1)
-
(b2b3
<
n(n
5 Since
n
> 3 we
get y
n(n it
follows
(n
-
1)(nl
Hence a272n
(n
I
-
-
are
1)2
First
=
ja272J-1
we
to
(n
1)1
-
=
the
prove
If
n
n
Tn(n)
=
for
+
b2b3
I) b2
1)b4)
(n
=
(n
+ ab4 for other, n -
I +
1)F2b2 1)
a2
some
n
21 + 1 and
Tb
an
=
-
1)
-
a
I +
(n =
E a
(n
=
-IL
2(n
-
n.
b5- E Onl
(T02
1 divides
!
-
72a2
-
n(n
(b2, a2T3)
=
-
=
n2(n
na2 +
and
=
pe)
-
-
(b2b3, a2)
-
-
_n
and
=
(n
to each
prime
Theorem 2.
morphic
(n
=
=
+
n
1)
-
a2F3
that
=
b2b3
tte,
-
2
-
=
(b3, F2a2),
1)b3.
-
(n +
-
Thus
1)b2F2
=
(n
-
1)a2b2b3
F 5)-
N. Since which
Ib4l
yields
=
n
and
ja2l
the contradiction
I)n.
following
a
suitable
m
n1 +
N.
l(b
+
T),
then
Bb
is
exactly
iso-
SITA with
3
following
the
Consider
Proof.
bo
By induction bk+1 G B if bk implies that
(b
bkTk
Cb
=
(3-3)
11)bk-1-
-
for
equality
the
47
Degree 5
of
Element
sequence
b, bk
:=
Indeed,
E= B
Nonreal
recurrent
show that
can
one
Faithful
a
each k
bk bk
=
We claim
> 0.
bb
=
(1
+
(21
+
that
1 + 1 (b +
1)
P
bbk
=
lbk
Ojai,
+
and,8i C- N U 101. Clearly Epi-I Oilail On the 12 (21 + 1) + EPj-j fli2lail. (bbk, bbk) (21 +1)2+212 (21 + 1). Therefore (6-b, bkTk)
where ai C= B
=
Z=
more
1) (21
+
hand,
other
=
1). Further(bbk, bbk)
=
P
p2i jail W.l.o.g.
we
assume
may
,3plapl
(21
6japl
Since =
1 and
In
T,,,,(21
=
x, y, xx
will
admit
Yx
V-y
=
fxy)
=
out
certain
those
1,
we
(21
for
all
1).
+
1) (1
+
=
On the other
1)
bk+1
that
which
1).
+
21(l
>
obtain
+
T. Then Bb
0 -< i <
m.
011apl.
=
=
G
a,
is
a
Nowthe
hand
Therefore
B.
linear
span of
isomorphism 0
immediately. be
interested
especially
in
those
integral
Ty
standard
If
n
=
n1 +
integral
z
GB
or
z
b2 + b2, b2 :
b2)
satisfying
this
group
odd
that
n
GT-algebras
standard
structure
G
1z,
(3.4)
nil.
[nl+l,
that
such that
elements
basis =
Some small
available.
Remark 2.
,3plapl
1)
21(l
>
have the property:
are
-2;
=
determine not
Pplapl
21 + 1 is odd and whenever
turn a
+
+
+
implies index such that b,
will
we
Then
Epi-1,31lail
1)al
+
follows
which
then It
.
lbi+(1+1)bi+l
=
1)
sequel
the
n
bobi
+
GT-algebras
(1
+
minimal
and
b,,,
1, bo,..., B,,,
1bk
=
m be a
Let
by 21
is divisible
bbk
Op.
F, pi-_1)3i2lail =
:?:
1)(1
+
1)2+ (21 +1)12.
<
bk lapl : 21 and, therefore, : (21 + 1)(1 + 1) which implies
Since
p
+
31
that
Pp
(21
=
f 3, 5, 7, 11, 171,
n
(3.4). which (3.4)
imply'
already
values
Lemma 5).
(cf. for
then
(3.4)
holds.
It
is
Even holds
property
problem counterexamples in are provided an
open
to are
Z. Arad et al.
48
Proof. already
If x, y,
(a)
2
+
If b2
(i) (ii) (iii)
(b)
[xyl
1).
:7 T3,
73, b2T2 =
(I
First the
1 for
=
1 + 1 for
=
note
prime
proofs
(a) By
n
=
21 + 1 is
=
>
=
of
b272
or
Icl jb2j
2(n
is
multiple
a
1)
-
(a) (ii)
equivalent:
are
equivalent:
(n2 +I)n 0
(b2 b3) b2 b3)
b4), b3 0 b4;
+
of
-==>
E
a c
for
n
all
which
(bA, b2b3)
<
is
1(b3
(c, b2b3)
(iii)
=,
are
2
calculation.
B.
divides
Lemma I there
1n(n
=
Supp(b2b3)\j1j.
E
c
yields 2
2
(i) and (b) (ii) Supp(bA) such
(bA,bA) lb2b3i
1
Oth-
the contradiction
+
(i),
==>
set
b2b3).
p
that
(n
1
+
-n
1 + 1. Now
that
(ii)
by simple
B.
n1 +
some e
Jb2b3jAC
lcj:! implies
(xy, xy)
n, then
prime number and that
a
conditions
(c, b2 b3) A
degree
of
is verified
B;
some e E
following
1C-2
that n
E
some e
1 for
n
For the
n'j.
This
conditions following + 1b5) b4 0 b5;
1)b4
n1 +
(e, b2b3)
Proof.
+
then the
=
erwise
that
then the
=
(e, b2b3) (e, b2b3)
(i) (ii)
elements
[n'+',
Then'
b2b3
If b2
basis
are =
Suppose
Remark 3.
1n(n
:7 - 1p,
x
implies
1cl
=
n
since
Icl
is
a
< -
multiple
n
2
1+1
of
n as
< 2n
above.
stated
The
implication
is clear.
Since A21CI + tt2jej c 0 e. Clearly, (ii) = , (iii). (1 + 1)2 n + 12n n and b2b3 (bA, b2b3), we get A 1 + 1, /-t 1, lei (1 + I)C + led G"B\Jej such b2b3 (iii) = - (ii) Lemma1 applied to u /-te supplies =
=
=
-
.
that
v
Comparing degrees yields v (b) Here LemmaI applied
(U, U) lul
(d, u) 1, Idl to
=
u:=
n
and
b272
-
(1
b2b3 n1
supplies
such that
v
:=
(c, b272)
>
(UU) lul
+
=
1-
I)e an
+ 1d.
element
c
C-
B#
SITA with
3
Faithful
a
Nonreal
Element
Degree
of
49
5
Rom
Icl follows
it
or
(ii) lei
=>
(i).
n.
In the first
=
Since
lei
that
assume
(w, w)
6
b272 =.nl
+
b2b
(1
(i) (ii)
b3T
(1
If
B
Cb
If
c
=
!,
Set
n.
+
+
c
=
73 b3
=
the
(1) 72b2 (2) 72b (3) 72b2
+
yields z
EE
n1 +
=
n1 +
(i)
Proof. nb2
n1 +
=
Set y
(n (n l(b6
b2(Cb)
b2C
b4T
-
-
=
(ii)
1
n1 +
=
/F3 b3 holds:
1)b61 1)b7; + b7) -
=
(1
n
odd, 6-b
6-b
then
,
=
T-2b2
=
ln
-
=
E
2n
=
may
nl and
Supp(w)
=
and
n
=
+
(b6772b2)b6
+
moreover,
NB
Cb
Tb,
then
one
(b7,72b2)b7-
Cb and (b6, b2 bb2 b)
=
1) b2.
Then
(I
1)b3T
=
L-fl-y.
(b6 72 b2 6b)
=
+
+
lb4T
=
(1
1
If
gcd(l + 1, 1)'= 1, V Supp(z) implies z C-
c
E
B, then Icl
=
1)2b2
+
From
and I
21n.
If,
-
some c G
then
2
2
(c,T2b2)
we
for
n1 + lc
n(n +1)=(b2b,b2b)=(Cb,72b2)=n
shows that
lei
Hence
Tb-
=
(b2b)T
+ lb2 +
Finally, Izi The equation
either
n
-
n.
-
or
=
NB.
=
-
21 + 1 is
n
cases
b3T
:=
lb2C
=
-
-
=
3
=
following
=
=
12n. By LemmaI there exists d 6 1. Comparing degrees yields 1, idl
76 b2F2 b373 T6 =7 b7 T7,
and
n
Icl
or
=
72 lb5; b5 b6 + 76, b6 0
b6 + b7, b6
moreover
n1 + lc
n2)tt-2 2n, we get b2F2 n1 + le, and we are finished. 2n b272 1-te n1. Then lwl w
T2b2 If of
=
M
lb4-
+
1)b2
((b2T2, b272)
!
1)b3
or
b2b3
is immediate.
-
that
+
1 and
v
(ii)
=
2n,
<
n
2n,
n2 -12n
-
Suppose +
<
case
1)n (d, w) l(d + e).
and
c G
=
=>
:=
Lemma4.
then
=
1(n2 2
=
such that
(iii)
lcl (i) lei
either
that
implication
The
juiv-1
:5
we
get
+
(1
y
+
=
lz
1)y for
+
lb4T
some
B.
+1(c,72b2)
2n and
T2b2
=
n1 + lc
=
6-b.
Otherwise
2(b6) 72b2). Hence
If
6-b
72b2 Tb,
=
=
n1 +
one
can
l(b6
+
apply
=
76) the
2(-b6,72b2) =
=
6-b, proving
previous
to
(c, b272) the
T, b3
=
first instead
21n.
part
of the
of b, b2.
assertion.
This
yields
Z. Arad et al.
50
b3 b3
bb
=
bb. Furthermore, since b2 b2 b2, b, b37 b4. This yields fb-3b2l
(n2
2
bb
=
of
instead
+
1)
n
(T3 b2 T3 b2)
=
(b272, b373)
=
7
(T2b2,T2b2)'I(b3b3,T3b3) -
-
-
) 72
to
4
3)
whence
:5
(b2F2 b2T2)'! (b3
=
(6b, 6b) =2 (n 2
2
b3 3)
3,
1
-
2
(i)
apply
can
n1j,
I
-
2
=
one
[nl+',
+
2
I)n.
Thus
(b272, b373) Corollary
and
(iii)
1
(IL
obtain
holds
for
n
2
+
v)
n
=
3. For
(72b2)(6b)
y
n21
(b6, X) (b6, x)
Then
Now we
algebras
n
+
=
n(/L
1
-
nl+lv,
+
(nI
+
Itb6
1)b6
+
n(v
able
Gb
(3.4) to [10),
Suppose
=
as
(1 the
was
Part
that
Gb and b
holds =
+
vb7)(nl
+
introduce
to
1)b7
+
-
n
=
21 +
1)c
*
binary
E
Gb. For
*
operation
d.
Then
*
to
g
*
all
1, 1
OT2-
B;
elements
d + le
c7c
g, h E
element
Gb If -
n
1 and the
>
1)
2 7
tt(mod
_=
E
3
b3
3
+
/-tb6
v),
+
vb7. Thus (iii)
Tc
=6_bj *
c
in
+
/-tlb6b7 so
section
holds
vlb7b6-
+
that
integral
For the
d is again Gb Of by setting
if if if
h g g
U
the assump-
standard
GT-
construction
we
2.
and that
6-b
=
Tb
111.
with d
c 9
h
b473) +12 b474
1),
(3.4)
N,
+
Set
of an
as 0 d, there exist degree n such that element Of Gb- We ex-
T
-
1, 1, =T
5, then with this multiplication, Of 9 G Gb is T-
inverse
(b3
b7))
+
vlb
4, Part II,
d,
c
I
for
+
l(b6
plb2+ 6
+
Gb\j1j
c, d
g
h
+
section
b2
fc
:=
G
basis
ln(tt
+
nI +
=
structure a group mentioned before.
I,
GB orv=b2+72,
v
(3.4) tend
n2 n
=
T2b2
means
7
are
=
From
(bTb7 T-2b2)
=
21. This
=
b3
3
7
3, consider
>
Gb
cd
(b7;T2b2)-
:=
O(mod 1), (b6 Y) _= n(p (b6) Y) yields the assertion.
satisfying
Lemma5.
Then
v
(b2b, b2b)
=
(b3 b3
=
b373-
=
and
_= =
want to refer
also
1)
+
(b272, b2T2)
=
(b2 b) (b2 b) W+ 1)T73 + lT4) ((I + 1) b3 + IN) + 73 + 1)2 b3 (1 l(l + 1) (73 b4 + 74 b3) +12 T74b4 (1 + 1)21. + (1 + 1)21 (b6 + b7) + 1 (1 + 1) (b3F4
x
tion
b2T-2
(b6,T2b2)
n(n
=
(b2 b2, 72 b2)
=
yields
I
Set /.t:=
2 we
(6-b) Cb)
=
Gb
is
a
group
with
unit
SITA with
3
Choose c, d
Proof.
(-c*-d) (c
dd d) d) (c ;-d)
Gb\j1j
E
Cb.
a
Nonreal
Faithful
with
01
c
Element
Degree
of
RomLemma4
(ii)
we
51
5
obtain
that
c 9 d, the same argument Gb and *Z 1 > 2. Weare now c * d E Gb Suppose that 6-b, yields (c o 1 * have definition x e I and x x we axioms. the to By group going prove I for all x (=- Gb. To prove the associativity law, we first mention that x * T e
=
=
Since -Z, whence
=
G
=
-
=
=
=
7
all
(3-5)
Gb)
x, y E
by Lemma4 (i)
and
(x*y)*"Y=x Choose x, y,
jx,
I
(a) (b) (c) (x
Y for
y
9
y,
-
(x
To prove
Wedistinguish
(!g*Y)ex=y
=
9
y)
z
*
(y
x *
=
following
the
allx,y
for
z),
*
that
assume
may
we
'(3.6)
Gb-
E
cases:
z
or
x
E B.
z
I
z
(xoy)
andY*
x*y--7orY=y*z,
x: TOz,x9y=A7andTOy*z.
*
(a) y) (b)
If If
(Y
X 0
Again
x *
+
=
Y
z
(Y y
=
0
-0
0
Y)
46
(3.5)
=-
z=
(X
0
is done
=
X(Yz)
(3.6))
and
(Y
X 40
=
z
+
(-Y
0
0
7))
=
X 0
T
1
=
=
Y)
Z.
*
similarly. yields
1(1
+
1)(X
(1
+
1)'X
O(mod 1)
all
for
0
Y)
* z
(Y
a
0
w
c
1(X
+
Z)
+
1(1
*
Y)z
+
=
1)X
*
((x y) z, xyz) Supp(x(yz))\jx*(y*z)j.
yields
2, the above equation
I >
(w,x(yz)) X 0(Y Z).
loz=
=
law in A
=
Since
=Fo(yoz)
T is similar.
Y)
0
e z
associativity
1)'(x
=
(using
T, then
X 0
=
case
The
(1
y
Z)
0
the
(c)
proof for
The
z.
showsusthatxo(yoz)
(3.6)
=F, then
x
9
(Y
0
1
=-
*
*
(XY)z Z)
+
1X(Y
Z).
*
=# O(mod 1) Thus
and
(xoy)oz
=
0
-
Y5 defined
gebra xx
A
=
yy and
simple
algebra T,,(n) Example 1 fulfill Zm+2 If b is as
Gx
in =
-
consequence
is the
subset
Suppose that X which is exactly
Proof.
The
c
X\jTj,
Lemma2
assumption which
yields
holds
isomorphic implies
means
b E X.
(3.4).
1 and the
Chapter
in
For x, y (E
T,,,
(n)
Example 1, then Gb
in
al-
table
11,
have
we
1, b,
Z3.
following
(3.4)
Remark 5.
x
defined
The table
Remark 4.
that
and that to
X C
condition
T.(n)
Cb
=
Gb. Thus
(3.2)
Tb.
If Bb contains
and contains
[bx]
=
b7b,
a
Bb
then
[nl+l,
of Lemma 2 holds.
n1j
table =
for
X. all
Therefore 0
Z. Arad et al.
52
Suppose
Lemma6.
6-b
(i)
that
n
2 1, b
41 +
=
(21
=
1)b2
+
2lb3.
+
Then neither
21c2
n1 +
=
nor
(ii)
b2b2
bb
=
21(b4
n1 +
=
+
b5), b4
T4- =,4 b5
=
=
T5-,
holds.
(b2
In each case, we have (6-b, 6-b) 2+ 1)n 2 (n Tb by Corollary 1. RomLemma4 (i) and (ii), it follows Tb 6-b. The support of (21 + 1)b + 2lb6, b6 :7 - b, and TA n1) E jc2, b4 + b5l contains a real basis element e 0 1 such
Assume the
Proof. b 2),
that
b2T
d
(2l)-1(6-b
:=
=
(e, d)
that
6-b
that
so
contrary.
=
=
=
-
=
1. Consider
the elements
2 (6-b) 2= n21 + n(n 1)d + l(n 1)d 2 ( b-2) (b2T) (21 + 1)2T b + (21 + 1) 21 (T6;"b + Tb6) + 1 (n b T2 1)T6 (21 +1)2 n1 + (21 + 1)2 21d + (21 + 1) 21 (-b6b + Tb6) + 1 (n 1)76 b6
X
-
y
=
-
=
b6
-
-
(e, x)
Then a
0(mod(n
-=
contradiction In the
b7b
and
b
Part
1, Theorem 4.9).
Remark 6.
Let
Icl
for
and
> n
Proof.
If
n
(e, y)
and
Tb implies
x
=
=
(21
1)2
+
bCbb)T
21
=
b0b
=
# 0(mod(n
21
=
case
1)),
-
0
y.
=
31 +
=
Westart
situation.
=
with n 3 it was proved in [L], Theorem 2.6, b2 + b3, b2 =76 b3 necessitates b3 b2 (see also [10], We are now going to study the corresponding general
commutative
0 T
that
1))
-
6-b
since
=
with =
all
c
21 + 1. If G
(u, u)
we assume
G
u
Supp(u),
(12
>
Jul
NB satisfies
then +
(12
(U 7U)
I)n,
then
=
+
n(l 1)n.
1) (U, U)
+
Lemma I
supplies
(1
<
1)2n
+
E B such
a c
that
Icl
By assumption
Icl
then
(u,u) all
=
=
Icl
=
elements
and
Idl
=
(1
n
that
b2
6 < 12
=
=
(u, U) Jul
>
and
(u, u)
<
1 is
prime
(1+1)2
Assume that
of
n, u
a
+
1)2 n,
to
I +
contradiction.
1,
so
that
A
n,
this
implies
Hence 0 <
=
=
+
1)b4
If
u
1
lu-Acl
:!
1c,
=
1 and
=
n.
Since
degree >- n, we conclude d : u Ac (-= the assumption. (1,2 +1)n, contradicting =
21 + I
is
odd,
that
either
1 is
lb5- Then 6 b = 6 n1 + l(b2 + b3) for b2, b3 of degree n such that bb2 b2b.
(1
1.
=
have
(U, U) n
(1
> 1
1+1
-
Since
n
=
elements
basis
!
I)nl-'.
of the support JcJ, whence
Theorem 3. and
+
>'
(c, U)
A:=
+
even
all
-
or
n
B 0
< 23
distinct
real
=
for distinct Proof. Assume that b7b nl+l(b2+b3) degree n such that b and b2 commute. Then (6-b, Tb by Corollary whence 6-b 1. Since b2 is real, =
=
real
basis elements
Cb)
(n 2
2
b2 also
+
1)n
=
62, b3 Of (b 2 b 2 ,
commutes with
T.
Hence
nb2
+
lb2b + lb2b3
=
b2(6-b)
=
(6-b)b2
=
nb2 + lbb2 + lb3b2
=
nb2 + lb2b
+
lb3b2)
SITA with
3
which
1bb3
nb + 1bb2 +
T) T7 74) 75
7
p
:=
nb + 1bb2 + 1b3b
=
=
-
(b3, b4 b4),
v
1b3b
=
-
7
53
is
74b
get
b4T
also
Degree 5
of
real). By Lemma4 (i) applied (1 + 1)T + 1b6, b6 7 T- Using (1 + 1)b + 1b',6 b' 6 b. By Lemma4 (b2) 74 b4) and v (b3, b4 b4) fulfill
(as b3
we
,
Element
nb + 1b2b +
b and
that
we see
-
(iii)7
(6-b)b
=
b2, b, b3, b4
of
instead
(i) again (b2 b4T4),
Lemma4
b(6-b)
=
Nonreal
Furthermore,
commute.
commutes with
*b3
shows that to
b2 and b3
that
means
Faithful
a
b4T4 n1 + /jb2 + vb3, K4b4- n1 + /-t'b2 + v'b3 1), (1, 1) 1 and 1, 0), (0, n (y, v), (/2, v) E f (n 74 b4) T4 b4) tt/2 + 1,12 (b474) b4T4) (b4 it 2+V2 =
=
-
(3-7)
-
=
n
i, i
For as
bi
12,31
E
real,
is
Supp(ui),
On + (ui, ui) 12 n
+
hj2)
(bi, juil
set aij
b
(I
(bib, bib)
=
(U2; U3)
and ui
=
1)n
=
(6-6, b?)
(b2 b, b3 b)
=
Supp(ui) does not (c, ui) is divided by JbI
(6-b,
=
=
n
2
b2 b3)
jej
+
=
element
an
n as
=
aij
< n
-
1
and
contain
Furthermore, otherwise
+
Clearly,
E NB.
bib- lb
:=
n
=
-
ln(Oz2i
+
1n (a32 +
(3-8)
C13i)7
(3.9)
a23)
1. [Since of degree n prime to Ibl. Now (3.8)
c
-
I is
shows that
a
contradiction.] If (U,, U,)
First
(12
+
n1 +
aijbi
>
+
+ a3i
a2i
we
-(Ui, 1n
-
then
ui
(1
=
-
1
+
n
> 1 + 2n
-
-
> n,
T
f 1, 21:
ji, jj 1)zi, Izil
for
show that
will
1)n,
n
Ui)
+ aji
n, qii
n
-
I and
therefore
b? z
(bib) Cbbi)
xi
+ =
ajibj (1b
+
=j2 b+j(j+j)(r,+ =
=
=
Tizi If
(U,, U,)
=
:
12 n1 +13 (b2
(1
zi
z
1) zi) (1T
+
+
1)7,7)
(1
+
+
(1
+
1)2Z,-F i
(1
+
1)277zi.
T)+(1+1)2
zi 77
+
b3)
+
1(1
+
1) (bZ-i
+
ZiT
+
b3)
+
1(1
+
1) (-bzi
+
Tb)
+
(ui, ui)
:
4
(3.10)
(Tbbi) (bib) 12 n1 n1 +
(12
and Lemma I
+
13 (b2
(b2,Tzi)b2
+ 1 +
supplies
1)n
+
>
a c
c
(b3,Tizi)b3.
(12
+ 1)n, Supp(ui)
(C7 Ui)
:
then
such that
luil
> 1 +
(1
+
1)2
n
by Remark
6
Z. Arad et al.
54
Comparing degrees yields
(3.8) last
that
deduce aii + aji equation of (3-10). we
zi
n
=
has
c
:=
degree
Finally,
1.
-
n
and ui
(i)
Lemma4
=
(I
1)zi.
+
(iii)
and
b4T4 54 b7b. By (3.7) we may assume w.l.o.g. that tL n1 + (n 0, i.e., b4T4 I)b2. Then (b4, b2b4) (b2, b4b4) v b2b4 (n I)b4 G NB fulfills Jul '= n. Wedistinguish the following main subcases: n 1, (Al) a22 n 2, (M) a22 (M) a22 < n 3, Note that, in the first two cases, (3.8) implies that (U2, U2) so that (3.10) applies.
p-ro-ve
Case A: v
=
-
n
I
-
n
=
-
From
the
and
1 and
-
-
=
-
=
-
-
(Al)
In this
n2
b 22
case
(n
+
1)2
-
n1 +
=
n
=
=
shows that
b2b
(I
=
+
v
I)T
b2(6-b) (b2b)T yields b2b3 In particular,
(b3,b 2)
=
applied
to
b4
=
nb2
I)n,
and
(bA,bA) + (v, v)
b2- From this
=
+ lb
2+ 1b2b3
1n1 +
=
2
(1
I)T-2
+
16-6
(I
+
1)75
+
1b3- Since b2b3
we see
(T, b3b)
=
that
a23
(T; U3)
1n1 +
=
=
(n
=
(b4,b2b4)2
it
follows
+
1(n
-
1))b2
(12+ (1+1)2 )b2 =
1 and that
n
(v, v)
+
b6
that
+
+
=
b, i.e.,
1b2b3
12 b3
+
1(1
b3b2 is real, this is also b5 0 b, b3, b2. Hence 0
+
1)T5
true =
Cauchy-Schwarz-Buniakowski
and the
for
(b3, b5)
b5. =
inequality
V3- yields
U3 and
(U3 1(1
I >
b4T4-
=
+
(U3) U3)
If
=
+
+ 1b. Now
=
=
1)b2
-
(b474, b2)2 (n- 1)2 n
i.e.,
n1,
=
(n
(12
>
1, then U3 U3) n2 + la23n 7
(U37 U3)
> -
(12 12 n
(bAb3b) I)n + 1(b5, b2)n 3
73--) +
1)n-
+ =
n
2
=
If =
1
=
=
> -
(b
2
1(1
b23) +
1, then using
9 > 6
=
(12
+
I)n.
1)n-
(3.8)
also
we
Now
(3.10)
obtain shows
I) Z3 for some z3 0 b of degree n and Ce23 + 033 n 1, i.e., lb and 1 b 32 + I)z3 n1 + 1(b2 + b3) =6-b. as Ce23 b3b [Clearly, 0 0 1+ 1 Z3 =7 b2 since (b, b3b2) (b, (1 + 1)b5 + 1b3) (z3, b3b) (b7 b3Z3) Lemma4 (i) applied to (b3) T7 73-) T) instead of (b2, b, b3, b4) yields Furthermore, 73-b 1 =7k 0 (1 + 1) b3 + 88, b8 0 b3 [Clearly b8 : k b2, since (z3, b-8) (z3, (I + 1)T + 1b) (z3, bb2).] Now Lemma4 (iii) applies and z373n 21 + g' that
U3
=
(1 + (I
+
=
=
-
=
(i) (ii) (iii)
g
=
g
=
g
=
(n (n 1 (b2
-
-
1)b2) 1)b3) + b3)
of the
-
following
=
=
-
=
one
=
=
=
where g has
-
-
forms:
SITA with
3
Consider
Nonreal
Element
(n
(6-b)2
=
21
55
b2b3 +12 b23 1)(b2 + b3) +12b 2+212 2 +12 +13 (n(n -1) (n 1) )b2 + (n(n 1) + 313)b3
=n
212 n)l
+
(b3b)(M3)
y
12 b +
+
n(n
+
-
-
-
(1b
=
+
(1
I)z3)(lb
+
(1
+
1),73-)
+
ZA +(1+1)2 Z3T3 (12 +(1+ 1)2 n)l +(13+ (1+ 1)2 %) g))b2 12(1+ 1) (b8 + 78) 1(1
1)(bZ3
+
+
-
+
(i)
In
this
I)Z3
(Z3, b2Z3) z3 (recall
case,
bq, bg :
+
(b2) Z373-) b2 0 Z3).
