AUTOMORPHISMS 0F INTEGRAL
GROUP RINGS

Thesis for the Degree of Ph. D.
MICHIGAN STATE UNIVERSITY
CHARLES FRANKLIN BRGWN
1971

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ABSTRACT
AUTmORHIISMs OF INTEGRAL GROUP RINGS
By

Char les Franklin Brown

For the most part, our attention is focused on a slightly
restricted subgroup of the group a of all ring automorphisms of
an integral group ring 2(6) of a finite group C over the rational
integers. This subgroup is denoted by M and consists of those
elements of d which are "normalized" in a natural way. Information
about 727 is easily converted to information about a. When the
comImtator subgroup G' of G is abelian, it is shown that Aut(G),
the group of all automorphisms of G, has a normal complement M in
720. Using this decomposition of 72a, we give, in Chapter 1, several
conditions which guarantee that all elements f of M can be
written, for all g 6 G, as f(g) - uq(g)u-1 for some a E Aut(G)
and some unit u of Q(G), the group ring of C over the field of

rational nuubers. Such an f is said to have an elementary repre-

 

sentation. Among the sufficient conditions for all elements of 71a

to have an elementary representation are: (1) G' has 1,2, or 3
elements, (2) G has at most one non-linear irreducible representa-
tion over the field of complex numbers, (3) G is nilpotent with
nilpotence class S 2, (4) G has a cyclic normal subgroup of prime

index. A key lama for the fourth result is that if f 6 71d and

Charles Franklin Brown

3,31 6 G are such that f(Cg) = Eél then the order of g and the
order of g1 are equal; 5;, 6'1 denote the class sums correSponding
to g and g1, reapectively, and this result is valid even if G'
is not abelian.

Chapter 2 contains several preliminary results about 73d. If
G' is abelian and, M is as described above, it is of interest to
know when every element of M fixes all class sums corresponding to
elements of G. This question has an affirmative answer if G
satisfies either of (l), (2) or (3) above. Although the question is
not answered, in general, it is shown that if M is non-trivial
then M contains non-trivial elements which fix all class sums
corresponding to elements of G. Without the assumption that G'
is abelian, we discuss when M - Aut (G). Some properties which the
elements of Aut(G) and m share are also described. f(A(K)) is
analyzed if f E‘na' has an elementary representation and K 9 G;
here A(K) denotes the kernel of the canonical ring homomorphism.of
Z(G) onto Z(G/K). Finally, the center of m is examined and shown
to be trivial if the center of G is trivial.

Examples are given in Chapter 3 to show that all elements of

na' can have elementary representations even when G' is non-abelian

and simple.

AUTOMORPHISMS OF INTEGRAL.GROUP RINGS

BY

Charles Franklin Brown

A THESIS
Submitted to
Michigan State University

in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1971

ACKNOWLEDGMENTS

This thesis has been realized under the very patient and
helpful guidance of Professor Joseph Adney. I gratefully acknowledge
his thoughtful suggestions, his pertinent questions and his encourage-

ment in difficult times.

ii

TABIE OF CONTENTS

Page
List of Tables 1V
Notation and Terminology v
Introduction 1
Chapter
1 ELEMENTARY REIRESENTATIONS FOR ELEMENTS OF 724 4
Section 1. Results from the Literature
Section 2. A Normal Complement for Aut (G) in 72a
Section 3. Some Sufficient Conditions for
Elementary Representations 3
Section 4. Order of Elements is Preserved by
Normalized Automorphisms; An
Application 12
Section 5. Groups with a Cyclic Normal Subgroup
of Prime Index 14
2 RELIMINARY RESULTS ABOUT 72a 18
Section 1. Properties of M n I 18
Section 2. When is m = Aut(G)? 19
Section.3. What is f(A(K))? 20
Section 4. How is m like Aut(G)? 22
Section 5. The Center of m 23
3 A METHOD FOR STUDYING M AND SOME EXAMPLES 25
Section 1. Introduction 25
Section 2. The Method 25
Section 3. The Examples 28
List of References 33
Appendix 34

iii

LIST OF TABIES

Table Page
1 Character Table for $6 30
2 Partial Character Table for S 32

7

iv

NOTATION AND TERMINOIDGY

G = a finite group

K.$ G = K is a subgroup of G

K s G = K is a normal subgroup of G

K' = the derived subgroup of K

Z(G) = the center of the group G

R(G) = the group ring of C over the ring R

Z(R(G)) = the center of R(G)

Z,Q,C = rational integers, rational field, complex field, respectively

L = the augmentation map from Z(G) onto Z given by L(2 agg) - 2 a
8 8

A(S) = { z (s-l)t(s)‘t(s) e 2(a) is arbitrary}, where s is a finite
868

subset of Z(G)

A = z - submodule of Z(G) Spanned by all differences ab-ba with
a,b e Z(G)

AP = A +-pZ(G), where p is a prime integer

Aut(G) B group of all automorphisms of the group C

a . the group of all ring automorphisms of Z(G)

72a - {f e d‘L(f(g)) . 1 for all g e c} = the set of all normalized

 

automorphisms of Z(G)
I = set of elements of a which fix 2(2 (6)) elementwise
Tu 8 endomorphism of Z(G) given by x 4 mm"1 for all x E Z(G);

u is a unit in Q(G) such that uxu.1 E Z(G) for all x E Z(G)

 

 

 

H B group basis for Z(G). That is, H is a subset of Z(G) such

that

(1) H is a multiplicative subgroup of the group of units of
Z(G)

(2) H is a free Z-basis for Z(G)

(3) L(h) = l for all h 6 H.

Let A be a finite group and a E A.

‘A‘, ‘a‘ = the order of the group A and the element a resPectively

Ca = the set of all conjugates of a in A

‘Ca‘ = the number of elements in Ca

C; = the element of Z(A) which is the sum of all elements

in Ca
Suppose x - 2 agg is an element of Z(G) and g1 E G. The phrase

8
"x sums to n on C