=
that
(13 +21(1+ 1)2 %) g))
+
n
-
5
212(1+ 1)b5
+
(n
Degree
of
the elements
b26-b 3
x
Faithful
a
n
=
Therefore,
1.
-
Hence
b3
b2Z3 get the contradic-
we
tion
n2 In
(ii),
case
(n
+
we
1)2 n
-
(b2)'X)
=
=
Thus
we
(A2)
In
n(n 13
(1
-
1)21
+
R 2
case,
1)2 n.
-
1)
-.
+
13
13
=
(b2) Y))
/2
V12
+
A
(0,
=
n
1) b2
,
so
(n
-
1)
=
2)b2
-
of
+
degree
212,
2
-
1)b 4
b2b 2= nb2 + 4 deduce
b7b4
(n -1)2
=
n
(n
b4
+
+
+
n
212
+ I
=
b2 = A b4
Therefore,
Using (3.7)
we
get
v)
=
(i) (tt',
and therefore
But this
means
commutes with
b7b4
(n
=
2 2
-
(n
(n I)b3.
-
I)nl
(n (n
-
-
Hence also
+
(n (n
-
+
commutes with
(b4F4, b4F4)
-
+
I)nl
-
b4
that
b2 Since
b4 is real. b4. Rom
1)b
-
b2
+
1)2 1, 0)
or
,
and commutes with
(n
b3 by (3.10).
n.
deduce fTom Lemma3 that
=
213
+ 21 >
-
commute.
that
313 +412
=
(b2) Y)
n1 +
1(n
1)
T4
=
element
>
-
1) +13
-
(AI). =
basis
a
Here b4 and
(n
(b2b4)b4
we
1) +12 (n
-
1) +12 (n
-
+
is
v
A2 + V2 (ii) (p', v')
real
n(n
=
have excluded
this
and b7
we
(n
+
n
=
(iii),
and in
(i)
(b2Z3, b2Z3)
=
get the contradiction
(b27 X)
n1 +
b 22)
(Z373,
=
b7
1)2 b2 (n
-
+
=
-
-
=
(b 4 b2)7 2
,
=
n2
1)(n
+
(n
is
b7b4 -
2))b2
+
(n
Hence
(bA, b7b4)
b4F4
(n 2-n+l)n, b2b4 (n 1)b4 =
-
1)(b2,
b 2)n; 7
-
1)b3
Z. Arad et al.
56
(b2, b2)7
and, therefore,
(b2b4)(b4b2)
x
(n -1)2b2+ 4
((n -1)2
n
b2b2 4 2 =(nl n 21 + n(n
y =
(n Since
(n
+
=y
x
(6-b)b
w:= =
=
2
(n
((1
2=
+
(n
nb3
+
1)(n
2))b3
(n
n(n
(n
1(b2
n(n
+
-
1(, _i)n)l
1)n
+
1(n
+
(1
(1n
+
n(n
2)
-
1(n
+
+
(n
2)
+
n(n
(n
+
2(n 2
1)(n
-
1)
-
2)b 2
-
+
(n
2)
-
1)2
-
b3
+
bg,
(n
+
1)(n
-
1)b2b3
-
2)2 )b2
-
+
1(n
2)b 2+ 1(n 2
-
2)2
-
Set
a23 =n-1.
b3)
+
1(n
+
+ +
1%, i.e.,
2)b2
-
1)nb3
+
1)2 )b3
-
2)b2
-
b3)
+
=b2+(n+
2))b2 +
n
I)b2b3-
-
b3))(nl
+
2)b2
+
1)b2 -
get bg =b3 and b2b3 +
b7)
+
1)
-
-
-
(n(n
+
+
2(n
+ +
+
2))l
-
-
+
I)b2)(nl
1)b4
-
bg 7 b2- Moreover,
2 b7
((n -1)3
+
nl+(n-2)b2+bg,
=
b7)((n
+
1)b4b7
-
b 27
i.e.,
1)b4
-
-
2)b2
(nl
2
+
n)l
1)(n
-
we
+
+
-
21+ (1n
n
2(n
-
(n 2+ n(n
=
((n
=
n-2,
=
+
1(n
1)b3b2
-
1))b2
-
+ lb
2 3
+
2
1b3'
Thus
(b27 W) Set
n(n
1n +
=
(b2) Z272-)
m
2)
-
2)2
-
(b2 bZ2)
Since
-
1(n
+
-
(Z2, b2 b)
=
7
1(n
+
1)
+
=
1+ 1,
1023
=
we
413 +612 +21-1. conclude
with
(3. 10)
that
313 +412 +21+ (1 i.e.,
1
m=
(ii)
--L, 1+1
-
In this
from b3
-
74b4 1
-
(b2 X2)
=
(b2 W)
7
7
=
413 +612
+ 21
-
17
contradiction.
a
case
Romn
1)2,M
+
=
(n
n1 +
(b3 74 b4)
=
=
7
b4b4. In particular, 1)b3 (b4 b3, b4), we deduce b4 b3
b4 is distinct 1) b4 + b's
-
(n
=
-
b8 =7 b4. Now
((n yields applies.
-
a33
=
1)2 n
+
n 21.
+
(1n
(n2+ 1(n ((1
+
1)n
-
1(b2
n(n
1(n
1n +
n(n
+
-
1)n)l
+
then
(b3277A)
=
(3.8)
=
A
n(n
(U3 U3)
implies
7
>
+
(12
(n
I)a33)
-
+
1)n
and
(3.10)
L Set
=
+
+
(b4b3, b4b3)
=
But
2.
-
Hence a23 2 ( b-) b2=(nl
w
1)n
+
2))b2
+
(1
(1n
+
n(n
+
-
b3))(nl
2)
+
21(n
-
(n
-
I)nb3
+
-
2)
2)b2 +
+
1(n
1(n
b3)
+
-
2)V2 + 1(n I)b3b2 + 1b32 2)2 + 1(n 1) + 1%+
-
-
-
1))b3-
Thus
(b27 W)
=
-
2)
+
1(n
-
2)2
+
1(n
-
1)
+ 1
=
413 +412 +31-1.
SITA with
3
Set
(b2 72-22)
m -==
7
(3.10)
with
(b2, TZ2)
Since
.
Faithful
a
Nonreal
(Z2, bb2)
=
Element
of
(Z2, b2 b)
=
Degree 1 +
1,
5
it
57
'follows
that
313 +412 +21+ (1
1)2M
+
(b2) X2)
=
(b2) Y)
413 +412 +31-1,
=
i.e., 13 +1 -.1 ,ra
Thus 1
2,
=
m
=
(b3,'72-Z2)
1 and
=
1(1-2) (1 + 1)2'
(1+ 1)2 n
(cf.
2
-
(3.10))
and
we
obtain
the
contradiction 44 <
Hence
(M)
13
+ 21 (1 +
(1
+
1)n
+
1) (b3, TZ2)
+
1(n
21(n
(n
n
3, then
-
1)2 n
-
< <
which in
contradicts
B:'b474
Case
(Tb)2 b2b2
x
y
(1 For
i
G
(bi, bb6).
+
11, 21
n
-
I
=
1)
-
-
1 must be odd
As in Lemma6, n
=
7
(b2) W)
(b4, b2b4) yields n2
=
n(n
=
+ 2 -
21
n(n
+
1)(b2
-
by Lemma6.
we
consider
n('n
I)a22
-
3),
3n +
:=
ri
(bi, (b2
+
1-b6)((1
+
=
8,
:=
+
+
+ 1 (1 +
b3)2),
This
will
turn
out
again
the elements
b3) +12 (b2
+
(Tb4)(74b) ((1 + 1)b 1)21 + 1(1 + 1)2 (b2 + b3)
set
=
=
-
(b3 X2)
=
37.
=
(b2, b4T4)
1)
-
> 3.
n
way.]
different
a
+
(1+1)2 (n
(b474, b2)2 (b2 b4, b2 b4) n(n + (n 1)(n 3))
6-b. [Then
=
2)
(M).
have excluded
we
If a22 <
-
b3 )2
I)T
1) (bb6
+ +
(bi, 76b6)
1b6)
TTb6) +12 T6 - b6 :!
n
-
1 and tj
Then
n(n
-
1) +12 ri
=
(bi, x)
212+1+12
=
(bi, y)
=
1(1
+
1)2 +21(1+ 1)t,
ri
13 +21(1+ 1)t,
t
ti
+
+
128,'
12S,.
Wewant to show now, that
In
tj n
particular
121.
> -
I
=
-
If
1
2 in
as =
1,
this
T-6 0
b
-
this
then
(3.11)
case,
we see
+
means
76
becomes 3-:! that
(3-12)
2
tj
!
E
Supp(uj).
First
we
show that
3 + ri = 4tj + si. Since si < 1. If 1 > 1, then (3.11) shows
Z. Arad et al.
58
1
(b 61
12).
21ti(mod
=---
odd.
Now
b2b)
=
we
Hence ti
show that
(-b6, bb2)
t2
=
12 n
0, ti
ti
for
2
Otherwise
2
31+1. Now (3.8) 2
!
+
<
n
4
n
odd
some
may
we
m
and 1 is
E N
that
w.l.o.g.
assume
yields
2
ln(a22
+
a32),
+
i.e.,
If
3,
>
n
we
a22 + a32
by (3.9),
(b
and if z3 15
=
=
(31
+
+
2Z3, b
76,
then
(b
+
contradiction
1)(1+ 1)
+
4
(12
=
contradiction
+
+
2Z3, b
1
2-
_
a22 + a32
n
ln(a23
:5
3. Hence
=
1)n
1 1.
If
3(a23
+
> n
-
n
(3.10)
and
2
-
a32)
+
=
n
=
3, then
=
(b3b, b2b)
(6-b, b2b3)
3(a23
-b6)
(b3b, b2b)
(65b, b2b3)
3(a23
+
finishes
This
cases.
a32)
I < a23 , whence (3.8) implies for i 3. If Z3 0 T6, then
=
applies
2b6)
both
in
.
41
Furthermore,
2.
=
a23 + a32
9 > 6
3
a
that
so
(U3) U3)
I
-
12 n
6
the
obtain
n
912+61+1
I +
a22 + 0132 >
proof
the
+
a32)
9)
a32)
9)
+
(3.12).
of
Now
(3.11)
reads ri
w
-=
+
si
n
(3-13)
1.
-
n-'(w, w), vi t(b2 + b3) 1n and Ivi I G 12, 3 1. Then 1w I
b6b n(vi, vj), i, i compute Set
=
bib
-
-
1
(bib, bib)
Cb-6
=
(1b
+
(bj2 b-b)
=
(n
+ 1 (aii
(b2b, b3b)
=
(1b
+
(b2b3, bb)
=
ln(a32
lb +
+ vi, +
Cb-6
aji))
fbr 6
For
tn.
=
+
V,)
(12
+
V3)
(12
lb
-
+
t2
6-6 I i, j 1 -
+
and rij
12, 3}
-
=
we
nii)n
n
+ V2) lb +
t7bF6
+
t2
+
N23)n
a23)
+
Hence 2 +
+ ceji
aii
a32 + a23
t2_1+r I
_I+
t2+r
,,7
(3.14)
.23
Moreover,
(2t2
+
)n
b3)
=
(t (b2
+
=
(n
1(82
+
+ w, t
+
S3))n
(b2
+
b3)
+
W)
=
(b6 b, b6 b)
(6-b, T6 b6)
SITA with
3
a
Faithful
Nonreal
Element
Degree
of
59
5
yields 32 + 83
follows
Now it
from
r2 + r3
t
"
(3-13),(3.14),(3-15.)
0222 + a33 +
3(a23
t
I +
-
that
+
a32)
a22 + a33 +
--
If
3, then 1
n
=
1
a22 + a33 +
the
We treat will
use
will
we
ueS
n
=
t
=
remaining equation
3,
t
separately.
cases
(3.14)
of
be
12,41
t2
n
11, 1
=
=
5,
t
t2
_
15, 1
=
X23
::
7,
t
+ aji
aii
implies
=
r-22 r-22
=
=
=
6.
6
I
=
-
that
r-23
=
t2-
t
I +
-
for
except
case
reason,
we
n
More
consider
all
=
15
3,5,91
possible
val-
explanation.
3 + 5
14, 6, 8, 12, 161
1 + Kii =
1
N.
3
and
+
K33 and a22 + a32
+
t2 +
=
15 + xii
5+ 8
and
7
E
N, ji, jj
a33 + a23.
=
=
11, 21
Now Remark I shows
'
K23
1
5 and a32 + a23
-7+
=
10.
a32)
+
16 + r-23
9+
7
We conclude
2 + X23 7
with
(3.13),
2(a23
+
E
N,
(3.14),(3.15)
that r2 + r3
=
(a22
=
82 + S3 + 28
+
we
precisely,
By (3.14)
a32 + a23
so
2 +
G
the contradiction
contradiction.
a
i +
I +
4. Here rii
=
1
and
1 n
41 + t
any further
+ K,,
3. Here K22 E
=
=
In each
given without
yields
equation
to derive
N. For this
will
2. Here K22 c
=
above
+
1
6 > 4
show that
=
I +
-
+
the first
Of X22, which
7,1
tI
and the
=
3(1
41 +
=
3(1
t
-
82 + 83 + 41
(3.15)
(a33
=
+
31 +
a23)
6-4 7
+
a32)
=
36
Z. Arad et al.
60
supplies
39. Now Lemma I
i.e.,
(c, W) Icl
Clearly,
Icl
whence =
n
37,
=
a
=
we see
that
15 and
w
=
=
9,
19, 1
t
=
11,t
23,1
where b2b
Proof.
6c + d for
5. Here X22 C-
=
39 7
7. Since
this
some
such that
A
=
means
basis
7
> 5.
Icl
implies
Icl
element
=
IwI,\-I
:5
2.
+
l(b2
=
This
If
n
+
15, 7, 9, 11, 13, 17, 251
1 + Kii
-
6. Here K22 E=
t2
Corollary 6-b 0 n1
<
But
6.
=
Supp(w)
d of
15,
w
105 6
=
degree
15.
=
7c
20,
<
Thus
contradiction.
t2
n
A
G
(w, W) IwI
:
<.jwjjcj-1
15 and A
': :
49,
and
a c
=
_
all
E
6+n..
N.
9
16,8,10,12,14,18,20,26,361
I +K"
21 + I
b3) for
2 +
and
2 + Kii
13,5,7,11,171,
distinct
real
basis
and
N.
b
=,4 T
elements
and
b-
=
Tb,
then
b2, b3 of degree
n,
bb2follows
immediately
from Remark 2 and Theorem 3.
El
3
Degree
3.3
Rom now i E
N>2
tables
on
we
are
SITA with
a
Nonreal
Faithful
Element
of
Degree
5
61
5 that
we assume
denote
elements
n
Tb. By di, 5, b :A T and that 6-b The following 15,20,25 respectively. explanation.
=
=
of order
any further
given without
ej, gi, two
Table
(6b-, Ci
bb
T
51 +
105
7
5a2 4b2 51 + 3b2 + b3 51 + 2c2 51 + 2b2 + 2b3 51 + 2b2 + b3 + N 51 + 2b2 + C2
125
51 +
8
51 + e2
45
2
3 4
5 6
b2 + d2
45 45
51 +
b2
+
b3
45
+ C2
b2 + b3 + b4 + b5
5.
c
45
H
-.4
T
1
5b2
125
2
105
17
5a2 + b2 4b2 + b3 3b2 + 2b3 3b2 + b3 + N 3b2 + C-2 2b2 + 2b3 +,b4 b2 + 2c2 2b2 + b3 + b4 + b5 b2 + b3 + 2C2 2b2 + d2 b2 + b3 + b4 + b5 + b6 b2 + b3 + b4 + C2 b2 + C2 + C3 b2 + b3 + d2 b2 + e2 C2 + d2
181
92
25
3 4 5 6 7
8 9 10 11
12 13 14 15 16
the
55
51 + c2 + C3
=
I and 11
55
51 +
Table
consider
65
10 12 51 +
to
65
9 11
Tables
75
imply that (6-b, cases following
CT)
G
which
85 65 55 55 45 45 35 35 35 25
25 25
25 25 25
145, 55, 65, 105, 1251 are
given
in
Table
III.
and that
we
have
Z. Arad et al.
62
Table
III
6-b
5(1
1
51 +
3
51 +
6
7
12 13
case
1)
d2
16
51 + C2 + C3 51 +
by
b2
+
+
b3
+ C2
b2 + N + N + b5
Theorem
105
3b4 3b4
65
5 of
+ +
2b5 2b5
65
the
b7) b7) + b7) + b7) + b7)
b8 b8 + b8 + b8 + b8 b6 + 2c4 b6 + 2c4 b6 + 2c4 b6 + 2c4 N + 2c4 +
+
45
+
+
45
Introduction
45
45 45 45 45 45 45 45
to
Lemma3 yields situation, (6b, Lemma 6 (i) shows that Furthermore,
105 in
possible.
2 is not
b2
51 +
=
b4
b2 + d2
15
covered +
n
+ C2
51 + e2
18 151 +
-
5a2 + b2
2(b6 2(b6 51 + c2 + C3 2(b6 51 + b2 + b3 + C2 2(b6 51 + b2 + b3 + b4 + b5 2(b6
17
Case 1 is
(6-5, 125
3b5 + b6 + b7 55 3b5 + b6 + b7 55 55 3b5 + C3 55 3b5 + C3
b4
C2
51 + e-2
51 +
14
n(n2
+
51 +
8
11
b3)
+
2b2 + b3 2b2 + 51 + 2b2 + b3 51 + 2b2 +
5
10
2(b2
51 + 2C2
4
V
5b2
a2)
51 + 4b2
2
9
+
our
this
book.
6b) =7 case
Since
105, hence 4 does not
occur.
3.4
Case 3
jb , b3l =_Lb, bl,
Otherwise we obtain two possiapplies. f b, T1. But the first is excluded by Thethe table algebras mE M2. Clearly, orem 3 if we T"'(5), 51+2(b2 +T2) N>3, contain nonreal elements b, b2 of degree 5 such that 6-b with this table algebras and b 0 b2,F2. Example 1 shows that there exist theorem claims The following that do not belong to this family. property that this is indeed the only exception. If
bilities
b2
=
b2, b3
then
Theorem 2
b3 and b3 assume that b2b =
=
T2
=
==
Theorem 4.
b4 0 b5
-
Suppose
Moreover
that
bb
assume
that
=
51 +
b, T, b2
2(b 2
+
b2 -),
commute
b2 :A F2, b, T, b2= 3b4 Then either pairwise.
+
2b5,
SITA with
3
Tn(5),
(i) Bb
for
Faithful
a
Nonreal
Element
Degree
of
5
63
3,
some m >
or
(ii)
Y5
Bb
Proof.
b2
Since
commutes with
5b2 + 2b2 + F2b2
6-bb2
2
b2 also
shows that
equation
b26b
=
72.
commutes with
the
5b2
--=
+
following
The
2
2b2
+
simple
2b2F2 fact
will
be used
sequel.
in the
If c, that
d,
de
u
-
=
(u, u)
Then
2c.
(u, u)
35 < 20 +
[del
I and
=
Jul
>
5 such
3d
either
=
b-
(3.16)
or
15 and
=
(de, de)
=
degree
=6-b, then [525251].
2 and e-e
=
of
elements
basis
e are
(c, de)
(b2,"jd) Set
b,
b and
(dd, b-)
=
25 + 20 (b2
=
7
d).
I and (u, u) 25, which (b2, 3d) (3.7). proves 2 and we obtain from Ce If e E B satisfies Cb, then (e, b2e) (b2, b7b) 51 + 2(72 + b2) and 2, then 72 (3.16) that (b2, T2b2) E 11, 21. If (b27 72b2) Remark 5 shows that (i) holds. Thus we may assume that
d =7 6-b then this [525]. implies [u]
shows that
equation
If
=
=
This
=
=
=
=
(b2, M) 2
(b2; 72 b2)
=
=
[b2 e]
1 and
[525251]
=
for
all
c- B
e
with
e-e
=
6-b.
(3-17) By
Lemma4 3 +
we
(b6, b5T)
have 1
10
Cb
=
T74b4
b474
(b 5T b4T)
b-4b
10
,
and
Tb)
57
b4T =
=
2b6, b6 :A
3b +
(b 2) 74b 5)
+
b. Hence
(-b2) 7,4b 5)
(3.18)
Wedefine
(6-b )2 T2b2
x
Y
251 +
(b2 Y)
=
18 + 6 ((b2)
=
36 + 6 (b6,
(b, b5T)
b2 instead +
we
=
7
5(1
(3.18),
(b2 X) 7
Since
and
of
a2) (cf.
=
b),
20 + 8 (b2,
2,
it
T2) -
+
+
8b2T2
+
+ 4jb-2 4b22 2
(-b4 b5 + 75 b4) + 4-b5 b5 18(b2 + 72) + 6(-b4b5 + T-5b4) + 4-b5b5. -b4 b4
451 +
Using (3.17)
20(b2
follows
+ 6
-
obtain
b272)
+ 4 (b2,
74 b5)
b5T)
+
b22)
+ 4 (b2
(b2) 75 b4))
+ 4 (b2
7
75 b5)
'72)2= 9
+ 4 (b2
1
75 b5)
(3.16) that (b2, 75 b5) 0 0. By Table III (with 2 (b2, b22) =A 4. Also b2 : -b2, since otherwise b2T2 to eb, we get the contradiction Applying (3-17) from -
Table
III). 45
4(b2) T2
-
that
we see
32 +
=
(b2b, b2b)
=
(b272) 6b)
=
25.
Z. Arad et al.
64
(72, b2)2
Thus
(b2, b22) --
b272
and therefore
(b2 Y) : 40 implies Cb2, b22) b2 + b7, b7 : b2) T2, 55 (b2,2 b22) 51 + b2 + T2 + 2b8, b8 78 0 b2. Hence
Now
3.
!-
3, then
=
b22
-
as
x
=
y
651 +
=
f 2,31.
E
7
=
(b272,
If
b2 7))2
44(b2
72)
+
16b8
+
4(b7
+
+
b7)
-
13(b2 +72)
101 +
=
+
=
x
and
(b2, X)
-b2
=
8b8
+
+
2(b7 +77)
=
3(-b4b5 +75b4)
-b5b5-
+
Thus 13
(b2, T-5b5)
yields
T-5b5
2,
=
=
3 (b2)
T4-b5
6-b
and
=
T-5b4)
+
+ 2 (b2
9(b2 +T2) +8b8 +2(b7+T7) equation
Romthis
follows
it
b7
that
(74b5
+
75 b4)
=
=
75 b5)
3(b4b5 +75b4)-
b8 and
T2-)
3 (b2 +
=
7
08
+
04T4 T5 b5) (Cb, Cb) 65 implies (b4 b5 T4 b 5) b474 j3b2 + 2b8 3b2 + 2b8l and therefore b2F2 2. contradicting (3.17). Thus we have proved (T-2, b2) 2 of instead b2 yields b) (with Finally,
TA
=
5
E
=
7
=
=
7
b 22
2b2 + b2 + 2b7, 51 + b2 b7 0 b2,T2 and b2b2 and (c, c) G NB, b2 % Supp(c)
Note that
Lemma 4)
But then
Table
III
=
=
c
b7b (cf.
=
b2
+ C)
Icl
=
+ =
(3.19)
10.
(3.19) also implies that b7 commutes with b2, b2 and b7 FartherI and (b2 7 Y5 b4) (b2) X) (b2 Y)) whence (b2, 75 b5) 4 b5) + (b2 T75b5 5 1 b2 72 E NB. The equation .
-
more,
40
3. Set q
=
=
:
=
=
7
-
-
that
that
(b2) 74b5)
/.t
T74b5
and that
(3.16)
/-tb2
+
7
-
(T4-b5)F4b5) yields
7
v-b2
=
0474,705)
and +
v
2bg
=(6-b,T5b5)
=
(T-2, F4b5) satisfy
:=
45
(p, v)
bg =7 b2 T2 In particular,
7
7
.
e 1(17 2), (2, 1)1 2, so (b5 T9 b4) =
7
yields
(b2, b9T9-) :A
(3.20)
0.
Wederive 651 +
40(b2
+
T2)
+
8(b7 +77
2(b7 Therefore,
one
of the
+
following
+
T7
c)
+
x
=
651 +
C)
cases
y
=
=
=
3(bg
holds:
40(b2
+
T9)
+
72)
+ q.
+
12(bg
+
79)
+
4q,
SITA with
3
bg bg b7, c bg 0 bq) jbq, T9-j
A:
=
=
B: Let
us
b2(b2b2)
5b2
=
A holds.
4b2
2
b2
+
b2b7
+
b8, b8 :
+
f b77 771,
=
that
assume
b7
=
b2b8
+
By (3.20) b2b7
we =
b22T-2
=
+
(b2b7)T2b7)
2b7
b272
a
E
=
2b2
+
NB with
b2 + b2
30
(b2b7, b2b7)
=
b2 (b272)
=
b2272 ,)
+
2b7b2-
f 1, 21.
c
b2
+
+
=
blo
2(b2
+
+
=
leads
(51
3b7
=
+
2b7)
=
30,
b2
b2
for
some
15 +
2(z, b7)
+
Z
2bio.
+
72 + z, 2-b2
+
T2
contradiction
to the
+
b2+ b2 (b7 2
+
2,
+
b2 + 2b7)
compute
we
=b22 2
=
=
which
+
(b7, b2b7)
If
+
(51
=
7
b272
65
(3.21)
I and therefore (b7, b2b7) 2 51 b7 + bio, bio 0 b2 72, b7 and b7 From obtain that we Supp(z). (3.21)
2
5b2 +
2b7b2
7
means
(b7, b22)
5
b2b2 + b2b2
+
+
blo,
=
2b7
+
2b7
+
b2
Again
Hence B holds.
=
2b2
+
+
blo
implies
+ b272 2b7T2,
+
b2b8
+
6b2
T2), b2
b2b8 This
b2b8
(b2 b2)7 =(b2, b2)9
=
2
=
This
contradiction.
b2b7 z
=
Degree
of
and
(bP22)
-
40
+
2b2
=
0
+
(b7, b2 b7)
have
2b2 + b2
Element
b7 and q2b8; b7 + 77 q.
-2
2b2 + b2b7
then
=
b2b7
+
2b2 + 477
+
c
Nonreal
Then
b2b2
+
Faithful
a
+
T7)
=
b272
+
2b7F2
2b7
+
b2 (b7
2,
(72, b7T2)
=
6b2
+
T-2
+ 2 b7 +
4b2 + 2b2 +
47b7
+
b2T2
b272
+
+
b2 (b7
+
T7)
2b7F2)
i.e.,
2b2 Rom (b2, we
b772)
+
(b7, b2) 2
=
+
77)
=
4-b7
+
2b7T2
(3.22)
-
(77, b272)
1) (3.20)
and
(3.21)
deduce
b7T2 b2T7 b2 b7
b7T7 Finally,
the
=
2b2 + T2 + 2b7) 2b2 + b2 + 2T7, 2 (b7 + T7) + b2 and =
=
=
77b7
=
51 +
2(b2
+
72)
(3.23) =6b-
identity
(b2b7)T 7 b2 (b777)
=
=
2b7T 7 5b2
+
-2
b2F7 2b2 + 2b2T2
+
2T7 2
+
=
=
101+ 5b2 + 6b 2 +
101 +
9b2
+
67b2
+
6b7
b7 +
+
2b72
2T7
Z. Arad et al.
66
yields 2
b7 If
we
set
T7,
b'
exactly supply thermore, [bb2l z E f b', T' J\fTj, Bb
=
equations
=
[52, 52, 5] (3.4)
X and b E
1, b', 71, b2 T2 I and (3.19),(3.23),(3.24) Example I (with V instead of b). FurFbb21 [ b_2] by (3.17) and [bz] [53, 52] for and V E Gb. Thus Lemma2 yields b E X, i.e.,
Bb2
X
the
as
7
given =
holds
jb',T1j.
Suppose
Lemma7.
Then
in
=
b, T,
b2 b3 7
If 73
b3
,
7
T3
that
bb
then Bb
=
11 b, T7 b2, b3 73 1 7
bb2
=
bb3
=
=
2
If b2
=
b3,
then
Bb
3b + 2b3
3b+T+T3 51+b+T+2b2 is
a
11, b7 T, b2l =
b2= 2 then
particular, (i), (ii) gebras of dimension 6, condition b which fulfills In
1, b,
Bb
and
5
table
subset
and 2b +
2(b
and
=
2b +
b22
=
51 + b +
bb3
=
T + 2(b2
b23
=
51 +
2(b
b2 b3
=
2 (b +
T)
(iii) of
and
B/Bb2 is
+
Iff.
3
by
T).
multiplication
is
given
by
2b3 + b2
T + b2 +
+
+
+
b3
b3)
T) b2
supply three homogeneous which contain 5, respectively, Table
To
given
T + 2b2
51 +
T7 b27 b3l
6b-
multiplication
bb2
4 and
by
given
4b2
51 +
1, b2}
is
+T3 + T3
=
=
multiplication
3b2 + b3
T + 3b3
bb2
If b2 0 b3,
37b + b + b3, b3 --7 b,
3b3 + 2b
=
=
b3T3 Bb2
and
7
b3
particular,
=
pairwise.
b22
In
2b2 and b2
51 + b + b +
=
commute
b2b3
(iii)
=
0
b_3
(ii)
(3.24)
2b2.
+
Case 5
3.5
(i)
then
37b7
=
standard a
nonreal
table
al-
element
SITA with
3
a
Nonreal
Faithful
Degree
of
Element
5
51 1 Tb b-b, whence b2 (b-b By Corollary 2 T. and with b 3T b and T3 also commute Therefore, b, b3 are pairwise commuting. 1 Set w bb2 2b, u 2, (b, b3T) By the assumption (b, bb2) 20 and Then lwl 15, Jul
Proof.
=
b 2-
=
b
-
=
b 37
b3T
-
-
b),
-
b2,
-
=
67
-
3
b.
=
b(bb)
b2T
b2
6-b
=
5b +
=
5b + 3b + b +
=
51 + 11b +
30
=
+
6b
+
47b
+
9b +
=
2
5b + b
+ 51 + b +
b3
b3T
+
=
Z + 2b2
51 + 11b +
=
2bb2
+
T + 2b2
b3
4b + 2w
+ 4b + 2w
N + 2w
3-b
-b3
+
2b2 + 3b3 +
+
Cb +
+
+ 2w
+ b +
u
U7
K3
=
T + 2b2
+ 51 + b +
+
(3.25)
u.
Define
(6-b )2
x
+
251 + 10b +
107b
73
b2T)
+
4(bb2
+
16(b
351 +
107b
251 +10b +
=
+
b)
+
+ b
20b2
2
2Cb + T2
+
6b3
b3 is
If
(a) (b)
(b3, bb2)
(a)
In this
3b3,
b2(6-b) (b2b)T i.e.,
2
4b2
2
4(bb2
24b2 + b3 + b 3 +
+
+
2
b2b) .-2
+ 4b 2
+b
+T3b+b3b)+ 6-b
2
22(b
+
T)
real,
not
20b2
+
then
it
+
3(b3
T3)
+
(3.25)
from
follows
3(T3b
+
+
b3b)
that
+
one
u =
bb2
case, =
=
3b3
=
b3; 2,
(F3, bA
2b +
3T3, b3T
+
(73, w)
=
=
5b2
+
b2b
26-b
+
3bb3
+
b2T
=
=
=
+ 2b 2= 2
101 +
2(b
(-b3) U)
=
663
b3
+
of the
+
b3
following
3-b3
b +
+
5b2 + 2(b +
T)
+
1
42
b3. Hence
T)
+ +
+
3(b3
+
2
=
=
=
=
b2b3
2
+ 2b 2
(b2 Tb3),
=
7
-
=
=
b3(6-b) (b3b)T
73)
3bb3i
101 + 3bb3 Since 3 (-b3, bb2) b2 + 3(b3 + T3) + 2b2 2 5 1 + 4b2 Furthermore, 3b2 + b3 + 73 and b 2 bb3
implies
+
holds:
conditions
yields
b2T)
b3T + b373
+
501 +
(i)
+
42
+
2
+
4(bb2
20b2 + 3-b + b + N + 101 + 2b + 2-b + 4b2 + 3b + T
+
bT2=(37b+b+b3)(3b+T+T3-)=96-b+3(b
y
+
b3b
+
b3T + 2b3b2
+
3-b3b
+
b3b
+
2b3. Finally,
5b3 b
=
2
37b
b3b
(b3b)b
2
+
=
=
=
b +
=
6T
b +
+
+
b3b
3b + + b 2= 3
3b2b
+
b3b
+
=
+
7b3
3b2 + Ilb3+
3b3T
b3b
3b2
3b2
+
4b3
+
+
413
+
2b3b2
b-3
ff b-3
2
+ b3
6b + b + 4b3 + 11b3 + 3b2
this
3-
Z. Arad et al.
68
b23
shows that
tiplication
(b)
First
we
(b2) X)
=
(b2; Y)
=
(b3) X)
=
(b3 Y)*
=
7
=
The derived
equations
verified
that
b 2) 2
24 +
this
completely multiplication
determine
mul-
is associative.
compute
24 +
8(b2, bb2)
4(b2,
+
=
8(b2) W)
+
b 2) 2
4(b2,
32 + 2 (b2) U) + (b2, b3T3) Tb3) + (b2, b373) 9 + 4(b37 W) + 4(b37 b 2) 4(T-3, bb2) + 4(b3, b 2)2 2 3 + 3(b3, bb3) + 3(b3 b3) + (73, Tb3) + (b3, bb3) + (b3 b3b3) 4 + 3 (T3-7 bb3) + 4 (b3) U) + (b3, b3T3) 20 + 6 (b3,
b2 b)
4(b3, bb2)
I +
+ 2 (b2)
=
+
=
7
(b2, U)
Since
3b+T+F3.
=
Bb and it is easily
in
=
2 (b2,
w)
(b3 U)
and
W),
I + 2 (b3)
=
7
4(b2 W) + 4(b2 b2) 2
=
7
we
obtain
(b2, b3F3)
8 +
(3.26)
and
1+4(b3,b2 2) =3(b3,bb3)+4(b3,W)+(b3,b3T3)The
yields
assumption 45
Therefore,
one
=
bb2
(M2, bb2)
of the
2b3
=
2b +
=
(Cb, b 2)2
following
+
Hence
10((b2, b2) 2
25 +
=
cases
b4, b4 =14- b, T3.
(3.27)
+
(b,
b 2)). 2
holds:
0 (b, b2) b2); 2 2 1 b2); (b, b2) (b, 2 2 2 07 (b, b2) (b, U). 2 2_ 2 Furthermore, (b3 bb2) (b2, bb3) implies
(b 1) (b2 b2) 2 (b2) (b2, b 2) 2 (W) (b2, b2) 2 7
=
27
=
=
=
=
=
=
35 <
=
=
(bb3 bb3) ,
=
7
=
(b373, 6-b)
=
b3T3) :! 1, since otherwise b373 b5 0 b6, by Remark 2 (a) contradicting Now (b,
(b2, b373) (bl)
In this
0. Now (3.28)
case
0
=
+
(b, b2) 2
becomes 1
=
(b, b373))
2(b +T) implies b32 3b5 1. Thus (-b3, bb3)
0 and
(b, b3T3)
51 +
=
=
(b2, b2b) (b2, w) (b, b3T3) (b3 Tb3) =
=
+
(b, b32)
=
(b, b3F ) 0
=
b373)
25 + 10 ((b2,
2b6
(3.28)
< L
(3.26) yield (b2, b373) (b3 w) and (3.27) reads
and
=
7
+
=
,
4(b3,b2 2) =3+3(b3,bb3)+(b3,b3F3)51 + 2b2 + b3 + 73, (T3, bb3) 0 and b3T3 (b3, b2)2 =,4 0, so that b22 51+b+b+b3 +b3. Write bb3 b3 +2b2+p and bb3 b+b3 +b3 +q, where p E NB, b, b2, b3; T3 -10 Supp(p), b2, b3 T3 % Supp(q) and IpI jqj-
Hence
=
_
=
-
=
=
7
=
=
SITA with
3
(b3) b2 b) [Recall that (b2, Tb3) 0 (Cf- (3-25)).] b3T3) 1) ((b, 2 =
(p, q)
i.e., 45 a
=
(51
+ b +
(Tb3, bb3)
=
(N
+
Hence p
20.
=
(bb3 bb3)
=
(6-b, b373)
=
,
2b2 +
q
=
follows
T + b3
Element
Degrpe
of
-1
((b3 U)
2
2
7
5
69
b3)
((b3
1)
-
that
+
F3,
T3-
p, b +
b3)
3b + b +
+
b3
q)
+
25
5 +
=
(p, q),
2b5 and
=
(51
=
1
=
7
It
(b 3F37 b 2)
Nonreal
(b3 W)
=
=
-
Faithful
a
T + 2b2,
+ b +
T + b3
51 + b +
+
73)
25,
contradiction.
(W)
In this
2
b2
case
(3.28) (b3 W) (b3) Tb3) (W) In this U) i.e., 2 (b2 ,
7
In
u
sequel
the
follows
it
case =
T
(b, b3T3)
=
(b7 b3T3)
=
=
7
51 + b +
=
becomes I
and
contradicting
that
2
b3
reads
b3 is real.
that
=
+ 2w
4b2,
+
(3.25)
Thus
previous
(b2, bb2)
=
3b3
=
from
the
(b, b2) 2) =
(b2, b3F3) (3.27)
yields
follows
But it
7
zero
(3.25)
with
we assume
(b3 Tb3).
must be
Now (3.25)
4b2.
2b2. Hence (3.26)
+
=
a
yields
=
0
that
case.
(b2 W) 7
contradiction.
w
=
b3
1
+
2
U
and
'
1 (b3 U) 2
1 +
(b3) W)
=
7
(b3, b2b)
=
(b2; Tb3)
=
(b2j U)
=
(mod 2).
0
Therefore,
(b37 U)
(ii)
If b2
b3, then (3.29)
=
b2 (6-b)
(b2b) i.e.,
b22
=
51 +
26-b
=
2(b
+
+
b).
determined.
completely
implies
5b2 + b2 b +
=
T2
2
=
+
Thus
b2T
Bb
=
2b +
2 2b 2
=
2b2 Wecompute
+
-
9b2 + 3(b
101 +
7(b
+
11 b, T, b2l 7
is easy to check that
It
(3.29)
(b2) U)-
bb2
+
2b2T
=
T)
+
If b2
9b27
multiplication associativity w
=
=
2
=
=
2
+ 2b 2
and the
2(b2+b3) and 0 b3, then (3.29) implies u T+2(b2+b3). In particular, 2b+2b3+b2, bb3 i.e., bb2 51 + + means b2 Moreover, T). 2(b 3
(iii)
T)
+
in
i.e.,
b22
=
=
(b2b)T
=
=
b2T + 2b22 2(b + T) + 7b2 + 4b3 26-b + 2b3T + b2T 101 + 4(b + T) + 9b2 + 5b2
b 2b 2
b2b +
2
+ 2b 2
=
51 + b +
b2(b2b)
+
T + b2
+
2b2b3
=
2b2b
=
5b + b2b +
+
b3- Romthis 2
+ b2
b3b
+
=
we
obtain
51 + 5b +
V + Cb
=
T + 3b2
51 + 9b +
+
5-b
5b3
+
is
b3+16U 2b3+b2) 2 (b, b32) (b3, bb3)
=
=
b2 (6-b)
b
law holds.
+
5b2
2b2b3 +
5b3
Z. Arad et al.
70
b2b3
i.e.,
2(b
=
T)
+
b2. Hence Bb 11, b, T, b2, b3l. The derived in Bb and associativity holds.
+
equations
=
multiplication
determine
51 + b + T + 2b2, b2 0 b, b 2 Suppose that 6-b 51 + 4b2; i. e., Bb2 b4 7 b, b, b3 and b3 =,4 b. Then R 1, b2 2
Lemma8.
=
0
3b4
=
+ b +
b3
=
Since
Proof. and
T.
b(6-b)
=
b2T
=
5b +
b2 +6-b
3b4T
+
b3T- b, Jul
=
bb2
w :=
10
-
Ju'J, JvJ
=
2bb2
+
C5 + b3T
6b + M4 + b3 + 2bb2
i.e.,
=
=
(6b
2
51
-
particular,
Case A:
=
this
u
=
2b5
u
=
b5
51 + 7b +
=
3b4T
3b4T
+
+ 51 + b
b3T-
b
-
-
T)
commutes with
b
Jwl
20 and
shows
b3
+
that
us
u
b4T
u :=
The last
15.
=
Set
T + 3b4 + b3 + 2b2 + T + 2b2 + bA 3b, u'
-
+
2bb2
Tb4
:=
-
3b,
v
Supp(u) USupp(u') USupp(v) USupp(w),
2b E NB- Then b
3b4 In
T, b2
b commutes with
Now we compute
+ 2w
=
equation
3u +
be
cannot
a
now
reads
(3.30)
v.
basis
element
and either
or
Case B:-
+
Case A. In this 65
(b,74b4)
i.e., x
:=
b4T74
b6, b5 0 b6 holds.
case
(bA b4T)
=
55 < 45 + c
is real.
=
=
there
Icl
necessitates
b4
=
(b2)74b4)
+
Lemma I
Hence
obtain
(74b47 Tb)
=
This
4.
=
25 +
10((b, T4b4)
immediately
+
(b2) 74b4));
74b4
yields
=
51 +
4b2- Set
51 then
-
(x, x) By
we
=
5
is
a
and
x
74b4 we
E
=
4c.
b4T4
-
25
Supp(x)
c
=
hand,
(b4T4) 6b)
=
25
-
(c,x)
such that
and Table III
=
51 +
we
25 +
(with T4-
4b2 and b4b2
Therefore, .0 (b, b2) (b4) b4T) 4 3b4 + b3, i.e., implies w 3b5 and v =
=
bb4
(-b4b4) 74b 4)
=
>
80. (Xlx) 1XI
A
This
have
40((b, c)
instead
of
+
(b2) C))
b) yields
that
conclude
b42
=
=
On the other
Cbb4)Tb4)
(u', u)
b2, i.e., Thus
(b4b4, b4b4)
=
=
(3-31)
b2b4-
2(b4, b5), i.e.,
b4 0 b5. By (3.30)
this
=
=
3b+2b5,b3T=
b+3b4 +b3 and bb2
=
2b+3b5-
(3-32)
SITA with
3
From this
Case B. In this =
(b, 74N)
i.e., BI:
(M2, bb2)
=
R 2
shows that
55
TA
Element
Nonreal
of
Degree 5.
71
obtain
we
65
which
Faithful
a
case
we
=
2)
10((b2,
25 +
b2
+
(b, b2)), 2
+
(b2774N)))
that
either
b24-
=
obtain
(-b4b4)Tb)
(b4T, bJ) (b2) 74b4)
+
4b2
51 +
=
(6-b, b 2)2
=
51 + 3b2 +
=
51 + b + b + 2b2
10((b,74b4)
immediately b2
3. This
T7-:
b7, b7
=
25 +
=
yields
or
B2:
b4b4
3b4 B 1.
=
This
+ 2w
3b5
=
3b6
+
+
becomes
(3-33)
V-
0 shows that b4 (-b4b4; T) b5, b6- It (b4 NT) (b4, b5 + b6) that that w we assume b3 2b5 + b3) b6, w.l.o.g. (3.33) may + b5) 3b4 b2 b3T) (b2, (b2 V) 3b4 + b5 and 1 (b2, (b3, b) W) (b3
Here
=
=
7
=
from
follows v
b3
+
(3.30)
Purthermore,
bb holds.
=
=
=
I
=
7
7
the contradiction
yields =
(b5, b4T)
(b2, bjb)
(b4, b2b)
(b4) W)
(T, T4b4)
(b4, b4T)
(b4,
(b47 2b5
=
+
b3)
0-
=
B2. Wehave I
Hence
we
3b5
+
(a) (b)
b3 b3
Let
us
v
assume
may
and
b5, b5, first
we
w
=
w
=
b3 + c and 3b5 and v
assume
1 +
We distinguish
the
b3 (al) b2 b3 (a2) b2 Here we get I (al) that b4 b3 b2). =
=
b3
(a)
that
(b31 C)
v =
=
means
b4
=
2c for
some c
Icl
E NB with
b3
+ 2w
10;
M5. Then
(b3, b2b)
=
(b2, b3T)
=
(b21 V)
=
(3.34)
2(b27 C)-
cases
=
(b3, b4T)
=
(b4, bb2)
(b4) C),
so
that
c
2b2
+
=
b2
But then
b + 2b2 + 2b4 which
becomes
=
+
holds.
b6)
b4. Then (3.30) b6 possibilities:
that w.l.o.g. following
the
obtain
3b + b5 +
=
T contradicting
b3T
=
b2T
the
72b
=
assumption
b
+
74
of the lemma.
+
b4 (recall
Z. Arad et al
72
(a2)
In this
b3
bb2
=
-
b2(bb)
=
(b2b)T
=
b2= 51 2
i.e.,
(3.34)
yields
b2. Next
we
case
2b
-
5b2 + b2b
26-b
b4
and therefore
(b, b2) 2
+
Y we
:=
T,,(3)
the
denote
X
f el,
wo,
:=
+
b3 In particular,
integral
4(b
101 +
=
(b4, b3b)
=
T)
+
b3
-
b commutes with
=
9b2
is real.
+
73)
4b3
2
+ 2b 2
+
_b3
Hence
(b4, T + 2b2
in the
as
+
+
3
+
previous
2b3) case.
Wehave
(Cb, b 2)2
=
3T
:=
..
25 +
=
already
N and T
E
Set
that
note
7b2+2(b+b+b
b2T
4. But this
m
T.,,(55).3
of
=
+
(b).
(b2, b22
+ 2b 2
contradiction
a
b3. First
+
2
b2b
(bb2, bb2)
=
Example 2. Let unit
T,
=
b2
=
compute
(b3, b4T)
=
consider
we
65
i.e.,
2b3T
T + b2
+ b +
1
Finally,
+
+
c
10((b,
implies
(b2,
eo, uo,
..
(3 )
fvi
wmJ7 where
el
is the
b 22)),
(b2,
b 22)
4.
where
u.,,
:=3ui;O
GT-algebras
standard
b 22) +
i
by
generated
the
mlUfeol
and
By V Y,
W,
of Y.
unit
is
eo
X and
and
respec-
tively.
(i)
X
ZP
Y
zj-
Vi + wi,
:=
Vi
:=
-
Wi; 0 :5 i <
ml
U
U
5eO ell is a basis for V Wsuch that (V (D W, X Y) is standard homogeneous table algebra of degree 5 with unit 1 and involutory automorphism 5 -+w := VV + WW,v E V, w E W. eo + el,
(ii) (iii)
h
:=
b:= vo + wo fulfills
1, hl is Moreover, X
H
Proof. following
First
note
a
6eo
bb
=
closed
Y/H that
=
the
1
-
=
-
5. 1 + b + b + 2h.
subset
with
nontrivial
=
9UiUj
(ii)
and
Clearly,
-V
phisms
10,
basis
vi+j
+
15eO
+
vi+j-m
set
idempotents
primitive
elements
multiply
of X
eo, el.
in
the
way:
ViVj
(i)
central
T,(5).3
.
.
.
and ,
ml
X
-W
Y is
a
basis
4vi+j+l (vo + v,) + 4vi+j-m-l for
of V and Winduce
and since
eo, el
are
real,
if if if
i+j i+j i +
< M, =
j
M,
>'M.
V E) W. Since the the
same
automorinvolutory on the index permutation
B is invariant
under
the
involutory
SITA with
3
automorphism _'tZzi
+
=
-V
:=
ziz i
+
vi+j
zP+ i+j
+
15eO +
4vi+j+l
zt Zi
=
7,
vi+j-,,,
+
+
+
4vi+j+l
zi+j 15eo
3z,-+i+j+l
This
=
+ 3z
be
=
wi
=
el)2
The follows
following Example
Lemma9.
GT-algebra
+
3zi
+
2zi-
associative.
5(eo
=
if
i
+j
> M,
if
i+j
if
i+j=m,
if
i+j>m,
M,
yield a complex that
(V
that
el)
4(5eo
+
Y)
W, X
E)
-
is
el)
-
Y)
W, X
table
algebra.
(XY)
function
algebra
valued
51 + 4h.
=
integral
an
the constant (D
<
2wi+j-m-l
-
+
H+
of H are
zPH
I
THRTJ
zP,
zi
I
zP
+
Z.
=
x
151
homomorphism of
is standard
and homo-
=
3-+h
=
=
6
zi-H
eo and 1
-
eo
and therefore
I
'
(zi-H)+ 6
=
-
=
7,
-vi
=
3
Ui.
proof. of
construction
3.3 of E N
basis
and A E T-n
-A
:=
Moreover,
0 < i < n,
is
a
certain
basis
for
a
polynomial
algebra
[32].
-
A-I 2
-vn-i,
=
=
i+j-m-l
(V
have
,\-lw.+. 2
is
+ el
to
we
n
WiWj
wn
Z+
idempotents
A with
Let
-
wi+j-m
+
means
(ziH)+
6 the
wo
-
-
equations
ml I
finishes
i+j+l
-
25eO
simple
-
This
+j
5.
10,...,
For i E
7,
already
degree
The central
el.
i
2wi+j+l
-
3el
-
extended
V (D W. But this
(iii)
2wi+j-m-l
+
and shows that
linearly
geneous of
z
Z+3-M-1
2zP 3z! =
the above
Moreover, can
wi
_
wi+j
-
+
4vi+j-m-l
Z_ i+j-m
(ii)
proves
< m,
=
+ zO+ + zM +
(5eO
i+j
lif +
+ wi+j-m + z-
i+j-m-l
+ vo + v.,,,
vi+j-m
5vi + 2
if
+ 2h
i+j
+
3h +
h
Then
+ wo + w,,,,
el
-
3z+
+
hzi-
ml.
73
2wi+j+l
+
4vi+j-m-l
vi+j
-
10....
E
5
Wiwi
-
5vi
+ 3
vo + v.,,,
zO+ + z,+,,
hzP
j
Degree
of
zi-+j+l
+
51 +
vivi
I
+ wi+j
3zp+ i+j+l
Zi+j-m
Choose i,
-
Element
Nonreal
+ wiwj
vivj
=
-W
ED
Faithful
a
an
?,
(A) 3
I +
+
A+1 2
'\-I
Wi+j the
R>I. f 1,
=
Wnj
table
Wi+j+l -
'\+I
linear
involutory
standard
WO,
(Wn n
The n-dimensional
2
WO) Wi+j-n-I
and
if if if
integral multiplication i+j < i+j=n, i + i >
defined by T mapping algebra automorphism. -
n,
n
=
1 and U7
Z. Arad et al.
74
Proof.
proof
The
[321 (for
'm
'p(b)
polynomial Wewill
Example
(3)
T-
follows
0')
=
make
Let
3.
lei,
the
:=,8b,,+,
now
:=
the
with
-
construction
,
+
wi+j
-3el
WiWj
if 7if if
wo + w,,,
2wi+j-m-l
-
of
Example
has to
,
in
out
Al
-
of
3.3.
factor
=,3b,,+,
Example as
3
2wi+j+l
-
-wi+j-m is the unit
3T,,,()
X:=
'p(b) in
as
one
one
of
w,,,I,
-
-
that
ab' instead
N and define
mG
wo,
similar
the
as
difference
+ Al +
a
pattern
same
only
-
the
ab'.
0
2.
Example
2 and Y:=
i+j < M, i+j=Tn, i + j > m,
of Y.
where el ated by X and
By V and W, we denote again the algebras generThen the statements Y, respectively. (i) to (iii) of Example 2 remain true. Moreover, T- (3) is not isomorphic to 3T,, (iv) 3Tn( E)3 Tn(3). M 3
(i)
Proof.
(ii)
and
real, z+ m-i
we have 1
X
Y. Choose
c
zj- zj t
=
z-.z-. "
11
ztz,-.
1 E X
=
Y for
X
all
i, j
i
l5eO
+ vo + vn
3zP+ i+j+l
3h +
z-
zi+j-m
+
3zi+j-m-l
vi+j
+
4vi+j+l
zi-+j
+
3zi-+j+l
l5eO +
+
ZP+j-m I
=
(5eo
h2 This
proves
Again, the be linearly This
proves
+
wi
=
wi
=
-
(ii) above
el
)2
are
zm-i, zi elements
-
of
2wi+j+l if
wi+j-m
+
-
i
+j
< M,
)if
i+j=m,
)if
i+j
> M,
if
i+j
< M,
if
i+j=m,
)if
i+j>Ta,
2wi+j-m-l
Zi+j-m-l
wi+j
-
+
-
2wi+j+l
zi+j+l -
+ wo
el
-
wm=
4vi+j-,n-l
+ wi+j-m
3zt+j-m-l
2zP 3zp =
+
3zi-,
+
2zi-
25eo
+
zi+j
+
equations
(V
to
E)
a
2wi+j-m-l
-m -1
1
+ el
=
5(eo
+
el)
+
(V (D W, X Y) yield that the constant complex valued algebra
and shows that
extended that
eo and e]. ::--
the
permutes
zO+ + zz;
51 + 2h +
5vi +
wm-i
wo + wm=
-
-
vo + vm + 3
vi+j-m
=
Hence
-
WjWj
-
hzP
vm-i
+ z,
4vi+j-,,n-l
hzi
+
3e,
-
+
-
V (D W. Since
Z,-.+i+j+l
+
vi+j-m
5vi
for
zP
Furthermore,
+ wi+j
+
=
basis
+ WjWj
zP++j
VjVj
a
10,..., ml. ml. Then
E
4vi+j+l
+
Y is
Y.
10,...
(E
VjVj
=
I
vi+j
X
Clearly,
W, X
Y)
is standard
4(5eo is
an
-
el)
=
integral
function
51 + 4h.
algebra.
table
(XY)
homomorphism. and homogeneous
x
f 51
can
of V G W. of
degree
5.
SITA with
3
proof
The
(iii)
of
is the
Faithful
a
Example
in
same as
Nonreal
Element
Degree
of
5
75
2.
IL, h, zt, zizi- defined that Assume i E T, (3) and ml. 10, si, via 0. Since both table algebras T- (3) are isomorphic are stan3 T,(A) 3 dard and homogeneous, 0 is an exact isomorphism. h2 and Clearly, 0(hi) T- (3)\j h2j with since ro and sm 70 are the only elements z of 3T,, ( )3
(iv)
We change
the
Example above by X2, h2, ri,
defined
by
2
in
Let
notation.
hl,
xi,
pi,
.
.
denote
us
the
elements
,
.
zp,
1, h, 3T,, ( a)3
elements
and the
qj
-
n
=
=
z
E
Supp(z-7)
M
follows
it
from
O(po)O(po)
holds.
0(po)
A:
either
that
O(po75o)
=
=
Let
=sm.
first
us
that
assume
A
case
that
Weclaim
0(pi) induction
For the
0(po)
B:
or
ro
5X2 + 2h2 + O(PO) + O(PO)
=
=
step take i
ri
for
all
E
10,
11
m
-
(3-35)
ml.
10,
i E
and
0 (pi)
that
assume
ri.
=
Then
ri
+
Since
30(pi+l) 0
+
O(qj+j) this
is exact,
the contradiction Next let
us
LpLj+j)
means
sm
=
assume
77o'
for
sm-i
=
E
10,
=
.
.
,
I
I
m
-
.
10,
i e
=
case
.
.
and
.
3rj+j
+
ri
=
(3.35).
o(p,,,)
In this
all
rori
=
proving
rj+j
O(T-o)
=
B holds.
case
step take i
For the induction
=
0(po)
=
that
O(pj)
0(po)O(pi)
0(popi)
=
+ si+,.
(3.35)
But
yields
rm. we
claim
that
(3.36)
MI.
,
assume
0(pi)
that
sm-i.
=
Then
Again
fact
the
But then
ro
A E xy
=
0
that
=
Tm6
R>o
and x, y,
'X+1
-\-'
z
2
Set
Proof.
(A
+
4
+
1)2
*9
z
x-z
A +
-
-
2
O(qj+j)
yields
is exact
be G
=
a
O(To-)
=
1.X 7+
O(popj)
=
sm-i
0(pi+,) O(pm)
such that
V,
A+1F. 2
=
=
2'+1
7 +
2
=
O(po)O(pj)
+ 3sm-i-I
so,
0
A-1 2
xT
x,
7 and z-z
Al + =
holds. 0
contradiction.
Suppose =
smsm-i
(3.36)
Hence
final
a
=
+ rm-i-1.
sm-i-I.
=
=
algebra.
table
standard
y. Then xT
2
u:=
+
O(po)
=
(U, V)
Let
Remark 7.
30(pi+,)
+
ri
that A
1 2
there
(x
+
7)
x-T.
Then
A2
_
4
1
1
(xx)7 (A
+ 4
=
1)
2
x7y A-1
A+ 1 2
u
+
2
xy'
exist
and
Z. Arad et al.
76
i.e.,
u
2
1'gl proving A2 + 1
A
2 so
(x, zT)
that
=
Proof.
If
(h, a7x-) (y, ah)
aTx)
b3 0 b, b4) T Bh 11, hj
b2
(ii)
jdx-j (h, a7a)
=
all
y E
B,
is
a
coset
and
if G
Yy-
=
E also
(bH)+Tb--H)+.
=
=
6b,
x-y consisting
3c, for
=
Now 3b 2+ 2bc
Since
H
set
:=
particular,
=
2c7b supply
b is nonreal
T(bh) (-bb)h
=
=
b(ch)
this
2-bb
+
=
c =
of degree 6-b-
is another
(bH)+G+
of
two
elements
(h, 6-b)
(y, y)
=
2,
B\jbj
,
3-bc
2cb +
c2
b2 and
5h + bh
=
101 +
+ T-h
for
c.
2(b
+
+ 2h
2
+ b +
b41
H
=
ch
(bH)+(bH)+,
(6-b, h2)
5
satisfying
2b + y for
=
=
3c+2b,
from bH consisting =
2G+ + 8E+ for
bh
=
then
some
H-
E+E+
some
y c
NB,
65
i.e.,
c(bh)
true
distinct
of degree
(bh, bh)
3b3
2b+3c,
=
G+G+
that
so
=
5 such that
holds:
H-coset
5 such that
=
and b 2
5 and bh
and
is also
T
of degree
following
2b + 3c and therefore
=
=
=
element
b-c
=
some c c-
bh
36c- +
we
4. In
from
=
basis
a
of degree
20 + y
If
degree
of
a
follows
it
Then the
jx, yj
x,y
(i) Wehave (bh, b) V Supp(y). Now
yields
51 + 4h.
=
51 + 2h + b +
=
subset.
Proof. b
bTb
that
closed
elements
=
1) (X, ZZ)) A,
-
=
that
=
two
xx
(A
4(x, ah) (ah, x) 0. For x =.a, this (x, ah) (h, dx-) 20 (ah, ah) (a7a, 51 + 4h) jahl. Hence a (I + h). 0 (aH)+ then
and
so
1b, cl for some nonreal 3h + c + Z!, c7c C2, b-c
=7 T
c
+
elements
16 that
=
=
=
If of
degree 4,
b4 =7 - b, where h is
=
bH
of
Suppose
Lemma 10.
(i)
(A
h2
h E B satisfies
aH+ for all basis
and
0
=
I for
=
=
obtain
we
0
that =
E B is
5 (h,
(a, ah)
yields
(XY, 27)
=
Romthis
assertion.
24.
x
=
of the
part
1.
-
(aH)+
then
1(aH)+1
A
(XZ, X7)
=
Suppose
Remark 8.
11, hj,
=
the first
=
ch
=
3C2 and 3bTb
Cc
=
(3-37)
3c + 2b. +
2c7b
=
T(ch)
=
6-b.
c(Th) (3-38)
Now
T)
+ 4h +
101 +
2(b
+
3-bc
b)
+
3(c
+
Z)
+ 13h
implies Te
=
M+
c
+'E.
(3.39)
SITA with
3
(ii) (i) yield
Set
(x Clearly, for
all
C,
=
u
(bH)+
:=
y)(Y
+
T)
+
=
c
and
v
u-u
=
10(l
for
and therefore
al
2.
=
Nonreal
G+
:=
h)
+
Ci.
H-cosets
Purthermore,
N<j.
E
v
vU
=
aiCt
uv
i
b +
=
Faithful
a
2
Set
uv
w
2(b +T +
+
Since
(b, vU)
2v
c
+
we
we
may
product
than
greater
degree
5 whose order
Now Remark 8
is
JE+l
yields
%
=
lei
jej,
=
(b
+
c)(x
+
y)
=
4. divisible by 5,'we conclude that IzI 80. Thus all contradicting ajiE+l < Jul of a E cannot consist divisible by 5. Clearly, two possibilities: that we obtain the following not
=
=
uv
=
G+
lG+l
(ii)
4.8
Proposition
Rom (3.42)
=
2v +
8g
2(x
=
+
y)
+
(3.41)
8g.
assume
either
y)
+
=
G+ =yH+ =y+yh
IG+l
(3.42)
2x+3y and yh= 2y+3x.
(h, x-T)
xT
3(x
[201) implies
of
deduce
we
that
=
=
IFy
or
2
=
(h, y`y)
and
Cb.
from
is distinct
(h, x-y)
=
3
=
Rom(3.38)
(h, Zix-). (3.40)
and
obtain:
For d c
1b, cl
and
(dz, dz) Using (3.41),
=
we
z
E
xx
=
yy
=
51 + 2h +
xy
=
yx
=
3h + b +
Ix, yj,
(cFd, z-z)
we
(51
=
=
x
+
2g,
cy
c
=7
+ 2h + b +
b,
have
(a, dz)
shows that
this cx
Next
a
degree
that
note
xh=
us
of
substituent
degree
5 and
10.
x+xh=xH+ (cf.
such that
E+. Then
Set g:=
First
a
lwl
=
=
=
E
contain
11
whose
degree
24
degrees of elements of E are single element of degree 5, so 5 Ifl (a) E fe, f 1, lei
(b)
5 cannot
Then
z
that
assume
..,
=
Ci =A H
have
Eli=2 aiCp.
=
of
and
(3.40)
bH,
640. By Lemma I there exists a j E 12, 80 and (w, w) wlw) element E contains that an := 8. > Suppose Cj aj IWI is not divisible by 5. Since bH and G consist of elements of
of two elements
77
Z).
that =
5
assumption
G =A
means -
The
+ y.
x
=
Degree
of
Element
=
y +
-
as
< 2
for
2g, bx
=
+
c.
b
U,
-
51 + 2h +
all
a
y +
c
c
+
Supp(dz)
2g, by
=
Z)
2g.
3b3
+
compute
b2 +6-b
b(:i-y)
=
3bh +
(bx)-g
=
y-y + 2g-y
=
=
51 + 2h + 8b +
51 + 2h +
c
+V+
T+
2jy-,
9c +
45.
and therefore
+
x
=
b4
Let we
78
Z. Arad et al.
i.e., b4
8b + b + 8c + 3b3 +
(cf.
Therefore
VT
-bc
=
b(Cb)
b4
=
7! +
+
T2
+
6b-
follows
Hence it
Let
( t-x, d E
T, cj!J,
now
us
6-b)
55,
=
Supp(xb)
whence 50
have
such that
(xb
=
B/H
quotient
assume
we
-
37b
=
T
+ 3c + b +
+
T+
10b +
=
c
this
yields
b3
'z and
+
a-O 20
-
x)
90,
!
a
(x, xb) 2
>
we
(3.41)
,
3
x
=
.1u, 6
y
Supp(g)
proved
x-x
Bb y-y
=
and
(xb, xb) supplies
I
=
a
e already implies Finally, passing to the Iv and z -1g) yield 6
d
=
=
=
6
uUm
0
Suppose that 6-b'= 51+b+T+2h, b 2 T. Then 2 51 + 4h and (A, B) is one of 3, b3 =/= table algebras of degree 5:
Theorem 5.
=
B
=
11, b, T, hl, 3T
B -2-'x ments
is
a
..
b 2=3h+b+T
( )3
T,,,
(3),
cosetofH:=j1,hj
VjVj
=
1, h,
+
wi+j-m
hvi
=
2vi
+
3wi,
hwi
=
3vi
+
2wi.
3Tm( )3
enumerated
b,
:=
3vi+j-,n-l
wi+j + 3wi+j+i 3h + wo + w,,
Viwj
B --- x
vo
wo,...'
means
that
1, h,
T-
(3),
M
vo
the
vm, wm such
basis
ele-
Ivi, wil
that
and
vi+j-m
(iii)
This
2m + 4.
=
+ wi+j+l vi+j + 3vi+j+l 51 + 2h + vo + vm
wjWj
=
homogeneous
andbh=3-b+2b;
dim A
be enumerated
can
11b, b3, b4jj
3b3+b+b4, following
the
=
(i) (ii)
=
bb.
=
Lemma I
=
that
so
contradiction.
(withA
and Remark 7
that
Thus
3. But then
!
-
Cb,
(3.43)
Since
holds
6b-
6c + 37! +
+ Z.
b is faithful.
as
that (b) (xb-x,xb-x 1XI (d, xb x)
x, A
bTb
(3.37),(3.38),(3-39),(3.42)
from
contradiction
a
-bc
=
5b + 2bh + b 2 +
=
bc
1, h, b,
T, b4,
2g-y. Since b3 0 b,
(3.39)),
:=
b,
+
+ wi+j-m-l
=
+ vi+j-m-l
2m, + 4.
vm, wm such
i
+j < M, i+j=m, i + j > M,
,
The basis that
+j < M, i+j=m, i + j > m, i
)if )if if
+ vi+j+l
3wi+j-m-l
dim A wo,...,
if if if
elements
Ivi, wil
is
a
can
coset
be
of
SITA with
3
11, hj
H :=
ViVj
+ wi+j+l vi+j + 3vi+j+l 3h + wo + vm
=
+
wi+j-m
I
ViWj=
b5T5
and If b5
2b +
=
=
Hence
T,
3wi+j-m-l
3wi+j+l
Let
+
vi+j-m
2vi + 3wi,
hwi
=
3vi + 2wi. we
3vi+j-m-l
+ ch
+
wi+'j-.-l
have h 2= 51 + 4h. Lemma10
3b,5 (b5 -7 b), b5 h Cb.
(i)
and
holds
79
5
=
b52, 6-b-5
2 3b + 2b5, b=
)if 7if if
i+j <M' i+j=m, i + i > m,
if 7if 7if
i+j<m, i+j=m, i +j > M,
provides
following
the
3h + b 5+75
=
(3.44)
assume
multiplication
C
=
cH+
=
1c, dj
=
by (3.44).
C+
1C1
=
(3.45)
b.
and E of
two elements
exactly
determined
completely
is
that
suppose that
containing c
Degree
,
51+2h+vo+wm
=
may
us
+ vi+j-m-l
+ vi+j+l
b5 :
H
of
=
then
we
+
wi+j
hvi
Proof. By Lemma8 equations: bh
Element
and
WiWj
=
Nonreal
Faithful
a
3(c
le, f I
=
+
d)
from
distinct
H-cosets
are
5. Then
degree =
C+
ICI
=
dH+
d + dh
implies
(3.46)
ch= 2c+3d and dh= 2d+3c.
Set eo
:=
-'+h 6
and el
eo
(1
eic
-
=
51-h. 6
eO)C
=
By (3.46)
-(C 2
-
we
have
(3.47)
d).
Therefore, Ce
=
(eO
+
e0ce
((c
+
d)(e
+
f)
+
(c
-
d)(e
-
f))
2
(ce
+
df),
i.e.,
ce=
df,cf
=
de and
(c+d)(e+f)
=
2c(e+f).
(3.48)
Z. Arad et al.
80
Let
that
us
that
now assume
E is the
uniquely
C
J ,75}
=76
determined
Lemma 10).
By (3.48)
c(b From Lemma10
(c, cb) Now it
follows
either
Using
Lemma10 and
Ivil
5
=
=
1wil,
..
vovi
vj
j
(vi
0,
E
:=
viwj
.
,
.
3vi+l
+
vi
we
=
m
-
.
+
1
=
=
+
b5),
so
that
f).
+
F5 (b5) Cb5)
=
(c7c, Cb)
=
define
we can
JB/Hj
m:=
,
=
By (3.48),(3.50),(3-51)
Take i,
4(e
2c(b
=
(3.49)
(b5
7
(d, 65)
C-tC
55, that
=
b,
wo
+
recursively
21
-
in the
(3.50)
H-cosets
following
Ci
=
fvi,
I,
wi
way:
b5) CO := JVO) WOJ) 2& + 8CP+l and
:=
=
3vi+l
0 < i <
+ wi+,,
(3-51) L
m
-
have
wivj,
11.
+e
=
CO+C-
vivj,
+ d +
d)
+
c+3e+f and 65 d+3f c+3f +e and 65 =d+3e+f.
vo
wiwj
c
(cb, cb)
and
(3.50)
f 0,.
i E
=
b5)(C
+
cb=
cb=
or
(b, 6-b)
=
(3.49)
from
b5)
+
E+E-+=(bH)(bH)+
and
(b
and
obtain
we
(b, c7c)
=
have
we
(bH)(bH)+
=
satisfying
(b+b5)(C+d)=2(c+d)+8(e+f) (cf.
C+U+-
satisfies
H-coset
wivo
(3.5 1)
From
wi+j)
=
viwo
=
and
wi +
(3.52)
vj
(vivo)
=
vi
(vjvo)
wj
(vivo)
=
vi
(wjvo)
3wi+l
we
vi
(3-52)
+ vi+,.
get
(vj
+
3vj+l
+
wj+j)
and
wj
i.e.,
vj
(vi
+
3vi+l
+
(3vi+l
+
wi+,)
vj+,).
Multiplying ing (3.52) yields
8vjvi+l
wi+j)
=
vi
(3vj+l
the first
=
+
wj+,)
equation
9vjvi+l
+
3vjwi+l
vi
and wj
(3vi+l
by 3, subtracting
-
3wjvi+l
-
(wj
+
+
3wj+l
wi+,)
=
+
vi
vj+j) (3-53)
(3wj+l
the second and
wjwi+l
=
8vivj+j,
i.e.,
vivj+l viwj+l
=
=
vi+ivj vi+lwj
=
wiwj+l
(cf.
(3.53))
=
wi+lwj for
all
and
i, j
E
10,...,
m
-
11.
+
apply-
(3.54)
SITA with
3
(3.52)
Rom (3.51),
wiwj
=
vivj
viwj for
By construction b5
A.
this
In
vov,,,
=
all
for
10,
E
have C,,,
=
vn(vi
=
=
vi
=
w,(vi
=
vi(3h
(51
10,...,
+ wo +
and
+ wi+j+l i +
j
<
(3.55)
m.
the two
obtain
we
show that
-
wn)
11.
possibilities:
10vi
=
viv,,,-i
vow,,,,
3vrnvi+l
+
3wn'vi+l
10wi + 6vi + 3wi+l
=
=
3h +
b5
+
75
+ vnwi+l
3vi+l
+ 6wi +
wvi
wiw,-i
=
b_5
=
Then
+
vnvi
=
=
wi+j)
+
=
m
wi+j) + vn)
+
3vi+l
81
5
+ vi+j+l
and viw,,,-i
+ 2h + vo +
(3.54)
and
3vi+l
Degree
of
Element
T5
Take i +
with
Co. Hence Uo; b Vo;
=
=
(3.44),(3.52) ml.
3vi+j+l 3wi+j+l
m}
...,
51 + 2h + b +
=
v,n(vivo) vi (vrnvo) w,,(vivo) vi(w,vo)
i, j
=
fO,...'
i E
all
+ +
wi+j
=
Nonreal
get
vi+j
=
vowi+j
Uo and w,, Uo and w.,,
case,
6-b
we
vovi+j
=
=
we
A: B: v,,
(3.54)
and
Faithfal
a
+ wi+1 + vivn,
+ wrnwi+l
--
+ viwr,,
+ vi+j
+ v,,,vi+l 10vi + 6wi + 3vi+l + wi+1 and 3v,,,,wi+l i.e., 3vrnvi+l + v,,,,Wi+l the + + Multiplying + (3.52)). 6vi 10wi (cf. + wnwi+l 3Wi+l vi+j 3w,,,vi+l the second and 3 yields first subtracting vnvi+l wnwi+l by equation =
=
=
+ 3vi
vi+j
+ wi
(3.54)
with
follows
and therefore that
and wivj ml with
wi+j-,n-l
=
wiwj +
B. Rom (3.44),(3.52),
(3.54),
E
75
and viwn-i
10,...,
m
3h + b5 + Take i E v,
(vivo)
vi(v,vo) w,, (vivo) vi(w,vo)
-
j
=
11.
>
wnvi+l
wi+1 + 3wi + vi.
=
Now it
=
v,vi+j-,
=
=
m.
-
we
obtain
vown
=
vivn-i
=
wiwn-i
51 + 2h + b +
=
T for all
vovn
i E
=
f 0,...,
b_5 MI.
Then
wi+j)
=
vn(vi
+
=
vi(3h
+ wo +
vm)
=
w,(vi
3vi+l
+
=
vi(51
+
=
+ + 3vi+j-,n-l vi+j-,,, for all + vi+j-,n-l + 3wi+j-,n-l wi+j-,n 3 T,,&E) Hence we conclude that B 3
vivj
=
vrnwi+j-,n
i 10'...' Example 2). T,n(3) (cf.
i, j
vnwi+l
3vi+l
+
=
=
+
3vnvi+l
10wi + 6vi + 3wi_ j
wi+j)
+ 2h + vo +
vrnvi
wn)
=
=
wnvi
+
+ vrnwi+l + vi+j
3wnvi+l
10vi + 6wi + 3vi+l
+ vivm,
+ Wnwi+i + wi+1 + viwn,
10wi + 6vi + 3wi+l + vi+j and 3vnwi+l + vnvi+l i.e., 3vnvi+l + vnwi+l the + 6wi + 3vi+l + wi+j 10vi + (cf (3.52)). Multiplying 3w,vi+l w,,,,wi+l the second yields first vnvi+l wnwi+l equation by 3 and subtracting =
=
=
wi+j-m
wi+j-
Tm-(3)
=
Applying also .(3.54) shows that wiwj vmvi+j-m vivj and therefore + vi+j-m+3vi+j-m-l wivj +vi+j-,n-l +3wi+j-m-l B m. > 3-Tm(sE) G Therefore, 0,---'mwithi+j --,for allij 3 0 (cf. Example 3).
wi+1 + 3wi + vi.
M
=
=
=
=
Integral
Standard
4
Faithful
a
Width
Element
of
Algebras Degree 5
with and
3
1,2
Bfinger
F.
Real
Table
1Department
of Mathematics
Computer
and
Science
University Ramat-Gan 52900, Israel Bar-Ilan
2Vogt-Groth-Weg
44
Hamburg, 22609, Germany
Introduction
4.1 This
(A, B) real
with
deals
chapter
L(B)
with
==
element
basis
Supp(b 2)\111
obtain
two
III
and
degree
b of
JbI
=
Theorem I. sume
Suppose fl,
that
B,,
ad
(A
-
=
3)
such that
=
Az-gxlxl,
x, y,
z
B,
cz
I
[b 2]
(b 2, b2)
[15,
proved by
51]
75
5252]
65
53
Suppose that (A, B) a prime integer
jai
a, d E B with a
=
=
d
=
Al +
Idl
=
U);
is
a
+
a
=
=
fl,
integral
Ivi
A and a7a
Ba
Theorem 2.10.
standard
such that
or
[331,
Blau in
H.I.
: =
a,
\
-
Al +
dj
(A
for -1
and a2
all
1)d. =
As-
algebra.
table 1
v
E
B\f 11.
Then either
Al +
(A
_
I)d,
2)d.
Case 1
4.2 In
was
al (that is, I)a + d, d'
ISuPp(b 2)1
(i.e.,
Table
A > 3 is
that
3
identity
(x, z-y)
=
21 [15 theorem
b G
types for b 2:
possible
1
The next
all
5 and width
(xy, z)
integral GT-algebras B* which contain a faithful
of standard
> 4 for
9 135- Using the basic
A,y-,Izl we
classification
the
case
support
1 of Table
of
b2
or
I it
not.
is useful
to
Wehave the
whether destinguish possibilities: following
Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 83 - 103, 2002 © Springer-Verlag Berlin Heidelberg 2002
b is contained
in the
84
Bfinger
Florian
51 + 3b + c, c G B5\f bl, (a) b2 51 + 3c + b, c C- B5\jbJ, (b) b2 (c) b2 =51+3c+d, c,dEB5\jbj, =
=
cod.
of the cases (a) and (b) we only deal with case (c). For a discussion [41) The proof of the next theorem that completely solves this case 2.3 of [41]. Some gaps in this proof that inspired by that of Proposition to two exceptional table algebras of dimension 4 and 5, respectively, will
Wewill refer
to
was
lead
be closed.
51 If b2 following
Theorem 2.
then
(a) (b)
of
one
B
=
cd
=
B
=
C2
=
de=
11, b,
Proof.
bc
d, d2
for distinct
real
basis
elements
of degree 5,
holds:,
cases
=
3b +
=
51 +
2d, bd c
b, c2
2c + 2d +
=
51 + 2c +
=
2d,
+ d + 2b.
3b + 2d, bd 2e + 2d + b, be 11, b, c, d, ej, bc 2c + 2d + e, 51+2c+2d, cd 2e+2c+d, ce 2b+2d+e, d2 51+c+d+2b, 51+b+c+d+e. 2b+2c+e, e2 =
=
=
=
=
(A, B)
Set
Case 1.
dj,
c,
2b + 2c +
particular,
In
+ 3c + d
=
the
w
bc
w
=
2b2, b2
C2
=
and
homogeneous.
five
3b. Then either
-
=
25 +
and therefore
is real
T2 0
3(c, C2)
w
2b2
=
or
E B
w
or
w
b3 + b4, b3 0 b4
=
b. Then +
(d, C2)
(b 2, C2)
=
(bc, bc)
=
65
=
Now
2c + 2d.
151 + 11c + 6d + dc
5c + 3c
=
2
+ dc
b2C
=
151 + 9c + 3d +
b(bc)
=
=
3b 2+ 2bb2
2bb2
yields 2bb2 In 5 or
particular, -
(d, C2) (d, dc)
=
Case 1. 1.
b3
=
35 +
we =
73 :A
3
have 2 (d,
(d, bb2)
(d, dc)
=
I
=
(d, bb2)
and
d, and bb2 10(d, d2) yields (d, d2) c,
(b2, bd)
2b2 + b
+
=
=
(d, bb2)
6b3
+
(d, dc).
that
Z
2.
=
2c + 2d + =
=
2N, N =.T4- = 6 b,
51 + 9d + 7c +
+
(4.1) (d, dc) : 5 (c, cd) 1 and (d, bb2) (d, de)
Since
either
-
=
=
=
2
-
2c + d By (4. 1) we have dc b3. Hence 45 (dc, dc) (d 2, =
=
1. Romthis
(bd, bd) Since
2c + 3d + de.
3 bb2) (4. 1) shows
3, equation =
=
=
2 and
d. Set
(b 2, d2)
(b, bd)
z :=
follows
it
=
=
5d + 3cd + d 2
=
51 + 7c + 5d +
=
2b3)
C2)
=
45. b 2)
(d,
b2d
2b3
+
that
=
d2 -51 -c-d.
=
=
=
+
b(bd)
2bb4
1, this
means
bd
Then =
b
2
+
2bb2
+
2bb4
=
SITA with
4
yields
z
40
2b5, b5
=
Thus
=
=
5(b2, be)
+
10(b2, b2C)
+
(b2, b2C)
(c2, bb2)
=
we
(bc)b2
=
b(cb2)
=
b2= 51 2 d
b4
=
(cb, cb2)
15(b2, be)
c
+ d +
x
-
so
b(b2C)
=
(bb2)C cb3
that
and 45
=
+ 8b +
Le.,'
bb3
yields
=
b4C
6b4 b2
=
(b4,
+
+ 2cd +
cb3
101 + 8c + 8d + 5b3 +
=
+ 2d +
2b4
11, b,
dj
(b2, bd)
=
=
-
ROM
2b57
c,
=
bd
and
bb2
=
=
2c + 2d +
have obtained
we
2,
=
we
b3
the
have
55,
101 + 8c + 8d +
5b3
+
2b5
101 + 8c + 6d +
4b3
+
cb3
(b, b2d)
(bb2, d)
=
that
we see
+
2b4
+
=
b2b2
=
b(bb2)
b2d
=
=
=
2, (b2, b2d) 2b6, b6
2b + b2 +
=
=
(d, b22)
76 0 b, b2
=
(bd)c
=
be +
b(dc)
=
2bc + bd + 2bb3
2b2 + b4
2bc + 2bd +
bb3
=
8b + 8b2 + 4b4 +
2b6. Therefore,
+
4(b6, b4)
2b2C
+
3b4C
=
=
7b + 4b2 + 4b4 + 2b4C
7b + 8b2 + 6b4 + 4b6
2b6. The equality =
(2b
=
(b2b,d 2)
+
b2
this b4 0 b6 In particular, 0- Wehave b2 + 2b6) -
=
10(b2, cb2)-
2b + 2 b4 + b2
(d2, b2)2 =(62, db2) 0
=
2b2+ 2b4C + b2
2b6
2b + 40
=
(b4, b2C))-
deduce
20 +
implies
=
2d + 2b5 + b3- Sincewe
202
=
30 +
=
+
from
(d2, b2)2 =(b2d, b2d),
=
ROMthis
9b2
=
10((b2, b2C)
+ 6d + 3 b3 +
b3 and b
2(b5,b3)
2C2
10 +
=
=
=
compute
we
b5-
+
(b, db2)
Since
Next
5
3
(a).
in
b5 0 b3- It follows
that
=
b5.
b2 c
b. Hence B
=
45 +
b3
+
2 and
b2b
+
2b3
10(b2, cb2)
+
I? (b4, b2 C)
2b2+ 2b4b
given
0 b2
Case 1. 1. 2. d
Moreover,
=
Degree 5 and Width
2b4, b2C)
3.
6c 3bb2 + 2b2= 2
constants
structure
+
(b4, b2C)
+
and
2b2
+
10(b4, b2C)
b2. Then b5
=
c
(b
=
of
2d +
=
+
(b2, b2 C)
conclude
Case 1.1.1.
implies
(bd, b2C)
=
bb4
and
(dc, bb2)
obtain
we
d,
c,
=
Hence 40
T5 :
=
Real Element
Faithful
a
+ =
2b6, b 20 +
means
+
2b2 + 2b4)
2(b57 b3) (d, b2 b4)
=
(b2d, bd)
20
(b4, b2 d)
=
(b4c, bb3)
=
(be, b4b3)
=
3(bb4, b3)
+
2(b2, b4b3)
=
30 +
2(b2, b4b3)
7
bb37
Bilnger
Florian
86
i.e., (b2, b4b3) b5 =A c, b3 and
(bb4, b2 b4)
we
b2b4
deduce
get bb6
shows
=
=
b(b2d)
=
that
B
=
Q
=
=
2bb2
+
2bb6
=
3bb4
+
2b2b4
b2 N)
=
2b2
bb2
2c +
+
=
2b3
2b3
+
(b37 b42),
bb4
b(64)
+
=
8b3
=
+ 6d +
2c 2+
=
4b3
+
4b,5
+
2bb6
101 + 8c + 4d +
5b3
+
6b5
b3
+
2b5 and from
=
5c +
=
2dc +
b3C + 2b5c
2b42
2bb4 +
=
C2
+ dc +
d2
+
265
263
=
=
d 2C
+
b4b2
=
3
=
that
=
2b42
c2
=
yield
IwI.
(bc, bc)
c2
51 +
=
51 +
c
5b3 + 8b5
64 + dc <
jbdj
=
c
=
+
25,
*. a
=
=
(d2, b2)4 c.
(64,
=
Hence
we
db 4)
have
c
(c, d2)
Then
(d, dc)
3 and
(b2, bd)
the contradiction
(b2 d2) ,
(bd, bd)
< 65.
Then =
(b 27 C2)
+ 3d
+ 3d.
151 + 8c + 9d + dc
(w, bd)
6b3 + 6b5)
+ 5x + 4d + 4v + 2c =
(d, bb2).
=
55
either
Case 2.1.
10
b3d
7
immediately
(w, w)
+
51 + 9c + 3d +
101 + 4c + 6d +
=
=
70 <
Case 2.
4c,
+
d(dc)
=
=
(d, dc)
(bc)b4
7b5
M5. Now
+
51+c+d+2v
2
(b5, b2 b4)
=
101 + 6c + 4d +
265
=
=
+
20 +
b3d
+
=
(d, bb2)
i.e.,
(bb27 Q)
d2. Thus (b, 64) 2 and 55 b4. This clearly gives us b5 b, b3 e, b6 f 1, b, c, d, ej, e : b2 and (b) holds. d
Case 1.2.
so
10 + 2 (d,
=
(b4, b2b3)
=
3b!5. The equation 2d2
=
=
2, (b3, b2b4)
=
2b5- Since
+
2cd +
(b4b2)C b4(b2C) yields imply
(b5, b2b4)
2bb6 + b5
+
(b27 64)
=
b5, b2 b4)
+ 2d +
b3
2c +
b3d
65
obtain
10 +
+
51 + 9c + 3d +
we
(2b3
=
4b3
2b3
(bb2)d
us
=
=
4c + 6d +
we
b2b4)
1. Rom (c,
=
or
15(c, c2)
25 +
(C, C2)
=
2 and
+
5(d, c2),
(d, C2)
=
0.
Wehave
3C2
=
5c +
=
151 + 9c + 3d +
Hence 30:5 contradiction.
+ dc
(d, bw)
=
=
b 2C
=
b(bc)
=
3b
2
+ bw
bw,
(w, bd).
But
(w, w).= IwI implies
SITA with
4
(c, c2)
Case 2.2.
5(d, dc)
(51
=
>,(c, d2)
4
i.e.,
=
=
of 1
d 2)
(c,
d, cd)
+ 3c +
(d, dc)
d2
2
cd)
,
that
so
b 2)
1,
=
(3b
=
we
(b 2, d2)
=
5 and Width
(d,
(d 2, C2)
=
of bd with
x
5], [5 4], [5',5']1,
-
(bc, bd)
=
(cd, cd)
constituent
=
5 1 + 4c, then
(bd, bd)
3 and therefore
=
Chapter 3 supplies a implies [bd b] (=- 1[4
=
d 2)2 5 <
(c,
=
(b
=
> 3. If
(b, bd)
0. Since
=
Degree
of
Real Element
(d, C2)
2 and
80 Thus
Faithful
a
+ w,
derive
87
3
we
have
bd) ! 15,
the contradiction
65.
=
E 175, 70} and Lemma I (x, bd) > 3 Now (b, bd) a (bd,bd) E 1105,85,551, =
0
contradiction.
Case 2
4.3
For Case 2 of Table
b 2= 51 + 2b + b 2= 51 + 2c+
(a) (b) (c)
we
only
treat
of
classification
2c, c E B5\fbl7 2d, c,d E B5\jbj,
Chapter
the
(a)
cases
(b)
case
Finally,
subcases.
elements
(c, cd) then
(c)
(d)
> one
B
=
cd
=
B
=
C2
=
dt
=
B
=
df
d
and
B
=
=
=
(b).
Wewere not able to
Theorem following solve will completely
Theorem 5
3
gives
give
a
answers
complete for
(c) by using
case
many
results
=
c, d
=
11, b,
dj,
bc
11, b, c, dj, 2c + 2d +
b, d2
c,
3d,
2b +
=
bd
2b +
=
3c,
C2
=
b2'
2c +
3b,
51 + 2d +
2b,
cd
=
+ 4d.
11, b,
c,
bc
d,tj,
2b + 2c +
=
d, bd
=
2b + 2d + c,
C2
=
51 + 2c + 2b.
2b + 2t +
bc
51 + 2d +
2c + 2d +
2t, cd 2d + 2h + t, t2 f 1, b, c, d, f 1, bc
=
51 +
c
2b + 2t + c,- bt d2
d, bd b, ct
2c+ 2h + t,
2c + 2d +
51 + 2c +
t, 2t,
+ d + t + h.
2b + 2f + d, bf c + 2d + 2f, f, bd 2f +2d+b, d2 51+b+c+d+f, 2c+d+2f, cf 51+2d+2b, cd 51 + 2c + 2b. 2b + 2c + d, f2 2 51+3c+d, cd 3b+c+d, 2b+3c, c 2b+3d, bd 11, b, c, dj, bc
d2 =
(f)
=A
c
distinct real basis 51 + 2c + 2d for pairwise Suppose that b2 of degree 5. Then (c, cd) + (d, cd) > 2. If we assume that (d, cd), (d, cd) if (c, cd) :A (d, cd) and that (c, c2) -. (d, d 2) if (c, cd) true: cases holds of the following
b,
=
B
C2
(e)
=
3.
d2= 51
(b)
Z
but the
Theorem 3.
(a)
main subcases:
following
have the
b2=51+2c+2Z,c(EB5,ZOC-
Wewill
of
I
(c, cd)
=
2b + 2c +
=
=
=
=
51 + 3d + =
1
=
[The tablealgebras
=
=
=
c.
(d, cd),
(c, c2)
defined
in
=
(c)
0
=
and
(d, d2). (d)
are
exactly
isomorphic.]
88
Florian
Proof.
Set
l3iinger be
x :=
20 +
(x, x)
20 +
(y, y) 0
c
b
(bd, bd)
=
(b 2, d2)
=
(x, x), (y, y) Rom (4.2)
(d, cd).
(A) (B) (C) (D)
(c, (c, (c, (c,
cd) cd) cd) cd)
=
=
=
=
one
obtain
(c, cd)
of the
following
(A)
51 +
I
yields
10((C' C2)
+
(d, C2)),
(4.3)
25 +
10((c,
2)
+
(d,d 2)).
(4.4)
G
d
=A (y, y).
35
f 15, 25,
Hence
45 1.
(4.5)
! 2. We can
(c, cd)
that
assume
>
holds:
cases
Case
(B) Applying Set
45 +
(4.3) cd
z :=
(z, z)
=
Since
and
(4.5)
3c.
Then
-
c
(4.4) 9(b2, b3) and
contradiction
the
c.
25 +
(d, cd)
+
51 + 4d.
=
Using (4.3)
(4.2)
51 + 3d +
(4.2)
(d, cd).
=
+ 3d.
c
(d, cd)),
4; 3; 2;
Wehave 0
Case
d2
=
we
Hence
+
=
(x, x) =,4
yields
10((c, cd)
=
,
(b 2, C2)
d
then
(b2 cd)
=
=
0
2b,
-
(be, be)
=
=
bd
:=
(be, bd)
(x) y)
20 +
Note that
2b and y
-
(cd, cd)
(c2,
=
=A d,
=
Theorem I
obtain
we
x
=
yields
cd
3b2 and
y
4c + d, 3b3. Now
==
=
30.
we
get
d 2)
=
[x]
[5 3), (C' C2)
=
25 +
15(d,
d 2) +
1 and
=
5(c,
d
0
2
Rom
(z, z) and
(4.4),
In each
Case
we
case
(C)
deduce
(4-2)
([y],
d 2),
(d,
immediately
Rom(4.3)
and
(c, d2)) yields
(4.5),
f 10, 201
E
we
a
E
J([5 3] 2,2),([5
3], 1,3),([52
51],2,0)1.
contradiction.
(c, c2))
([x],
obtain
E
([5 3],2),([5
2,511,0)1.
5 1 + 2d + 2c. Now (4.2) reads 3 (e, y) 3e, jej 57 C2 (CI) x so that (d,cd) G 10,31. If we assume that (e,y) : - 0 and (equiv10(d,cd), 3e and (d, d2) x 3, then (4.4) and (4.5) yield y alently) (d, cd) 0 contradicting (4.2). Hence (e, y) (d, cd). Since
Case
=
=
=
=
=
=
4b + 6e + 3ce
i.e., then
3ce
(4.4)
=:
=
9b +
2cb, + 3ce
2y,
shows that
c(cb)
=
we see
that
(d, d2)
=
=
=
=
y
C2 b =
4, i.e.,
=
5b + 2db + 2cb
3f for d2
some =
real
51 + 4d.
=
f
13b +'6e + 2y, E
Now
B5\f ej. we
But
compute
SITA with
4
(cd, ed) for-
(c 2, d2)
=
some
real
g E
a
be
=
3c +
gc
us
=
g
derive
=
b,
Case Z :=
so
e
2
3d +
B
(C2) x C2-5172d,
=
2c2
+ 2cd
b 2C
=
5c +
=
101 + 4c + 4d + 3be
3gc
=
2
3gc
c(cd)
=
2c
=
101 + 13d + 4c +
b(bc)
=
2b
=
2
89
3
2c+3g
=
+ 3be
+
=
c2d
=
=
5d + 2cd + 2d
2
6g
2g. Rom
4g
3c 2+
=
11, b,
=
f,
2e +
dj
c,
T
e
=
d
e
(cb)e
=
by
jej
and
=
5
=
as
4g
6c +
implies
multiply
=
f
=
stated
+ c
in
3e2,
and
(a).
0. Set if 1, (c, c2) d (d, 2)d- (c, d 2)C. =
d 2- 51
w:=
2
2be + 3e
elements
=A f =7,
(d, cd)d-2c
=
Theorem 5. This
and the basis
=
cd-
v:=
c(be)
2cg
d2 and therefore
=
that
6g
Hence cd
65.
=
5 and Width
2g. Now
151 + 6c + 12d + we
Degree
of
(51+2c+2d,51+4d) B5\ICI. The equation
101 + 4c + 4d +
gives
Real Element
=
101 + 13c + 4d +
yields
Faithful
-
We compute 9c + 6d + 2v + 2z
-
b 2C
=
b(bc)
(4.6)
101+4d+4c+2be+bf 2be+bf =5c+2d+2v+2z. and
(be)2
(2b
=
+ 2e +
f)2
b2C2
(51
=
=
4b2
+ 8be +
+ 2c +
2d)(51
4bf
+
4e2
z)
+ 2d +
+
4ef
+
f2 (4.7)
=
2 251 + 20d + 10c + 5z + 4d + 4cd + 2cz + 2dz.
Counting
c's
in
(4-7)
10 +
yields 4e 2+
(c,
4ef
+
f2)
=
d 2)+
4(c,
2
5
(z,
z
+
(4.8)
V).
(C' f2) 0 4 since 65 > 25 + 10 (c, f2) + (e, f2) is even. Clearly, (f, Cf) Hence 5 (C' f2) E 10,21. We consider the three (fC, fC) : (f, Cf)2. (Z' f2)
Hence
=
=
possible
subcases.,
(C2. 1) (d, cd) (C2.2) (d, cd) (C2.3) (d, cd)
=
=
=
2; 1; 0.
Case
(C2.1)
(4.2) (4.4)
implies (2e + f, y) 0. Set yield (d, d 2)
Here
we
have cd =
=
45
=
=
2c + 2d +
w :=
(cd, cd)
d 2- 51 =
h,
h E
[y]
20 and therefore -
(c2, d2)
2c. =
=
h. Equation B5\fc, dj, 25 and [5 2 51]. But (y, y) =
Rom 25 +
(z, w),
Bfinger
Florian
90
we
obtain
z
=
w
=
2t,
B5\jc,
t G
dj,
Using
t.
the
law,
associativity
we
derive
2c2+2dc+ch=
101+8d+4t+4c+2h+ch=
2dt
i.e.,
2h + 3d +
=
ch,
2ct
=
101 + 5d + 4c + 4t +
=
2d
=
101 + 5c + 4d + 4t +
2
+ 2dc + dh
Next let d 2C2
(51
=
2t) (51
+ 2d +
(2c
the
26 +
2
cd
=
5c +
2c2
Let
us
h2
h)2
+ 2d +
of
occurence
4((t,
h)
+
(C' t2))
4((t,
h)
+
(d, t2))
first
multiply we
that
assume
0
+ 2c +
=,A (h, ct).
(4.9)
2t) 4t2
401 + 24c + 24d + 16t + 8h + 4ch + 4dh + h 2.
=
and d in the expressions
c
(c, a)
=
stated
as
b
have either
(c,)3)
h
a
and 3
8(t, h)
28 +
h 2)
2
=
=
(d,h 2),
yields:
(c, h2)
+
ct
=
[ch]
us
=
2c + 2h +
t,
[5 2 dh
52 =
a
=
=
51]
=
101 + 6d + 9t + 6h +,2c
c
=
=
51 + c i.e., t2 like Bh multiply =
(bh, bh) means
=
[bt]
(b2, c
+ d + t + h. in
h 2)
(c) =
with 65
(d,h 2), i.e.,
=
and the elements
(bh, bh)
of b. Since first
(4.10)
Then
h.
5 1 + 2c + 2 d
(c2,h 2)
2h +
2
=
(4.10)
h 2).
(d,
case
of Bh
(b 2 h2)
=
=
,
done,
we are
65
in the
contradiction.
54
t
dj
c,
In the
+
(c, h2)
implies
11, h,
=
5 2].
[5 3
=
h2
i. e.,
(4.10)
Bh
8(t, h)
28 +
h instead
with
[bhl
or
(ch,ch) shows
(d,,3)
h. Then
=
second case, Lemma2 yields Let us now assume that
(c,
a)
Therefore,
(b)
in =
='(d, t
b 2.
51 +2c+2d=
=
+ 2d
2ct,
and 26 +
2dt
compute
us
=
(cd)2
Counting
(cd)d
=
251 + 18c + 18d + 20t + 4h + 4dt + 4ct +
,8
2+
2dt,
2h + 3c + dh. Hence
=
(h, dt)
a
5d+2d
=
and
101 + 8c + 4t + 4d + 2h + dh
i.e.,
(dc)c=dC2
[dh].
=
45
=
(c2, d2)
=
means
2 f [5 52 51), [102,51]1.
[bh]
Bh
=
we
=
7t + 4d + 4h + Hence
=
[53,
(d, t2)
and
=
ch
2h + d +
=
Finally,
c2t
=
we
2t, compute
5t + 2dt +
2t2
2t2,
f 1, h,
now
=
(cd, ed)
obtain
c(ct)
of b. Let
But
=
2d + 2h + t.
+ 2ch + ct
h instead
(C' t2)
implies b 2. Now
Using (4.9),
+ 2t and dt
2c2
=
5
21
2
c,
us
d, tj
and the
assume
and
yields
that
(bt, bt) b E
=
elements
b
of
Bh. Then
(b 2, t2)
Bh contrary
=
to
45 our
SITA with
4
either we
(d, cd)
(C2.2)
Case
y
=
2f
[y] (d, d2)
have
yield
(C2.2.3)
Hence
2.
(d, d2) (d, d2) (d, d2)
(C2.2.1) (C2.2.2)
(C2.2.1) It B5\jc, d, bl.
Rom (4.2)
1.
=
+ g, g E B5 [5 3], so that
2(h, w)
15 +
20 +
=
bg
b(bd) bg
d 2)
(bf,
45
(be, ec) (be, c2) follows
yield
=
In
10, so that particular,
(4.4)
and
(4.5)
possibilities
+ 2d + 2h for
c
=
=
(bd7 bf)
=
=
10 +
=
20 +
some
real
h E
(cd, cd)
(c25 d2)
(b 2, f2)
25 +
Hence
we
> 4
we see
which
(fC, fC)
=
that
=
(C7 f2)
=
=
(4.12)
=
(4.13)
=
+
2(e, cf),
(z, w),
35 +
10(C' f2)
+
=
=
w
=
0.
b. If 2k
If
(4.15)
(d, f2),
(4.16)
=
z
=
=
so
z,
(C7 f2)
10(d, f2)
51 + 2c + 2k.
(4.17)
2k for that
that
=
+
(Z' f2)
some
real
0, (h, v) (4.15) gives a 0, then (4.8) and
we assume
the contradiction 25 +
(4.14)
-
and therefore
k
have
f2
(g, fd),
2(e, ec) + (f, ec) (z, be) (z, be 2c).
(h, v) 0
supplies
y2' C2)
+
20 + =
(z, z) 0 10, =
(g, f c)
10 +
=
=
that
2(d, f2)
(4.11)
=
=
(4.17)
20 +
=
(be, cf)
=
101+6c+9d+2v+2w, 2w,
2(c5 f2) + (g, fC) 2(e, f d) + (d, f2) 20 + 2(h, v),
(2b + 2e + f, ec) (be, 51 + 2d + z)
=
b 2d=
=
d + 2v +
d}. Our aim is to prove that k and and (4. 11) yield h
85 <
Therefore,
(v, v)
(bf, bf)
=
from
contradiction.
(4.9)
c2)
2(h, z) ='(bf, 25 +
B5\fc, (4.6)
following
bf
that
(bf, cd)
then
=
B5\16, A 9 B5\je, f g
4h +
(bd, fc) (be, fd)
It
f, y)
+
Wehave
101 + 6c + 8d + 4h +
k E
h.
10.
(4.6)
from
(2e
obtain
we
=
e
the
g EE
=
follows
b
91
3
+ g, g E
=
=
implies
+ u, u E NB, e 0 Supp(u) 54 3 by (4.4). Furthermore,
obtain
=
=
=
(d, d2)
we
1, y 2f 1, u 2g, 0, (U7 U)
=
y
or
Degree 5 and Width
of
Real Element
Bh, which clearly
Hence b E
assumption.
Faithful
a
< 65.
=
Bilnger
Florian
92
From45
(i)
real
(4.12) (e, fd)
B5\jb,
1, (f,fc) (b,fc) f 1. Wedistinguish =
=
fc=
2weget
the subcases
2
=
w
t
=
(4-6)
Furthermore
d,.hj.
(4-15) yield (g, fc), i.e.,
and
Here
obtain
suplies
9f +6b+6g+4e=5f bc + 2bd + 2bh
=
and
t E
B5\jc,
h + p, p C
v
(i)
(fc,fc)
=
some
2h
v
(ii)
(f2,C2)
==
+ 2t for
2f
b +
be
h +
'k, (g, f d)
fc
g,
=
2c + 2k + h.
=
5f
+
and h =7 k Rom (4-6)' we b+2f +2g and fd 2b+2e+g. This yields =
=b 2f =
+2cf +2df
6b + 2e +
7--
=
=
2g
+
b(bf)
20h,
bh=2f +2g+e, 2
101+6c+8d+4h+bg=2b 101 + 9d + 6c + 6h +
bg
=
2
(cf)b
b
(bf)c
=
c2
(bc)f
=
2bf
+
bg
=
b2 d
b(bd)
2bg
=
51 + 4c + 8d+ 8h + 4k
+ 2dc+ 2he
=
51 + 4c + 4d + 2k + 4h + 2hc
+
2bf
=
5d + 2cd +
2d2
2h,
d + 2k + =
+
2k,
2bf
+
2ef
+
+
f2
51 + 4c + 4d + 4h + 2k +
=
2ef,
ef =2d+2h+k,hc=2d+2h+k, 101 + 9d + 2c + 6k + 6h
101 + 2c + 7d + 2k + 2h +
dk
=
d + 2k +
2C2
=
P so
(c, k2)
51 + 2c
that
e2d
(cd)d
cd2
5d +
2d2
+ 2kd
c2
+ 2hd
=
+ ed + ch + ck
ck,
h + 2k + 2c.
particular, =
d2
2cd +
51 + 7c + 5d + 3k + 4h +
In
c(cd)
2dk,
5c +
=
=
2h,
51 + 9c + 5d + 5h + 5k
ck
+ 2ch + cd
t
:=
-
bk
2d -
=
2g
=
(k, ck)
=
2
V. F'urthemore, -
(t, t)
(k, dk) (g, bk)
(d, V), i.e., 2 (k, bg) =
2e fulfills
=
(bk, bk)
-
40
=
(b2, k2)
-
40
=
25,
(k, be)
(e, bk),
SITA with
4
whence t
11, b,
=
5 1 and therefore
d, f I and
c,
Real Element
Faithful
a
obtain
we
k
b, following
the
e
=
=
=
51 + 2c + 2d
be
=
2b + 2c +
bd
=
2b +
g
d, h
=
f
=
3
93
Thus B
table:
f
2b
=
cf d2
c,
5 and Width
+ d
bf =c+2d+2f c2 51 + 2d + cd
Degree
multiplication
b2
2f
of
=
2c + d +
=
2f
=
51 + b +
2f
+ 2d + b
f
+ d +
c
df =2b+2c+d
f2 Hence
(d)
(e, f d)
=
holds 1
(4.12) E B5\jc,
and q
(cf,-df)
y2' cd) 0
(4.13).
=
=
=
the previous
b
imply
Hence t
6b + 2e +
5f
(b
=
(51 I
+
=
observations
=
t c=
0. Furthermore, (4.15) we obtain (k, w) 2 and g =7 b, e, we have (g, f d) < 2 (b, f d) 1 h + q for some real (h, w) whence w 2b + e + g + r for some real r E B5\f b, e, g}. In
From
Since by (4.13). yields (g, fd) d, hj. Hence f d
=
particular,
an& t
(ii).
true.
51 + 2c + 2b.
=
+
2f
+
2t, df)
+ 2c +
e, g,
rl
also
2k,
and k
=
show that
10 +
2(t,
2b +
2c + d + h +
p)
0
t
p.
Clearly,
0
k
+ g +
e
r)
=
2(k,p)
20 +
=
0
e
h. Now
by (4.14)
and t
b 2f
+
:7
g
+
2df
by
Wenow compute
r.
2g
+ 2bh
=
be + 2bd + 2bh
=
b(bf)
=
=
5f
2cf
=6b+9f +6t+2e+2g, bh
i.e.,
=:F
-
yields
2f
(bf)c
=
(bf)c
=
d + h + bt
previous
equation 25
a
Finally,
+ 3t.
=
b
2
+
c2
the equation
2bf
51 + 4c + 6d + 4h + 2bt
+ 2bt
51 + 4d + 4c + 2k + 2h +
+ 2dc+ 2hc
=
(51
Since
k + p + he.
that +
4
4c,
=
(h, he)
=
51 + 2d +
(h, bt) (c, h 2)
2k)
=
=
(t, bh)
=
whence h 2
(h 2,c2)
=
3,
=
2p
+ 2hc
we
obtain
by
the
51 + 4c and
(he, he)
=
85,
contradiction.
(C2.2.2)
We list
20 +
some
that
equations
will
2(e, ec) + (f, ec) (2b + 2e (be, c2) (be, 51 + 2d + z) 10 + 2 (e, ec) (z, be) --
=
=
be used in the
+ =
+
sequel:
(be, ec) (z, be), (f ec),
f, ec)
10 + ,
=
(4.18)
94
Biinger
Florian
5d + 2cd + 2bd
(be, f c)
(bf, C2) (2b
(bf,
=
b(bd)
=
2b2
f, f c)
(4.19)
2bg,
+ be +
z) 2(f, bd) + (bf, z) (bf, z) 20 2(bc, f) + (C' f 2) + 2(e, f c)
51 + 2d +
+ 2e +
=
b2d
=
=
=
+
=
(e, f c),
(4.20) (be, ed) 2 (be, e)
(be, 26 + d + v) (bd, e) + (be, v) 25 + (be, v) (bd, ec) 2g, ec) 2(bc, e) + (e, ec) 20 + (e, ec) + 2(g, ec), =
=
+
=
=
(2b
+
25 +
Rom (4.19)
we
+
e
(v, v)
(0, d2)
=
=
2(g, ec)
+
.1 35 +
(z, w).
(4.22)
obtain
(p, be) Assume that
(C' f2)
(v, z)
Hence
is odd.
(cd, cd)
=
(4.21)
=
=
=
z
is
+
r
all
and
s
v
Supp(be)\jdj.
p E
reading
0. Then =
for
even
(4-8)
equation =
s
+ t for
(4.23) modulo 4 shows that distinct
pairwise
elements
0 B5\jc, dj. But then (4.8) yields (C' 62) (c, ef) and it folthat lows from (4.21) (v, be) is odd contradicting (4.23). Hence (C' f 2) 2. Reading equation [521 and modulo 2 if [z] (4.7) modulo 4 if [z] 1 Let 1 5 5 2c us z. 10. + now assume that + f2 ], [101 ] I yields. 1 [51, (z, z) Then (4.6),(4.18),(4.20) yield
r, s, t
E
=
=
=
=
40 + 4 (e,
ec)
+ 3 (f
,
ec)
=
(z,
=
(z,
bf
2be +
2z + 2v + 5 c +
2d)
=
20 + 2 (v,
z),
10 0 (be, z) and (e, ec) (f, ec). But it follows from (v, z) this means [v] (v, be) is odd. In particular, [51, 51]. Now 10 10 is even, (v, z) implies v (be, v) (v, v) (be, z) z, so that Hence z 2k for some real k E B5\jc, a contradiction. dj. Now (f c, f c) 2 and (b,fc) I yield 45, (f,fc) (f2,C2) (51 +2c+2k,51 +2d+2k) for 2t It real b follows t that some c E + + 2f f B5\fb, 1. f
whence
(4.21) (z, z)
=
=
=
=
that
=
=
=
=
=
=
=
=
=
=
=
=
=
=
Hence
(bf, C2)
=
(bf,
(be, f c)
=
(2b
(k, bf)
> 2.
(bf,bf)
7
6
=
(k,
f,
2k)
b +
(b 2, f2)
=
<
k
=
we
obtain
Thus
c, d and
=
2t)
+ 2c +
:
bf)
2f
2(f, bd)
+
(51
2.
2be +
=
=
+
2(k, bf)
20 +
=
2(k, bf)
=
4(e, t).
hand,
On the other
(k, bf) (k be) =76 0 and since by (4.23). Therefore, shows that
51 + 2d + + 2e +
(k,
(c, be)
=
2d,
51 + 2c +
(k, bf )
=
2, (d, be)
4k + 2v + 5c +
2d)
2.
1,
2k)
=
From we
4 +
45
(4.6) we get 2 (k, be)
have
2(v, k)
=
SITA with
4
which c:
gives s
E
v
k +
=
Rom (4.22)
h, h
we
B5\lc,
c
obtain
0
(C2, dk)
=
(51
+ 2d +
(cd, ck)
=
(2c
+ d + k +
h, ck)
Therefore,
b
us (d, k 2) ! B5\16, f 1. But
45 a
that
means-
+ 2k + 2h.
Real Element
Faithful
a
=
3.
2k, dk)
d, kj, (k, w)
=
be
2(C2 k)
5 and Width
d 2)
95
2(d,
k
2)
(ed, k)
k and bk
2f
k
2)
+ 2e +
s
=
2(d,
=
25
for
some
then
(bk, bk)
=
k2)
2
(b
,
=
(51
2c, k 2)
+ 2d +
10(d,
25 +
k
2)
55,
contradiction.
(C2.2.3)
First
we
compute
(bf, C2) (bf, 51 + (2b + 2e + f, fc) ==
(bc, fc)
=
25 +
(v, v)
(cd, cd)
=
2d + =
z)
=
2 (bc,
(0, d2)
2(f, bd) + (bf, z) (bf, z) 20 + (C' f2) + 2(e, fc) =
f )
=
(z, w), (v, v)
25 +
+
(e, fc), (4.24)
(z, w),
=
(4.25)
20 + 2(e, ec) + (f, ec) (bc, ec) (2b + 2e + f, ec) 10 + (z, be), (be, c2) (be, 51 + 2d + z) 10 + 2 (e, ec) + (f ec), (z, be) =
(bf C2) (2b ,
(bc, f c)
=
=
(bf,
,
z) 2(f, bd) + (bf, z) (bf, z) 20 2(bc, f) + (C' f2) + 2(e, f c)
51 + 2d +
+ 2e +
f, f c)
=
=
=
that
[bf]
51 + 2d +
=
z
[57
5
2(d, f2)
52]
2
and
(4.27)
Applying
=
now assume
us
(4.8)
65 > 25 +
=
=
0 by (4.4). Let (d, d 2) yields that (z, v) is odd. This pairwise distinct r, s, t G B5. Applying Furthermore, (4-6) becomes 2be + bf 1[51,5 41, [5, 52 5211 and
Note that
(4.26)
=
=
so
bf
Now
.
+
,
3
2c + 2k + d and
=
(k,
=
d 2) +
2(k,
=
Degree
of
means
(4.8) =
(d, f2)
=
=
c
(bf,bf) It
E
follows
and
s
(c,
e
0.
z
2)
s
Then
for
+ t
(C' ef ). Hence [bf] G
=
0
=
145,851, from
Therefore,
+ 2z.
(e, f c). (4.27)
(C' f2)
that +
r
again yields
> 2.
bf
=
2r + 4s + 2t + 5c + 3d.
(b 2, f2)
yields
v
+
f2
(4.7)
that
be
2c + d +
=
v
and 20 Hence
(e, f d)
=
2
=
(be, bf)
(f, fd)
and
=
(b
2
thereby
ef) f
2(d,ef). d. Now and
(4.25)
yields
the
contradiction 45
=
(fd, fd)
==
(f2 d ) ,
=
(51
+ 2d + z, 51 +
c
+
w)
=
25 +
(z, w)
35.
l3iinger
Florian
96
(C, f 2)
that
We conclude
f
2 and
=
2
51 + 2c +
=
10. Then (4.25) yields 15 (z, z) 10. Furthermore, (v, v) 110, 20} whence (z, w) 0 and (C' e2) we obtain (c, ef ). Therefore,
that
assume
=
=
=
bf
2be +
45
=
2(d, ef) gives
=
2b, k =7
z
=
from
even,
51 + 2c
i) (k, v) (4.6)).
(cf.
(4.25)
=
=
But
obtain
we
(bf bf
=
,
now
2d, ef)
+ 2c +
(b2, f 2)
=
=
(b2, ef)
(be, bf)
=
30,
the contradiction
(51+c+w,51+2c+z)
=
=
(z,w)
25+
=
35.
real
(k,
+ 2
1
=
,
=
=
=
=
=
(k,
bf
2be +
4k + 2v + 5c +
2d)
=
4 +
2(k, v) (4.28)
1
Then h ':=
us
z)
51 + 2c +
that
Hence bk
gives
(51
=
3, and
=
(d 27f2)
0, 1.
=
2(d, ef)
some
(4.24) + 2k)
(k, v)
Hence
=
k (E B5\fc, dj- It follows from (4.6) that (k, bf) is V' f 2) (51 + 2c + (k, bf ) : - 0 and from (bf bf ) 2. Note that 1 and 45 that (k, bf) 2, (d, be) (c, be) (k, be) :5 2. Counting 4's in equation (4.6) yields
2k for
2(k, be)
6 >
2d,
+ 2c +
and be= 2c+d+z.
(d, ef )
=
implies
c, d
(51
=
+
=
(df,df)
55 < Hence
2(d, z)
2(c, ef)
(e, df )
us
(cf. (4.7)). Let us now jwj ! (z, w) (v, v) G 10 (4.8) supplies (z, v) reads Now (4.6) v z. z
4z + 5c + 2d and since
bf =c+2z
have
we
=
=
45 +
=
=
=
=
(k, w)
(b 2, dk)
=
(51
+ 2c +
(bd, bk)
=
(2b
+
v
(k, d2).
2d, dk)
+
p)
=
real
+
B5\jb,
p E
=
e,
c
+ 2k + 2h
f 1. Applying
get the contradiction
we
d 2)
2(k,
10 +
2k, bf
2c + d +
=
some
Thus
2(cd, k)
=
2f
B5, be
+ p for
2f
=
+ u, 2e +
e
k E
-
2e +
(p, u)
2(k, v)
=
+
2(k, w)
=
20
< 15.
1. But then lbe-2c-d-ki 0. Then (4.28) shows that (k, be) (k, v) c+2k+2s where r,s (4.6) yield v r+s, be= 2c+d+k+r, bf e, then jpj (possibly equal) real elements of degree 5. Set p := bk 2f 10 and (4.25) supply Supp(p), (u, u) b, e, f
(ii)
=
=
=
and
-
=
-
5
are
=
=
10,
=
(b 2, dk)
(51
(bd, bk)
(V, V) (2b +
Hence
r
P1 + P2 for
(c2, kd) 2 (d, k2)
s,
2d, dk)
+ 2c +
e
2f
+ u,
(v,v)
+
10
=
distinct
PI
=
(51
+ 2c +
=
(cd, ck)
5
=
e
=
P2 E
p)
+
=
(z,w)
+
5 +
-
=
2(k, d2) (p, u)
and the
B5 Moreover,
2k, kd)
(2c
2(cd, k)
=
2(cd, k)
+ d + v,
ck)
=
:5 5 +
above we
+
IpI
=
+
(2k, w)
=
C2)
k 2) =
equation
=
20.
(z, w)
15.
shows that
have
2(d,
> 2 (k,
2(k, v)
2(v, k)
+
2(d7 k2)
p
SITA with
4
Case
k 2) ::-
(d,
Hence
a
Faithful
2. From this
Real Element
obtain
we
(b2, k2)
=
(51
+ 2c +
(kb,kb)
=
(2f
+ e+pl
(C2.3)
In this
(4.2)
case
Degree
of
5 and Width
3
97
the contradiction
2d, k 2)
> 45
+P2,2f
+
+P2)
+pl
e
=
35.
means
Y)
(4.29)
0
=
Wecompute
(bC)2 b2C2
i.e.,
3 )2
=
(51
+ 2c +
=
251 + 20d + 18c + 5z + 4v + 4d 2+ 2cz +
Counting 10 +
4e 2 +
(d,
Weconsider
(C2.3.1) (C2.3.2)
=
c's
g +
h,
z
=
2g,
g E
may
whence d 2
=
c
+
4bx + X2 . both
on
4ef
+
12 +
g, h C-
B5\fc,
f2)
yields
2(z,z +v) 5 + (z, d 2))
=
2dz,
51 + 12d+ 10c+ 5z + 4v + 4d 2+
sides
2
and
v) 5 ((z,
B5\fc, dj, dj, 71
=
at the first
assume
=
w.l.o.g.
(ed, cd)
51 + 4d.
5c bf 4g or be
2be +
+
f2)
(4.30)
d 2).
4(d,
+
=
=
=
54
g
h
g.
equation of (4.30) v 3g. Hence
we
(c,
obtain
4e 2
(c
2 ,
d2)
=
+2(d,
25
d
2)
(g
+
+
h,
bf
2
bf
b(bc)
=
2b
=
101+4d+8g+2h+9c,
+ 2be +
=
=
b
2
C=
be we have either 8g + 2h. Therefore, c + 2h + 2g. In the first cd, bf 3g
+
2c +
Using (4.29)
also
=
(e, bg)
+
(f, bg)
the second leads 65
=
(e, bd)
+ f2)
d2),
5c +
=
(cd, be) +
=
2(e, de)
to
(g7 be) a
(g, bf
6.
contradiction:
(cb, de) +
+
(f, de)
=
(2b
=
+ 2e, +
10(e, de)
f, de)
+
5(f, de).
2c2
2c +
case
contradiction 5 >
+4ef
=
Furthermore,
101 + 4c + 4d + 2be +
=
=
X2
two subcases
=
65
the
f2
201 + 8c + 8d + 4bx +
=
z)
+ 2d +
and d's
+
z
we
+ 4bx +
2d)(5
(c, 4e2
4ef
(C2.3. 1) Looking 0 and
X2
+
8be+ 4bf + 4e 2 + 4ef +
2cz + 2dz.
i.e., bf
4b2
(2b
=
+ 2cd
2g
we
+
h,
obtain
Bilnger
Florian
98
(C2.3.2)
2 (mod 4), so that equation of (4.30) supplies (C' f 2) 0 (mod 4). By equation of (4.30) implies (d, f2) this means (d, f 2) 0. Therefore observation we obtain
The first
2. C, f 2) the previous
=
(C' 62
+
ef)
+
20 +
Moreover,
=
The second
=
(g, v)
=
(v, v)
(d, e2
and
(cd, cd)
=
ef)
+
(c2, d2)
=
3 +
=
(g, v)
10(d, d2)
25 +
=
(g, d2)
+
(d, d2). (4.31)
+
10(g, d2) gives
+
us
(v, v) (d, d2)
Therefore,
+
(g, d2)
(c, e2
4 + so -
(d, by (4.32) that
(h, d2) =7
e
2
e
(v,,v)
(e, de)
0 and set
a
(bc, ed)
=
ef)
+
=
(g,
+
=
h be the
Let
15.
this
d
we
obtain
2+ ef)
(d,d 2)
+
2)
+
(d,d 2)
0, (g, v) (unique)
1
=
5,
>
(g, d2) + (d, d 2) and of Ig, dj satisfying
=
element
(f de),
0
and
e
(4.32) (4.31)
with
(d,
(g,d 2)
+
ef)
+
5, (c, e2
'f
+
-
(g,v)
3+
=
(f, de)
+
10(g, d2).
+
Combining
> 1.
(e, de)
5 >
10(d, d2)
5 +
=
then
,
de
+
ae
=
3f
This
.
means
10a +
5,3
Hence
(a,,O)
(4-31))
=
G
we
-
(be, cd)
h
a
1(0, 5), (1, 4), (2, 3) 1.
0. Set
u :=
(be, be)
+ 45
Then
JwJ
and
e2
=
that
Then
w
=
But
v)
(be, v)
20 +
=
(c, be)
as
-
d2
-
(d, u) JwJ
=
=
=
50,
a
=='10 and 2k for
(h
51
+ u, ad +
20 +
:
The
2cl
-
(c, e2)
2 and
=
w)
Jul
and
of
(4.33)
h).
0
=
=
(Cf.
let
2(h, d) i.e., c, gj, 3, (C' f2)
B5\fd, (d, ef)
k E
Now
=
=
2 and
w)-
(a, u,w) < (2, 3). (a, 0) that
assume
(h ', d)
(h
+
that
assume
2e +
(4.33)
,
+ u,
(h+u, w).
de =
=
Then
ad.
-
first
us
us
+
2),
=
Let
Finally, =
e
51
becomes 60
contradiction. reads 40
2-
=
25 +
=
e
10(d,
(d 2, e2) a(h, d) + (h
(de,de)
=
25 +
w:=
(4.33)
some
51 + 2d + 2k.
=
=
h and
-
the choice
by
0
15
(b2, e2)
=
(a2 +'32 )5 25 +
(Recall (1, 4). (d, d)
2c +
have 35 <
whence
(be,
=
=
Hence h
3f
,
(d, f2)
d2 =
=
0
=
(ef, ef)
=
(e 2, f2)
=
25 +
2(k, f2
-
51
-
2c) : -
d,
u
=
51 + d + =
the contradiction 55 <
+
45.
3k
3k,
(c, k) yield
35.
SITA with
4
(D)
Case
(c, c 2) (c, c2) (c, c 2) (M)
(D I) (D2) (D3) where
(DI)
Here
f
real
(4.2)
From
possible
following
obtain
we
values
0
bf =
=
an
have c2
we
(be)
2v, 3
3c+
=
2
=
(x, x)
10(c, c2)
15 +
=
2)
problem
open
5 and Width
(4.5)
and
3
99
supplies
the
whence c
in the
45,
=
so
theorem.
that
x
3f for
=
some
compute
we
=
(x, x)
3bf v
b(bc)
=
3h for
=
+ d + 3h.
b 2C
=
+
c's
in both
expressions
53
(c, b2C2)
=
(c, (bC)2)
(bf, bf)
=
(b 2, f2)
2d)(51 + d + 3c) 4b 2+ 12bf + 9f2. 3f)2
B5\jc,
h E
dj.
2v,
Thus
we
obtain
compute
we
+ 2c +
101 + 13c + 4d +
=
real
some
Next
(51 (2b
Counting
Degree
is stated
it
as
51 + d + 3c and
=
First
3c + 2h and cd
b2C2
c
of
(d,d 2),
=
remains
101 + 4c + 4d +
i.e., bf
(c,
of
Real Element
3, 1,
B5\jbj.
E=-
Faithful
a
251 + 15d + 25c + 8cd +
6C2
+ 2d
2
=
yields
9(C' f2)
44+
=
Since 65
(d, f2)
that
we see
2(d,
29 +
whence
d 2)
(d,
=
3,
yields Thus
C2
=
given
(df we
-
3f df
(df
51 + d+ in
B
-
3f )
-
,
obtain
=
3c, cd
d
that
so
45 +
f2
3, i.e.,
=
2)
d
51 +
b2
j1,c,d,bj, =
3f)
-
c
=
30. But this
=
+ 3d.
c
(d,b 2C2)
2
3f, df
51 +
=
=
c+d+ 3b, d 2
10(d, f2)
35 +
=
Romthis
(d, (bC)2)
=
+ 3d
=
(df, df)
f2
51 +
c
35
=
(d 2, f2)
=
that
Now
.
implies [df 51+2c+2d, =
follows
it
3f]
-
=
[15 51), i.e.,
=
2b+3d, bd
be=
These
+ 3d.
75
are
the
d f. 2b+3c, equations =
=
(e).
(1)2)
Put
that
(x, x)
z
:=
d2
25,
51
-
so
d
-
that
x
-
=
and
c
2e +
f
v
101 + 9c + 4d + 2z + 2v
b(bc)
=
2b 2+ 2be +
i.e.,
bf
5c +
=
=
cd
:=
c
-
distinct
for
d.
-
real
2c2
e,
+ 2cd
follows
It
f =
e
B5\jbj-
b 2C
101 + 4c + 4d + 2be +
5c + 2z + 2v
=
2be +
from
(4.2)
Wehave
=
bf,
(4-34)
bf
and 4b 2+ 8be +
(2b (51
4bf
2 + 4e +
+ 2e +
f)2
+ 2c +
2d)(51
251 + 15c + 15d + 5z +
=
20
(bC)2 +
c
4ef =
+
f2
b 2C2
+ d +
+ 4cd +
(4.35)
z)
2d2
+ 2cz + 2dz
Minger
Florian
100
and d's
c's
Counting
both sides
on
of the last
(a) 5 + 4(c, e') + 4(c, ef) + (C' f 2) 13 (b) 4(d, e2) + 4(d, ef) + (d, f 2) (Recall
(x, y)
that
2
=
(Z Z + V) 2(d, d2) +
5
(4.36)
1
+
=
,
yields
equation 2
(Z'
5
d2).
+
V
0.)
Reading these equations
modulo two,
(C' f 2)
that
we see
(d, f 2)
and
are
odd. 20. Then z 2k for some real assume that (z, z) (4.36) (a) modulo, (4.5) yields (C, f 2) =- 3 (mod 4), 51 + 3c + d. Since 1 c (be, f) (c, bf), we have z := bf and c V Supp(z). Furthermore,
Let
us
=
=
=
(z, z)
5 +
(f b, f b)
=
60. By Lemma (1) of Chapter 3, there is an (z, z) 3. Therefore, (z,z)lzl-l [z] E j[45],[54],[53,5']j (a,z) contradiction. a C80, 501, 1100, (z, z)
i.e.,
that
f
2
we
and
(U,U)
so z
=
f
b + g for
(e, b2)
Thus
=
The
c2.
Set
(f2'
2)
C
35 but this
=
(f bd)
=
(e,
I g E
(4.35)
read
f
:=
c
-
b
modulo
f
-
two
such means
obtain
we
NB, Then Jul
E:-:
=
20
65,
which
(51+c+d+z,51+c+d+z)
(f be) B5\fb, c, dj. ,
=
2d)
51 + 2c+
=
only possible
is =
(c, bd)
=
,
we u
=
=
15
equation =
real
If
10.
=
=
Therefore,
some
0
so
c.
=
-
z
(fC,fc)
=
(u, u)
shows that that
+ d +
c
S u pp (u)
b, f V
10+
(z, z)
that
see
51 +
=
=
Supp(z)
c
a
=
Thus
MB, JzJ
E
+3c+d)
(51 +2c+2b,51
f2
that
so
-
(f2 b2)
=
dj.
k G B5\fc,
=
Reading
=
when
(c, be)
=
=
(d, be)
=
2 (d,
0 and
(e, c2)
e) =
45
[u] E 1 [15, 51, 51], [15, 101 (b, C2) (b, z) and hence =
particular,
In
=
=
(e,
have
we
(b, cd)
(b, v).
=
51 + c+ d+
b+g)
=
(e, g)
that
i.e.,
=
(be, c2)
=
(g, be)
2 +
i.e.,
(be, ec)
(g, v)
=
=
(g,
1 +
=
20+
(d2,d- b+g)
+ 2 (e, +
ec)
ec
(v, v)
+ =
ec)
(bd, e)
+
5c + 2b +
2(e,
10 +
(v,v)
e)
(be, e)
2 + 2 (e,
=
2(g, v)
2 (be,
g).
+
+
(e, g)
-
(c, ec)
(e, 2) b
=
+
(be, g)
Rom (4.34)
2v)
2be +
On the other
hand,
+
(cd, cd)
=
=
(C2 d2) ,
=
=
be)
=
(be, g),
we
30 +
(d
4 +
have
4(e, ec)
-
+
4(e, g),
-
+ b + g, d
and as (d+z,d+b+g) (d+b+g,d2) 51 -cl :5 Jd2 (d2 -51 -c,d+b+g) =
(e, g)
+
10 +
obtain
we
(g,
2g
ec)
20 + 2 (e,
=
15
2),
=
ld+b+gl
15. Weconclude
-
that
(v, v)
=
25, [vj
=
[52, 511
and
(d
+ b + g, d
2)
5.
(4.37)
-
SITA with
4
Real Element
=
v
45
=
(be,be)
=
(b 2, e2)
(ce, ce)
Thus
[de] 45
r
51]
(de, de)
=
finishes
This
In
4 for
x
such that
(=-
c7c
=
we
(d 2, e2) ,
=
that
following
(for proofs
Example
1.
c,
and
+
e
(e, bc)
=
2
=
2(d,
25 + 2
(d, e2)
=
(d
2
=
2d + b +
g)
25 +
2(d
e
2)
=
40 +
(c, de),
=
:!
(4.35)
=51+2d+b+r, (e, ed) yields
+ d + b + g, 51 + 2d + b +
c
integral
Y6
example
Z,
algebras
table a
faithful
several
explanations
1, c,Z, h,
g,
basis
In order
introduces
and further
with
nonreal
classified.
were
2
we
=
51 + 2h +
C2
=
3-c +
c
ch
=
371+
2c
cg
=
3h+g+77
dg-
=
7! +
=
51 + 4h
=
3-c +
=
3c+Z+g
=
51 +
3g
c
,
d + b +
(r, g)
so
g)
+ 7!
+ g -
+ 11
+"E + 2h
=
hj c7c
=
51 + 2h +
c2
=
3-c +
ch
=
2c + 7! + 2h
h2
=
51 +
c
c
+Z
+ h
2(c
+
Z).
=
element
11} c
that
=
and
of
35.
'
c7c.
jxj
degree
the classification
bases distinguished to Chapter 3):
refer
2g
c
L(B)
to state
771
c7c
gg
f 1,
=A
d
contain
92
=
=
101
0
hg
Y4
(b, ce)
25 +
h2
(ii)
=
2 (e, de) (d, e 2) (e, cd) (4.37) yields the contradiction
51+2h+c+Z
theorem, algebras
(i)
2(d, e2)
+
3
Therefore,
0.
=
p2) =(be,e)=1wehavee
have
3, standard
=
(51
5 and Width
proof.
B\111
the
2)
and hence
the
Chapter all
e
Finally,
9.
52
=
=
=
[52
(0,
=
2(c,e 2)
25 +
=
showsthat(d,e 2) =2.Since(b,. r E B5\fb, Furthermore, c, dj. [52 52 51] and therefore [cel 45
Degree
of
clearly means (g, v) 1, e =7 g and (e, ec) =g+2e and be= 2c+2g+e. Now
But this
yields
Faithful
a
of table
! 5
(iii)
Bfinger
Florian
102
If Y5
1,
gl
Z, h,
c,
c7c= 51+2h+c+Z
C2
=
3Z+c+g
ch
=
2c+2g+h
h2
=
51+c+Z+h+g
=
Z+
cg
2(h
2= 51 + 9
hg
(iV)
2(c +7!)
bb
=
51 + b +
=
51 + 4h
b2= 3h
3T,,
(5)3
vjvj
T,,, (3)
=
1, h,
I
wjwj
+
vi+j-m
hvi
2vi + 3wi
hwi
=
3vi
=
51 + 4h.
wjwj
+
.
.
.
11, h,
:=
vo, wo,...,
vi+j + 3vi+j+l 3h + wo + v.
=
vm,
hvi
=
2vi
hwi
=
3vi + 2wi
h2=
7if if if 7
i+j i+j
,
i +
)if 7if if
i+j i +j
jxj
--
+
< Tn, =
M,
> M, < M,
=Tn'
j
>
wml,
+ vi+j-m-l
if )if if
+ vi+j+l
+
3vi+j-m-l
+ wi+j-m-l
i +
,
,
< M, =
j
M,
> m,
,
+j < M, i+j=m, i + j >
i
3wi
51 + 4h.
Suppose and
i+j i+j i +j
+ wi+j+l
3wi+j-m-l
+
vi+j-m
Ill
)if 7if if
+ vi+j-m-l
3wi+j+l wi+j 51 + 2h + vo + wm
viwj
=
+ wi+j-m-l
3wi+j-m-l
+
L(B)
11 VMI WM
I
2wi
wi+j-,n
Theorem 4.
T
+ 2b.
3vi+j-m-l
+
=
T-M(3)
=
2h
+ vi+j+l wi+j + 3wi+j+l 3h + wo + wm
h2
vjvj
+ b +
3T
vo, wo,
wi+j-,,,
3Tm( )3
=
T+
+ wi+j+l vi+j + 3vi+j+l 5. 1 + 2h + vo + vm
viwj
(vi)
Z)
+ h.
h2 bh
(v)
g)
+
hl,
11, b,
Z4
=
+
2(c
that 4
for
(A, B)
is
all
B\111.
x
G
a
standard
If there
integral is
a
table nonreal
algebra c
with
E B and
a
SiTA
4
real
h G
one
of
B5 such that
c7c
We are
Suppose
Theorem 5.
T2m(5)j
Ta
Set
Proof.
b,
that
bc
B,
then
exactly
5 and Width
exactly
is
3T,(L ) 3 T,,,,(3),
satisfy lbl isomorphic Example (1)).
B
G
c
Degree
of
of the introduction
101y Y42 Y5 (cf.
E NU u
(c)
case
Then B is
51 + 2c + 2-c.
(c +7!),
51 + 2h +
Y4, Y52 Y62 Z4,
able to solve
now
Real Element
Faithful
=
algebras example (1).
in
=
a
the table
given
b2
with
=
103
3
isomorphic
3T,,&2)3
of this
to
T-M(3)
section.
b, Z 54 c and 5, T Icl to one of the table algebras =
=
2b. Wehave
-
(b2, C2) (bc, Fc) (u, U) 10(C' C2) + 10(-E, C2)
20 +
=
=
(4.38)
and
(u, u)
Since
2, then
c7c
!
[bxl that us
JxJ
u
we
(b
=
2 ,
C7 C)
20(c, c7c).
25 +
=
(4.39)
from
obtain
(c, c7c)
that
G
(4-39)
11, 21.
(c, c7c)
If
m is
=
necessarily
even
that
(c, c7c)
Chapter
E
for
d,
distinct
3
yields
b E=-
B,'
and
we
b is real
as =
(4.39)
1, then
E,B5\fbl. 1[52], [517 511, [101] 1. Let us
e
=
=
Now Lemma 2 of
assume
2d +
[x]
=
=
=
[53,52].
now =
(bc, bc)
=
P. Now Theorem 2 of Chapter 3 yields B, In particular, m G N U 101. B, is five homogeneous. For b2 so that (bx, bx) dchave x-T 65 and (b2' XX)
=
10 and
=
15,
=
we
conclude Let
Jul
some
therefore
whence
(u, u)
51 + 2c + 27!
=
Tm(5) for x E Bc\fbl
20 +
Set
e
first
x
(u, u)
implies c7c
:=
look
51
-
at the
-
c
case
-
25,
=
Z. Then
[x)
=
[52],
B5. Then Theorem 4 yields that B, is one T-M(3) of the table algebras 3T,,, Y4, Y5) Y6) Z4, 3Tm 3 T,,(3), 3 defined'in we see that B" is five homogenous. example (1). In particular, Our aim is to apply Lemma (2) of Chapter 3 in order to prove that b E Be.
i.e.,
x
2h for
=
For this
reason
some
real
h E
the have to prove in all cases that for each x E Bc\fbl of bx is divisible element y of the support by 5. If
we
degree
of each element
this
not
100.
divisible by 5 and as jyj > 4 clearly degrees of basis elements we deduce in any case, since But this is not possible
x
that
(bx, bx)
the definition
:! 2. Hence table algebras
is
by
the
basic
our
(y, bx)
then
case,
assumption
on
(bx, bx) > (y, bX)21yl > 25.4 e B,\11, bj has degree 5 so nonreal whence (c, xT) as c is
=
b
=
h
)
Let
Set y
or
finally c2 C -
-
that
a
=
(b2, x y)
=
we
=
obtain
25 +
B
yields
that
By (4.38)
we
=
B is
20(c, t-x)
g). that
(x, x)
=
10.
have
< 65
Looking at either Y4 (and
B,.
a:=
(Z,
c
2) 7
0.
ccc-.
45 so
=
asume
us :=
particular
of the
Y5 (and b
is
minimal
2. But
=
(C7,c d-)c
now
(4-38)
c
2, c 2)
yields
+
a2 )5
the contradiction
+
(y, y) 10
=
(2d
+ e,
2-d
+
j). 0
of Primitive
The Enumeration
5
Commutative
with
Association
Non-symmetric
a
Valency,
Schemes of
Relation
Most 4
at
Hirasaka','
Mitsugu
IDepartment
of Mathematics
and
Computer Science
University Ramat-Gan 52900, Israel
Bar-Ilan
2Depattment National
Taipei,
University
Taiwan
Introduction
5.1
(X, G) be (X, g) is
Let
Then to
of Mathematics
Taiwan
all
find
an
hypotheses
about
W.J.
numbers
intersection
the whole structure
and to determine
[62],
scheme in the
association
regular digraph for each regular digraphs which might
g E
a
Wong classified
all
or
sense
of
G. It
is
be
an
an
induced
[641 an
element
where X is finite.
problem
interesting of G under
certain
subgraph of the graph,
(X, G).
of
primitive
permutation groups valency 3, [541, [551 W.L. Quirin and C.C. Sims of primitive transitive permutation groups with a 2-orbit gave a classification of the above results, it is known that the of valency 4. As a generalization of a fix-point stabilizer of each primitive transitive cardinality permutation non-trivial valency (see [37]). group is bounded by the minimal In 1981 L.Babai [261 proposed the following In
with
a
2-orbit
Conjecture the maximal
non-trivial
1.
of
(L. Babai) valency
elements
transitive
and in
Let
of G is of G.
(X, G) bounded
be
by
a
primitive function
a
association
of
scheme.
the minimal
Then
valency
of
seems to be quite However, this conjecture open, and we need to make the or comproblem easier, for instance to assume that G is a P-polynomial to impose some restrictions numbers. If G is or on intersection mutative, relation of valency and contains at most 4, commutative a non-symmetric then Conjecture 1 is true (see [46], [47)). In this chapter, commutative association schemes we focus on primitive whose valency is small enough. We observed relation with a non-symmetric series of such schemes with growing I GI are translation that all known infinite the following schemes in a sense of [36, pp. 661 (see Example 2). This validates
Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 105 - 119, 2002 © Springer-Verlag Berlin Heidelberg 2002
Mitsugu
106
Conjecture
(M. Muzychuk)
2.
If
scheme.
ciation
(X, G)
is
In this
chapter
4,
Hirasaka
translation
a
moreover
which
the
is
give
be
primitive
IGI
commutative
is
large
asso-
enough,
then
scheme.
Conjecture
that
a
and
non-symmetric
association
prove
we
we
(X, G)
Let
g E G is
the enumeration
scheme of 2-orbits
2 is true
of all
of
a
valency of g is at most schemes, each of group on an elementary
if the
such association
permutation
abelian
group. Another motivation
schemes with all
erties:
a
of this
chapter comes from research of association which satisfy the following points, propall relations of odd valency are connected;
number of
prime
relations
non-trivial
If an association scheme is a translation scheme over a non-symmetric. of association scheme cyclic group of prime order, then the whole structure determined is uniquely elements. Asby the minimal valency of non-trivial schemes of this type are called sociation cyclotomic (see [36, Cor. 2.10.2]). schemes with a prime However, it is still open to determine all association and there are only classifications number of points, if the cardinality of the of the are immediate point set is small enough. The following consequences of this chapter: all association enumeration schemes with a prime number of points relation of valency 3 are cyclotomic; a non-symmetric containing scheme with a prime number of points there is no association a containing relation of valency 4. non-symme ric In contrast to the situation of the previous paragraphs, any primitive scheme with a relation of valency 3 is a P-polynomial, symmetric association which is proved in [63]. of distanceCombining this with the classification regular graphs of valency 3, we see that- there are exactly four isomorphism to the following classes corresponding graphs: a complete graph with 4 points; a Coxeter a Peterson a Biggsgraph with 10 points; graph with 28 points; Smith graph with 126 points (see [36, pp. 1791). In this chapter we use notations of [64] which are slightly from different are
'
used in the Introduction
those
Let X be
a
the empty set. Wedefine 1x Let r C X x X be given. x
E
X,
define
we
Definition
(i) (ii) (iii)
Ix
E
Wedenote
xg
following
E
:
=
called :=
pl99,
a
We set
r*
X1 (x, y)
c
The pair conditions:
of X x X which does not contain
partition
I (x, x) I
E
x
:=
X 1.
J(x,y)
I (y,x)
E
rJ and,
for
each
rJ,
(X, G)
is
called
an
association
scheme
if
it
G;
For each g E G we have g* cFor all d, e, f E G and x,
(X, Y) are
ly
:=
( [64]).
I
the
satisfies
xr
book.
to this
set and G be
finite
E
f
y
E
X,
lxdnye*l
is
whenever
constant
-
the number
the
G;
I xdnye* I
intersection
which
is called
with
numbers the
valency
(x, y)
C
f by
of G. For
of g.
pf, d
and
each g E
Jpf,,dj
G,
we
d,
e,
f
G
GJ
abbreviate
(X, G) (X, G) (X, G)
Wesay that Wesay that We say that
where G'
connected
[64],
1 011.bwing
I
we
=
f P,d for
all
Schemes
d,
f
e,
symmetric primitive G f 1x 1.
is
107
G.
E
if g = g* for each g c G. if, for each g G G' the graph
is
:=
f if Pde
commutative
is
Association
Commutative
of Primitive
The Enumeration
5
(X, g)
is
-
define
product
the
If
de:=
of two elements
Pfe
GI
C
e
E
G to be
7' 01-
d
For each (x, y) E X x X, we (x, y) by r (x., y) Let (X, G) and (Y, H) be two association isom4phic to'(Y, H) if there are two bijections such that (x, V) E g if and only if (O(x), 0(y))
denote
shall
d,
(5.1) element
the
of G
containing
-
each g E G. For each g
G,
E
for
that,
0 E
X
:
p(g)
(X, G)
Wesay that ---+
for
is
Y and p : G --+ H all x, y E X and for
'
define
we
adjacency
the
f Note
schemes.
d,
all
E
e
matrix
(X, y)
if
1, 0,
as
follows:
EE g
otherwise.
G we have
PfdeAf
AdAe fEG
(see [27, pp. 53]). It GI is a subalgebra algebra of (X, G),
follows
with
table
Example
We denote
scheme then
Given g
(X, H)
E
be
an
G, a
ation
it
we
shall
g
the Bose-Mesner
orbits
L (X,
write
L(X, g)
is
a
ordering
g)
for
the
set
of association
poset with
respect of partitions
the set
on
(see [28])
scheme
association
schemes with about
on
a
non-symmetric
intersection
numbers
(see [46], re
*lation
of
..There
exists
a
to the
schemes
paxtial
of X
maximal
order
X which
x
element
following
the
[471)
(see [201, [331)
are
valency
as point example frequently
set
commutative primitive of valency at most
relation,
already
the elements appears
in
and the structure
known.
in
These results
associ4.
of
Some a non-
show that
gives a great amount of restricscheme, and makes it Possible of an elementary abelian group. this chapter.
at most 4
of the association
the whole structure
characterize
The
it
-
refinement
symmetric relation a non-symmetric to
by JA_' I
C spanned call
g),
results
tion
we
and A be
set
X and the
which is unique (see [60, pp. 18]). is to determine The goal of this chapter
L (X,
over
scheme.
association
finite
g E H. Then
with
where -< is
might
a
formal
the
[27],
a
an
of
the Bose-Mesner
spanned by
space
a finite transitively group acting of A on X x X is an association if X ts a finite by 2-orb(A; X). Furthermore, group algebra of 2-orb(A; X) is isomorphic to the Schur ring sums of the orbits Of Aid,, on X (see [58]).
Let X be
1.
Then the pair
X.
on
of
the vector
C). Following
by CG. In the sense of [20], CG is a standard basis JAgI g E GJ. The following distinguished
denoted
algebra example
integral is a typical
that
kat(X,
of
Hirasaka
Mitsugu
108
Example 2. Let Fp,,, be a the symmetric group S,,+I, dimensional
product of Zp' for some odd prime p by of which on ZPn is given by its natural n-
the action
Then
representation. scheme with
association
(p, n)
semidirect
2-orb(rp,n;
Zpn)
is
primitive of valency
(3, 2).
=
For each g E G, the ordinary
shall
we
denote
just
write
product
A,
by A
B
A,
of
g instead
of matrices
commutative
a
relation
non-symmetric
a
for
as in (5-1). product of two relations from the theory of standard are results following
and
convenience
B in order
-
unless
+ I
n
distinguish
to
between the The
bras,
which
(H-1.
is
connected
a
isomorphic
rank
at most
The
following
with 3. Let
a
(i) (ii) (iii)
=
an
immediate
(H.I.
Blau)
Then
maxgEGx Pgaa* a
-
relation
3
-
1X,
al
! 2 for
Although
a
properties result
the
Theorem 2.
Let
non-symmetric the following:
e
E
in
-
ai
JgJ
this
-
Xa2
a*
-
fg
E
G,ng
g c
GJr,
relation-
are
(X, G) relation
a
9
of
=
an
determined
chapter
our
mingEGx
3 which is not
-'g
=
necessarily
=
ai-1
a2:7
3 and
a*;
(5.2)
+ ai-2
is
31
=
31
U
13 1xJ; -
(e, f)
then
E
scheme
association
[46],
in
shorter
J(d, d*), (d*, Ix)J.
shall
we
than
one
satisfying
prove
given
in
the
[46]
the same
The
main results:
be
a primitive of valency
x C3; Zp) for some p of Zp and C3 : Aut(Cp);
2-orb(Cp tion
proof
enumerations
a
C
Theorem 1 is
in
scheme
association
3. Assume
Then
a.
non-symmetric
as
since
following
d,
some
-
:=
:=
Jaiji>o
Ifpdf de
commutative
1;
=
=
=
a
of valency of valency
2a* + a2 for some a2 E G with For each i > 2 we set
a
where ao
(i)
be
relation
following:
ai+1
(iv)
(X, G)
Let
have the
we
of Theorem 1.
consequence
non-symmetric G be a non-symmetric
connected.
alge-
[33]).
than 3.
is 1
E
table
Blau).
connected
a
integral
(see [20],
relation
greater
Corollary
schemes
association
M
Introduction odd prime
for
Let (X, G) be a commutative scheme association 3. Then CG of valency 3. Assume MingGGx ',g to one of the following: O(C3, H) for some abelian group H of 2; or 0 (S3, Z,2,,,) for some m 4 where 0 is defined in the to this book. Furthermore, then m is an 'if (X, G) is primitive
Theorem I with
reformulated
are
_=
commutative
3.
Then
(X, G)
1 mod 3 where
association is
Cp
scheme with
isomorphic
is
a
regular
to
one
of
permuta-
2-orb(C'
(ii)
G' Z2) for P
4
V
Of TV,2 of Let -K be
Z3) p
of
index
group
Let
(X, G)
(X, G)
with
a
The
Then
(i)
If
(ii)
E
a
G'
where
Cp
If
G'
c
a
is
(i)
Since Ka
==
X is =
Ka
Cp
a
be
a
valency
in Section
I
Example 2.
in
Cayley graph
the
E
is
1
5.3 and with
a
2-orb(Cp
-
Ma
=
1, and
inductively
minimal we can take finite, are distinct. 1, jxjjo<j<,,,_j
some
xi
to
n
E
(-r); Zp) for some prime -x. : Zp -+ Zp, x i-+
xO E X.
be
a
Then there element
unique
Z>O such
that
from
xO
-=
exists
a
in xi-la. xn-
Since that
primitivity
Cay(Zn, 111)
a)
[20,
from that
n
is
desired.
as
Let
valency
a
(=-
G be with
It
follows
exists
a
1.
Ka
Weconclude
(X, a)
=
from
unique
unique element in c Z>O such that
distinct.
p
-
Then there
n
prime
andr
We conclude
(1XiTO
=
x
of Zp
permutation
Gx be with
We define
Zp) for
2-orb(Cp;
-
=
a
a
Scheme.
association
2-orb(Aut(X, a), X) E C(X, a). Since 2-orb(Cn; Zn) 2-orb(Cn; Zn) has no relation of valency greater than 1, it follows implies Primitivity j2-orb(Cn;Zn)1Prop.. 5.81 that L(X,a)
(ii)
enu-
5.2, with non-symmetric
of Zp;
(X, G)
2 then
regular
Therefore,
of
the
Section
2 in
or
commutative
(X, G)
permutation
=
a
(X, a)
prime,
of E
2
or
primitive
1 then
=
regular
a
E xoa.
Ka*
n,,
with
Let
unique.xi
as
following:
with
odd p where
Proof.
(X,G)
Let
have the
we
valency
of
case
Theorem 4.
is
scheme. Wegive
association
of
relation
of valency 3 relation non-symmetric of valency 4 in Section 5.4. relation
5.2
Fp,2
3 where
a
commutative
primitive
a
subgroup
2 ;
Of Fp,3-
2
be
of
meration
a
109
scheme with association commutative primitive to either Then is isomorphic (X, G) of valencY 4. where -I-pl,3 is a sub-' some odd prime Z3) for p .
be
2-orb(Fp',3;
or
3where C2 >1 G is P
and E C K. Wedenote
group
relation
non-symmetric
2-orb(F'p,3;
>
Example
in
as
Schemes
-
(X, G)
Let
Theorem 3. a
finite
a
is
odd prime p'>
some
by Cay (K, E)
K
over
odd primep
-Pp,2
2 where
for Z2) P
2-orb(Fp,2;
(iii)
Some
,
index
Association
Commutative
of Primitive
The Enumeration
5
xi-la xO
from
=
Then, by (i) there
X2 G x1a, -
=
2.
[20, Prop.
xn-
fXi-21Since
-
that
no
=
Wedefine
finite,
X is
Ka
2 and
=
is a
Since
a
non-trivial
a*.
Let
we can =
a*,
1,
a)
-
Cay(Zn, 11,
-
11)
relation
(xO, xj)
inductively
that
primitivity
(1xi 10
5.81 jxoj.
take
xi
c
to
a.
be
minimal
1xi1O
are
Mitsugu
110
2-orb(C,,,
Therefore, -r
-4
x
:
The
chain, it
all,
of
First
define
shall
we
and characterize
is not
valency
chain
is
where
odd prime. Since other than 2, it
unique
a
primitive
desired.
as
cyclic
as a
primitivity.
assume
Hence
scheme with
association
C(X, a)
E
an
valency
of
(-r);Zn)
x
of elements
sequence
a
is
0
3
closed
a
to
necessary
commutative
2-orb(Cn in L(X, a),
valency
of
case
n
relation
non-trivial that
a), X)
that
primitivity
scheme
association
commutative
2-orb(Aut(X,
c -_
from
has no (-r);Z,,) [20, Prop. 5-8]
x
from
5.3
(-r); Z,,)
x
follows
It
-x.
2-orb(Cn follows
Hirasaka
a
of
X, which
is
called
In order
to
do
group.
that
(X, G)
non-symmetric
relation
we
connected
suppose
a
this, is
a
of
satisfying
3
min
ni
1EGx
(5.3)
3.
=
Theorem 4 shows that
commutative scheme with a association any primitive relation of valency 1,2. greater than 2 has no non-trivial of (5-3) Hence the assumption is natural. We have already determined the schemes isomorphism classes of the Bose-Mesner algebras of such association
relation
(see
valency
of
1)
Theorem Let
of X is
elements
(Xi Xi+2)
of
length
length
(xo,
of
is defined
as
Wefix
and here with
ng
a
ndne-
Proof.
j. In addition
is
E
G and say just that similar
chain
d
Pfe*
[27, 7
xi,
pa2 aa
(xo,
(xo,
xi,...
0 < i <
,
j
-
xj)
of
1, and
in Corollary I (ii). edges, namely the above, the length of (xo, xi) as
number of its
to the
chain
of an a-chain for convenience, g-chain also hold for each g c- G results. 54 g*. Weneed the following preliminary
d,
e,
f
a
instead
statements
E
for
G we have the
r' d
following:
--
d Pe-fKe.
56].
pp.
1. It
=
,
xi)
be
a
follows
chain.
that
xi,
of any
xi,
length,
xi+1)
is
a
and there
Then, by Corollary exists a unique
I
(ii)
and Lemma 1 (ii)
there
xi+1
and
A sequence each i with
2 where a2 is
-
the
as
for
f
=
(xo
Let
j
a*. a
EfEG Pdenf;
See
paa2a*
xj)
xi,
E
0 < i <
defined
is
a:7
3 and
=
(xi, xi+l)
if
I.
For all
=
P1der'f
a-chain
3 and g
=
na
a-chain
each i with
an
claim
we
Lemma1.
(i) (ii)
an
c a2 for
7
The
*
G be such that
c
a
n xi-la2,
E xia
chain exists
of
length a
closed
i + 1. Hence
chain
since
we can
JXJ
construct
is finite.
a
where ai
of
a
on
i.
length
the
is
n
xi)
(xo,xl,...
Let
Lemma2.
Use induction
Proof. shall
prove
that
with
0 <
each
j by Corollary
First,
r(xo, xi+,)
1(ii).
(xo, xi)
xia
hypothesis,
Xi
2),
-
follows
from
Second,
we
=
forces
and
rbr(xo,y)
r(xo, Xi-2) a
(5.2)
=
-
ai
=
we
x0
< n==
and ai-I we
1
-
have
we
1,
fxi+,,
y,
zj,
xi-ja*
that
so
n
Nr(xo,y)
n,(x,,,x,+,.)
=
I
rlr(xo,z)
i
=
jXi-2)
r(xo, xj+j)
ai-1
=
=
(xo, y),
r
Y)
Z1
r
n,(,,O,,)
Since
3.
=
(xo, z) I by
the induc-
a*r(xo,
xi-1) 1(iii), =
r(xo, y)
+
a*aj-1
=
and
3
by
the
inductive
f r(xo, y), r(xo, z)j C a*aj-, r(xo, y) r(xo, z). Then, since =
hypothesis,
we
and a*
r(xo, aj+j the case of
xi+,),
=
INr(xo,xi+,)
have Kr(xo,xi+,)
Kr(xo,z)
fact
the
6 and
3-lx+ai-2+ai
r(xo, z).
+
desired.
as
Kr(xo,y))
r(xo, xj+j)
+
that
consider
=
r(xo, z).
+
of
fKr(xo,xi+l))
r(xo, y)
we
r(xo, y)
r(xo, xi+,),
=
case
6 E
rbr(xo,z)
(5.2)
+
r(xo, Xi-2)
aj+j
the
Kr(xo,y)
7
=/=
I
ai-1
-
as
i
I
2,
aj+j aj-2
r(xo, y)
+ aj-2
desired.
Wemay assume Kr(xo,y) we have ai aaina*ai-,,
Since
aj-2-
=
=-
=
Kr(xo,z)
by (5.3).
that
n aaj aj-2
have
Since 1x E
since
ai
=
have
*
2
aj-2
=
=
a*
a*aj-1
(i). r(xo, xi+,)
by Corollary
E
On the
I
0 Ix,
Z_
=
=
-
Thus, either
for
11, 3, 61.
C
n,(x.,y)
=
jr (xo, xi+,),
loss of generality. y without from (5.2) = a. It follows
hand,
2. We
-
lemma holds
of this
=
that
E
have a-a*
2. Assume 3 < i <
1,
=
statement
aa* and aaj r(xo, xj+j) Cz ar(xo, xi) (=we f aj-2, ai 21. r(xo, xi+,) Assume the contrary, Weclaim r(xo, xj+j) i.e., r(xo, xi+,) aj-2. a* a, it follows either Since r (y, z) c r (y, xi) r (xi, z) z E yaj 2 or z
other
n
r(xo, xi-I)a* a*r(xo, xi-1) r (xo, z) 1, we have
1 E
Since i
a
by the inductive
Finally,
a*a,
length.
<
Then
from Theorem I that
case
Corollary =
from
follows
-
ai
consider
r(XO, Xi-2)
Since
hypothesis
the
-
(xo, y),
r
a*
i.e.,
1 < i
ill
follows
it
a*ai-I
since
IT (XO,
7
(xo, xi)
r
=
if i
:=
Weconclude
consider
we
or
=
Also,
It
of
Schemes
with
i
shortest
the
if the
aj+j
=
j :5 i. Weset
a
It
of length
chain
a
chain
is trivial
It
fKr(xo,xi+l) aaj
be
closed
Association
r(xo, xi).
=
tive
Commutative
of Primitive
The Enumeration
5
of Z7 Xi-2
E
xi-ja*
n yaj-2
or
z, xj+1
c- xia
n
yal-2.
al-2. yal- 2
=
(E
Hirasaka
Mitsugu
112
implies
This
2 <
ai-2
Pai-2a
paaai-21
2 <
or
c'ontradicting =
Lei
(zo,
Zn)
zi,
r(ZO, Zn-1)
be
generalized
Lemma2 is
a
xo)
'r(zn-1,
=
1(i)
Corollary
since
_a*
and Lemma1(ii). by commutativity paaai-2 Paa* -Paaj-2 By the'above claim, we have r(xo, xi+,) aj-2 aj+j the proof. -
closed
a*
=
r(ZOi Zn-2)
and
following
to the
2 Let (xo,xl,... Corollary Then r(xo, xj) same length. i j =- 0 mod n.
of the shortest
chain
=
completes 0
length.
r(Zn-2
=
This
-.
=
Then
ZO)*
5
=
(5.4)
a*.2
corollary.
Xj) and (yojyj,... yj) r(yo, yj). Furthermore,
chains
be two xi
of the if and only if
xi
=
-
Proof. n
By
Lemma 2,
1. Assume
-
j
we
have
r
(xO, xj)
Weconclude
> n.
r
=
(yo, yj)
for
each
j with
(5.4)
that
from Lemma2 and
0 <
j
<
n xn-2a2)
X0, Xn G xn-ja
I by Lemma 1(ii) and Corollary implying xO xn since pa pa2 1(ii). a2athat the starting This implies point of any chain of length n coincides with Hence r(xo,xj) the end point. for each j with 0 < j :5 n. In r(yo,yj) r(xn-l,xl) particular, 7'(Xl,xn-l)* r(xo,xn-2)* a2. This implies =
=
=
aa
=
=
r(xo,xj) have
we
=
Since there r(xo,xj-n)r(xO, Xj-in) r(xo, xj)
exists
=
=
=
=
i E
Z>o suchthat
T(YO, Yj-in)
proof of the first statement. Since jxjjo<j
=
r(xo, xj)
<
n-1,
desired.
as
This
=
r(xO5 xi-in)
follows
statement
if
from
the
one.
Definition
2.
chains
with
2 <
define
a
(yi, yi')
(i) (ii)
If
Let
L
j :5
relation
E g
for
L is
a
=
n
line,
=
(yo,
(L'I- L") Proof.
(i)
E & and E 6,
is
an
Wedefine yi-ja
Yi-2
n yja,
and
.
:=
.
yj)
(Y1)
-
-
a
be two yl, 5 yj) For each g E G we E if and only if ..
.
=
1.
be
Y27
there
(yb,
=
line
n. if j by defining (L, L')
-
,
.
and L'
L is
-
a
,
chain.
Yn,'Yl)
exists
a
there
exists
Then is
have the
line
and
chain
L'
a
unique
we
following:
(L, L)
d; from y6
G
starting
6,;
L'
=7
(L, L")
immediate which
yj)
.
of chains j
f yj
inductively
yz -,an yja* 2
.
0 :5 i yl,
L
then -
If (L, L')
yl,
the set
on
For each yb E yo a such that (L, L') E
(iii)
(yo,
We say that
-
each i with
Let L
Lemma3.
in
r(yo, yj))
the
completes
first
j-in
0 <
=
then E
54
a.
of
each i with
well-defined
and a*2
It
a
unique
L"
chain
such that
-
consequence
y, for is
a2
(5.4)
by Corollary remains
to
Corollary
and
1 < i <
j
to be
1(i)
show that
and
L'
2.
unique element
a
yi,'_1 :=
0 (y 0......
yj
since
yj)
is
The Enumeration
5
a
chain.
suffices
It
I
yz 0
=A Y.42
Note yj
construction.
by
yj+j
since
Then
construction.
yi', 1(i).
Corollary
contradicting
r(Yi1Y42) n N+2a*
(yO,
=
0 :! ,
G aa
113
i :
j
2.
-
la*,a2}
=
0
and a*2
by
a,
and
yj
and
have n
Yi+2a,
contradiction
This
Schemes
i with
E yja2
E yja
yj+j
each I
since
Yi+2
we
Let L' Now we prove (iii). a*. Hence there exists (yo, yi)
r
i
some
for
(=- a2
I
for
a*
E
Association
Commutative
(Yi Yi +2)
show that
to
(Yi Yi+2)
Otherwise
of Primitive
yl,...
(ii).
proves
yj).
,
Then
we
have yO
=
yo E y6a n yoa2 c y0a
By (ii),
there
exists
remains
d. It
unique chain L"
a
proof
%
%_
Lj
Lo, Ll,. it (Li, Lj+j) .
.
,
)
P:=
(yi, y /) E a2 by induc1, which completes the 71
1),
2.
-
=
Ln) j
<
o :
j), y(i
0 < i
<
j
-
and
1,
is
a
of
chain
(Li, Li+2)
G
&2
In
chain
of lines
n.
exists
a
plane
y
-
:=
(0, 0)
1, j
-
for
y(j,n))
(y(j,0),y(j,1),... could
where the coordinates
n
=
Lj
and
be read modulo
n
by
convenience.
1),...
,
y(i
k, i
-
-
k))
is
a
for all i, j, k
chain
n.
y(i+l,j),
we
<
< j plane if j
a
0 <
have
r(y(i,
y(i,j+l) j), y(i
E
y(i,j)any(i+l,j+l)a* a* for + 1))
1, i
+
and all
=
i, j with
y(i+lj):7 0 <
i, j
<
n
1.
Wewould like
i, j
Lj)
(Lo Ll,.
We say that
=
k <
by Corollary 0 <
E
I < i <
implies by paa2a*
each i with
(Lo,Lj,...
i, j,
Since +
(L', L")
such that
-
(y(i,
Lemma4.
Proof.
is
2. We set
0 <
for
0 < i
Lj) Ln)
j with
each
Corollary
y(i, j
yj")
a we say that particular, By Lemma3 (ii), (iii), there 2 that Ln from Corollary It follows Lo. of this section, remainder we set
.
For the
with
,
.
d, (ii)
c
obtained
is
be lines.
E a
each i with
(Lo, L1, (Lo, Ll....
for
.
0
Let for
.
(iii).
of
lines
each i with
71
z
of L"
Uniqueness
i.
on
for
E=- a2
yLl, y ' 1), (yi, y , y '))
((yi-1,
Since
(yo", yj',
=
show
to
(yi, yi') tion
lyl I.
-
n.
to prove
that
r
i) j. Wemay assume (i, j)
=
y (i
(y (i, j),
-
2, j
2))
-
=
i. e., r (y (i, j), y (i 2, j loss of generality without
Assume the contrary,
-
(2, 2)
-
a2 for
2)) by
all
=
i, j with
a* for
some
construction.
Since y
it
follow
sume r
from
y(2,2)
(y (2, 0),
y
(1, 0),
Corollary =
(1, 0))
y(1,0) =
y
(0, 1),
1(ii)
y (2,
that
2)
c oa n y
y(2,2)
G
We may jy(1,0),y(0,1)j. r(y(2,O),y(2,2)) a2 I (ii).
follows
by symmetry. Corollary a*, contradicting It
(1, 1) a*,
=
as-
Hirasaka
Mitsugu
114
jy(i
Lemma5.
with
i, j
0 <
k, j), y(i, j
+
< n
Proof.
We give
the
similarly by construction,
n
1
Assume the
chain.
3 < k <
k with
distinct
1)
and
-
E
ak)
(ak-1,
Since 3 < k
<
n
3,
-
2,
=
same
without
Ka
=
Corollary y
(2, 0))
r
=
3 and y (1, that y (1,
0) : 0) (y (0, 2), y (0, 1))
1 (ii)
y
loss
k
i, j,
we can
the
k) I I
-
for
there
prove
k
=
definition
of
< 3
for
at
least
are
and end
1, a
some
two
Hence
points.
Since y(1, 0), generality. 1 (iv) that Corollary
of
from
ak)
E
and hence ak G oa n
J(a2, a*),2 (a*,2 Ix)l. =
a*
=
a*.2
a
implies
This
y(k, O)a*.
(0, 1) and y (1, 0) =A y (0, 2), it 2) and y (2, 0) : = y (0, 1). Thus,
a*, contradicting
=
all
clear
is
2
-
starting
(ak-2,
and
0 Ix,
y (0,
=
n
2, it follows
y(O, 1), y(2, 0), y(O, 1), y(O, 2) Since
since
It
Lemma4 that
the
y(O, k)
have ak
we
(0, 0) only,
=
construction.
k
from
k with
for
distinct
are
by I ly (k, 0), y (0, k), y (- k,
J(a, a*), (a*, 1x)l
e
by
by Corollary
I
k)l
-
1.
(i, j)
for
i. e.,
=
-
k, j
-
-
case
is
so
length
y(k, 0) (k, 0) a*k
n y
oa
n
cases
follows
3. It
assume
may
y (0,
n
the
other
contrary,
of the
chains
for
for
assertion
-
we
proof
the
k), y(i
+
and I < k <
I
-
0
a*.
This
follows a
=
r
from
(y (1, 0),
completes
the
proof. For each i with
Lemma6.
ai
subscripts
where the
ajai a
and Pa'jaj
> 2.
Thus, Corollary
it
remains
2. If
ai
to
=
ai*
1,
read modulo
are
(i, j, k)
=
have
we
ai* =7
a2i
n.
(i, 0, i) ly (0, 2i),
y
(i, i),
y (0,
0)), r(o, y(O, -i))}
a2i
ajai
=7 a
only.
11x, ail,
=
If
ai
=
0 a
then
and hence a2i
=
a2i
3. 1x
(n, 3)
=
I
Assume the contrary,
have a2i
=
r
(o,
y (2i,
i.e.,
0))
=
r
1 Let (YO Y1 Y2 7 107 O)l fYilo
Proposition
7
y0a, n
and
=h a by =7 ai, as 0
Lemma7.
Proof.
are
f al, a2ij
desired.
we
-i) I
2 and Lemma4 that
show that
then
-
+ a2i,
y(i, i)), r(o, y(2i,
jr(o,
=
n
2ai*
=
&om Corollary
follows
It
a
ai
-
Lemma5 for
Applying
Proof. distinct.
of
I < i <
=
n
7
i
n
=
3i
(y (2i, 0), o) -
-
-
then
,
yn) be (X, a)
for *
a --
i E Z>O. By Corollary 2, Lemma6. 0 a , contradicting
some =
chain
of length
Cay(Zn, E)
n.
where
If n =-
1 mod 3
The Enumeration
5
'Assume
Proof.
r(yo, yi)
(yo, yi)
r
for
a
=
a, and hence a2i
=
(Y6 is
an
ai2=
length
(yo,yi2)=
ai
also
is
n,
=Jyj+j,
I
Yj+i2
Yj+i,
i
Z'.n
E
and
2
Ka
each
for
implies
have
we
ai
that
j with
y0a
j :5
0 <
Zx n
c
jyj,
=
n
distinct.
are
Yj2j,
Yj,
It
and hence
implies
1. This
-
2 that
of the shortest
ai-chain
closed
11, i, i2j
3 that
Corollary
from
follows a
Hence
=
Then
115
Yj2)
Yi) Y2i)
Since the
have
we
by
7
1 mod n.
Lemma6. This
2. It by Corollary length of
i a.
=
Corollary
from
follows
yja
length
of
a-chain r
0
i
some
a2
=
Schemes
Association
Commutative
of Primitive
X
1Y40
Since
Corollary
n
JXJ
=
completes
This
Theorem 5. a
Ir (yo, yi) 10
=
G is
c
a
isomorphic
Cay(Z,,
(i) (ii)
< i
<
11,
-
j:jr,,(y.,yj)
I +
=
follows
it
be
< i
<
prime 1,
some
each k with
11
I mod 3.
I
(iii)
and
n
some
k, jk),
+
y(ik,
jk
oa
n
is
JEJ
P
E
by
3
is not
=
3;
Zn}
is
a
connected
=
k, jk
-
k)j
:!
that from primitivity n2. This implies that n is
Lemma7.
O)j, 0)},
0 jy(1,
Wi) 0)10<j
-
follows
JXJ
X and
with
coincides
k), y(ik
+
ly(ik,jk)li7i
ZnJ
If
odd prime p > 3.
ly(ik,jk)li,j
1
-
E
which
the
ak since
to
jy(ik
=
scheme.
graph (X, a)
association
-
1) 1) for
2, Lemma 4 and Lemma 5. It
If oan
=_
1 mod 3 where E < Zx with
=-
p -
1 < k <
respect
ik)ak
prime,
-
3, then
of valency
by Corollary odd
n
commutative
primitive
a
non-symmetric to one of the following:
E) for
For
10
element
y(ik,
If
Corollary
0
(X, G)
Let
component with
an
from
proof.
the
Cay(Zp', 1 (1, 0), (0, 1), (-
Proof.
n
2 that
jy(1,
then
then
we are we
done
claim
that
by Proposition
1.
fY(i)j)1O
Assume the contrary, distinct. y(k, 1) for some (i, j), (k, 1) E i.e., y(i, j) Z2. Wegive the proof for the case (k, 1) (0, 0) only, since we can prove the for other cases by construction. assertion 2, we have By Corollary similarly =
are
=
n
1x This
tinct,
=
r(o, y(i, j))
implies ai in particular
=
y(-j,O) y(i,O), (o, y(i, 0), y(2i, 0),...
aj*. By o
Applying tradicting the map
Lemma 5, y
=
that
C= ,
r(o, y(i, 0))r(y(i,
(z-
(i, j) with
oaj
y(ni,
0
0))
is
ajaj.
j, -j) I are disf y(i, j), y(i + j, 0), y(i 2 (i + j, 0). It follows from Corollary is Since n a prime, -=h y(i,O). y(-j,O)
a
n
-1)1),
=
-
y
as
ai-chain
closed
I for the ai-chain, Proposition the assumption. Therefore, y (i, j) 1-4 (i, j) (E Z2 gives us
Cay (Z2, 1 (1, 0), (0, 1),(-I,
0), y(i, j))
we
have X
of the =
fY(iij)1O
isomorphism.
desired.
shortest
length
IYU, 0)10<j
distinct.
n.
con-
Then
to
0
Hirasaka
'Mitsugu
116
Proof of Theorem Let
and
following
Wegive the either
of
O(C3, Zp); 0 (C3, Zp2)
First
of
all,
CG -_
then
2-orb(Aut(X, 2-orb(Aut(X,
(X, H)
3,
=
-<
follows
and hence
like
Weset X of
valency
that
d
=
e
od
the
Z2 and
:=
implies
Uf
=
we
x,
,
have ai
j
(i, j), (j, i)
Since
unique.
is
G
aj
e
=
+
E od n
od n
(i, i)aj-i
of Therefore, e: obtain
f
to
(-i,
i
we
C-
i), (i
-
(i, j), (i
-
asso-
relation
no
f
e,
(-j, f
Pai
aj
i
=
=
-
==
E od n
pj.eaj
(j, i)j
j, -j),
as
E
G such
Paj. fai
j) I
1.
Observing
(i, o)aj, =
P_
Paiaj
0.
It can be easily checked i, -i), y(-j,i fy(j j), y(ij)j. oe and p > 3. Applying Theorem 5 for e E G we (-i, -j) (X, e) _- Cay(Zp', f (1, 0), (0, 1), (- 1, 1) 1) since IX I p2. Thus,
oe
that
=
-
-
:
=
-
contains
(i, j)
regular
a
R< -r
-
=
j, -j)
-
exist
j, -j),
Paiaj
ZP2)
Then
e.
from
conclude
i), (i
-
have
G contains
G there
-
implying
(0, j)aj,
we
2-orb(Fp,2;
shows that
(o, (i, j))
3. Let
+ ai+j,
1(-i,j
a), X) above
e* since
Aut(X, e)
where
n
that
(5-5)
each d E H
i, -i),
i)
-
for =
-
have
Theorem 4 shows
Z2.) P
isomorphic
(0, 0). (5.5)
that,
3 and rf
-
we
.
that
that
C3. Then
x
above observation
-
(X, G)
o :=
I (i, j), (j, i), (i
(i, j), (-i,
=
=
case
that
to prove
Aut(Cp)
2-orb(Aut(X, a), X), as desired. Aut(X, a) -Pp,2 Wewrite 2-orb(Aut(X, we have (X, H) G C(X, a), it follows from the
the
6. This
=
By (5.5),
and
Cp
-_
from
schemes.
ciation
from Theorem 5
where C3 :5
JXJ p is a prime prime squared then CGis isomorphic
CG ,_ 0 (C3, Wewould
-Pp,2
or
(X, H) is maximal, and hence unique. (X, H). Then, by the above observation
that
(X, G)
C3
Aut(X, a)
case
It
(X, G)
have
Since
short.
for
observation Let
we
the
maximal,
is
a
=
C(X,a).
G
m
from Theorem 1: If
JXJ p2 is 0 (83, Zp').
or
3. Weconclude
=
observations If
consider
we
a* and r.,,, of C,
to either
consider
we
a), X) a), X)
minlEGX ni Second, as
a:
Aut(X, a) is isomorphic Pp,2 is as in Example 2.
that
to
2.
G be such that
E
a
F-+
subgroup
Aut(X, e) (-i,
-j),
R with
Aut(X, d)
:
R is
a
<
R
C2 P
-
X
CP2.
(S3
Since X
( 'r))
Sylow p-subgroup
unique
of
Aut(X, d).
Since
-Vp,2=
nAut(X,
h)
<
Aut(X, d),
hEH.
it
follows
Since
completes
(X, e)
that
d G H
-
G is
the proof.
=
Cay(Z2,P oe)
arbitrary,
this
.
Hence proves
(X, e) the
is uniquely determined. uniqueness of (X, G). This
0
The
5.4
non-symmetric (07(1, 0, 0))rp,3
E G is
the
orbit
commutative
a
Furthermore, [471 shows that relation a non-symmetric 3. valency 1, 2,
(X, a)
Let
be
as
any
of
primitive valency
-Pp,3,
-
4 has
(X, H)
a); X)
2-orb(Aut(X,
:=
of H'
elements
the
of
relation
non-trivial
no
scheme
association
commutative
We set
Theorem 6.
in
Aut(X, a)
Since
short.
(see [471).
result
scheme. If association primitive to is then the isomorphic graph (X, a) of valencY 4, where Fp,3 is as in Example 2 and o := (0, 0, 0). be
with
for
117
4
following
the
(X, G)
Let
Theorem 6. a
obtained
already
Wehave
valency
of
case
Schemes
Association
Commutative
of Primitive
The Enumeration
5
characterized
are
as
follows:
(i) (ii) (iii) (iv)
A
4; (o, (x, 0, 0)) rp, and Ka. lax I x c Zp" I where ax 6; bx I x E Z'P I where bx (0 (X7 X7 0))rp,3 and Kb. Jcxy I xy =A 0, x = 6 y} where cxy := (07 (X, y' 0))rp,3 and rcx. z)(z y)(y x) 0 01 where Jdxy,J xyz: 0, (x 24. := (0, (X, y' Z))rp,3 and rld.,., 3
S C D:=
dx,,
(X, G)
Let
Then
scheme.
we
C(X, a)
E
each d
For
Proof.
=
(i)
rf
=
is
C-
H
exists
in
[47].
G such that
E
(ii). (X, H)
ax
AUC
C
-
cvz
is
Jx,
since
H' by
dxy,
Since
If p C(X,
Now we prove
Note cy-x,z 0 cv,z-x axc,yz. from Lemma 1(i) that
since
f
d
=
U
e
=
dxyz
(i),
E
zJ
y,
there
are
we
=
E
3 then
D
a).
d,,,,
Let
0. Hence
=
E V
E c,,.,-,, cy-,,,,, and p > 3. It follows
axcvz,, are distinct
(5.6)
+ cy,z-x.
no
of H'
element
have H
=
H'.
which
Therefore,
contains
(X, H)
proof of (ii). unique. maximal, d E H Since (X, G) - (X, H) by (ii), we prove (iii). Finally, union of elements. of G. For each e G Gwith e C f we have pe. disjoint by (5.6). It follows from Lemma I (ii) that completes
implies
121ke,
of two elements
of
and hence Ke
valency
=
=
is
the
-
=pea.c,,
and
moreover
+ cy-x,z
arbitrary, This
and hence
This
and
f
12.
proved
dx,yz properly.
L(X, a);
in e,
is clear that we may assume p > 3. It have dxvz Then we (X,H'). (X,H)
Thus,
association
commutative
primitive
a
element
G there
-
be
following:
have the
AUBUCCG; (X, H) is a unique maximal ne
12;
=
Lemma8.
(i) (ii) (iii)
=
-
-
-
G is
a
I
Pc*" axer,,,,JKe-
12.
12 in G. This
Hence d E H
completes
-
the
G is
proof.
a
disjoint
union
0
Mitsugu
118
It
easily
is
checked
Proof of Theorem If p
3 then
=
CUD, dC-D andaE A,
For allcE
Lemma9.
Proof.
Hirasaka
(X, H)
since
3.
0 and G
D=
=
A U B U C by Lemma8(i), with (X, G) -- (X,
(X, G) E L(X, a) d.,.,,, E H G there
and
We may
=
< 1.
Z3). P
2-orb(l'p,3;
--
Assume p > 3. Let for each Lemma8 (iii), n,
havePadc
we
nf
=
12.
all,
we
shall
exist
-
assume
u
f
e,
(x,
:=
H). By d.,yz
G such that
E
z)
y,
desired.
as
without
E oe
=
eU f
loss
of
generality. of
First a
show that
uniquely
is
oe
determined.
Nowwe consider
triple
(w,
v)
a,
U oc)
E
A
x
(od
x
ul)
-
cEC
such that
i.e.,
u,
r(o, u)
v
E odnwa If there
exists
f since,
p'e-a.
-
=
2
otherwise =
pc-.
=
ea
Lemma9. contradicting The following is a set of
((0,
y,
z),
a,,,
(-x,
((x,
y,
0),
a-,,
(x
PC-ar.-Ir" e
such
y
x,
-
z, y
-
a triple (w, 2, and hence
such =
PCW.
=
Aacw
then
r
(o, u) = k
e,
7
triples: z
((x,y7x)7az-x,(z,y,x));
x));
-
-z));
z,
-
=
v)
a,
((x, 0, z),
a,,
((x,
ay,
z),
x,
(x
-
(y,
y, -y,
x,
z));
Z'
-Z),
z
-
y));
((x,y,y),a,-y,(x,z,y)).
Thus,
f(-X)
Y
X,
X), (X
z
(Y' Replacing
(x,
y,
z)
E oe
f(-Y'
(5.7)
(-Z'
X
and
(5.8)
Secondly,
Y,
z
-
Z' Y
prove we
shall
e =.
Since p W,
:=
>
3, there
(wi, 0, 0).
x,
we
x)
z,
Z), (z the
-
exists
z), (z, or
wi
E
z,
x,
y)
(z,
-
of
(X, e) r(z,
u
+
Z'
Z' Y
Y) I
or (x, x), (z,
z, x,
(5.7)
of.
C
y) y)
of
E
E oe.
applying the Again, applying
and
E oe. we obtain
Z), (Y
X
Y
uniqueness
=
x)
y,
(y,
Z'
X, -X,
show that
r(o, u)
X), (X,
Y,
obtain
Y), (Y
X
-
I
by (y,
argument for them, the same argument for (y, same
Y), (X
Y, -Y'
Z) (Z'
X,
X), (z
-
Y,
X, X
z
X,
Y,
-
-Y)l
-X), C
of
(5.8) -
oe.
is
isomorphic
z)
Zx such that
for r
each
(o,
u
to
z
+
E
wi)
Cay(Z3P) oe), i.e.,
ZP3.
(5.9) A U B U C where
that
We claim
r(x, y) y + wi)
if
there
+ wi
x
v,
(07
'=
r(o,
=
G
n
xaw,.
+ W1,
U
W2)
0)
written
(y
(X
r(j:
as
E
contradicting
I
U
+
u
+
7
8jWj)) 3
r(E
siwi,
SjWj)
i=1
z),
+
u
e.
W2) E e and (07 07 W3)- Since SiWi (8i (E ZP
+ 81W,
i=1
r(z,
E v
:=
1,2,3
r(o
SiWi)
WI)
exists
3
U+
siwi,
there
119
+ W2i Y +
2
2
+ W1, y +
+ wl)d*,
Ei=
as
Schemes
e,
-
and W3
Wj))
+
aw,
=
" W2, W3 E ZP such that
exists
r(o, U))
aw,.
-
(X
then
u,
=
e
.
(X + W3) Y + W3) E e where W2 is a basis of Z3 z E Z3 is jWi}i=1,2,3 P P7 above claim, the we have 1, 2,3). By e
x
-
E
xaw., Lemma9.
Similarly,
and y f Since
e
=
(x + wi, n (y + wi) e*. This implies
Otherwise
Association
Commutative
of Primitive
The Enumeration
5
desired.
Finally,
we
shall
0. Assume the
for
each
0.
Gn D
show that
that,
Weclaim
Zpx
G
w
and each g E G
h 1. It follows Pa,,,g Lemma9. contradicting =
Applying
above claim
the
(u,
jr(o,
v,
w))l 11u,
jr ((), (U, Assume the some
I ju,
If
jv, wl claim
=
f
E
G
wl
n
Ix,
e, v,
H,
-
that
Pga,*, h
we
obtain
zjj
y,
we
2}
',: :f
=
n E)
pha,,,gKhlr-g
=
=
0.
2,
(5-10)
show that
Weshall
for
for
have awg n E) Lemma 9 we have
H, by
-
Then,
h E awe, n D. and 8(iii) from Lemma 1(ii)
i.e.,
contrary,
w)
v,
wj
v,
Zp'.
E
n
jx,
0 without
y,
zj I
w))
y,
V,
G
-
E E
Z3 1, P
H since
x
=
=
11u,
u
(5-11)
0. v,
assume
follows
It
n -D
G and
may
we
generality.
E
W)
duvw
1, then
=
of
loss
r(o, (x,
that
by (5.10)
i.e.,
contrary,
(u,
W)) I (U,
V,
x
wj =
from
(5.10)
0
Again,
w.
n
u
jx,
zJ1
y,
and
ly, zj
and the
n
above
by the above
claim, du,w a
=
dxvw
E
r(o, (x,
w))a,-,y
y,
n D
=
0,
contradiction.
I Ju,
if
G
-
v,
wj n Ix, y, zj I r(o, (u, v, z))
H. Since
r(o, (u,
v,
z))
C-
G
-
H. It
du,w
=
G
0, then
a,-yr(o,
follows E
the
by
(u,
y,
from the
aw-,r(o,
(u,
v,
above
z)) by
claim
the above
n D=
claim,
y,
z))
we
E
have
that
above claim
z))
r(o, (u,
0,
second and final 0. Combining the first, commutative is a unique primitive we (X, G) steps, together scheme in L(X, a) which. is not (X, H). This completes the proof. association a
contradiction.
Therefore,
conclude
G n E)
=
that
0
Index
a-chain,
fusion
110
P-polynomial, B-graded set 2-orbits
of
fusion
106
of
fibres,
(G; X),
schemes, 8 8 subalgebra,
10
table algebra, generalized 6 group-like, 1 GT-algebra, V GT-algebras, GT-homomorphism, 4
5
1
-
adjacency matrix Ag, 107 16 algebra A,,,(sgn), 14, algebras F.,,, F' Arad, Z., V Arisha, H., V association scheme, 4 association schemes, 4, 106 M,
L., 105 basic relations,
Herzog, M., V Hirasaka, M., VII G., V Hoheisel, V homogeneous coherent algebras, homogeneous coherent configuration, 2 homogeneous GT-algebra, table algebra, homogeneous integral V homogeneous table algebras,
Babai, basic
5
6
sets,
graph,
Biggs-Smith Blau, H.I.,
106
integral integral
108
algebra, algebras,
Bose-Mesner Bose-M6sner
5,
107
VI
F., V
Biinger,
Cayley graph of E algebras, D., V Chillag,
over
centralizer
K,
109
4
107 commutative, Coxeter graph, 106
cyclotomic
scheme, 106
association
coset,
Kawada, Y.,
V
C-cosets, line, 112 3 linear,
3
left
Darafsheh, M.R., V degrees of (A, B), 2 graphs, distance-regular basis, I distinguished double
2 GT-algebra, algebra, 2 intersection numbers, 5, 106, 107 107 isomorphic, ITA, 2, V V Iwahori-Heeke algebras,
table
3, 6
VI
Miloslavsky, Muzychuk, non-singular, order
faithful, Fisman, fission, fission
4
2
E., V 9
Frobenius
3
VI table algberas, P-polynomial Peterson graph, 106 2, 107 primitive, association scheme, 105 primitive transitive primitive permutation groups,
schemes,
1
(A; B),
of
'
106 enumeration, Erez, J., V exactly isomorphic,
V., V M., 106, V
105
8-
algebra,
I
Quirin,
W.L.,
105
4 2
Index
126
algebra
quotient
of
(A, B),
3
13 strong, strong element, 13 symmetric, 2, 107
Rahnamai, A., V real, 1, 2 105 regular digraphs, 3 rescaled, 4 rescaling, right coset, 3
table table table tensor
subset
generated algebra, I subset, 2, 3 9 product,
the width S-r i ng,
6
trivial
6 Schur ring, Schur-H 'adamard,
C., V Scopolla, simple quantity, Sims, C.C., 105 SITA, , VI 2 standard, standard integral
Druck:
-
Verarbeitung:
2
valency, valency
of b E
fusions,
B, I
9
105 of g,
106
6
Wong. W.J., 105 wreath product, table
Strauss
Schdffer,
algebras,
Offsetdruck, Grtinstadt
VI
M8rlenbach
Xu, B.,
V
10
by b,
2
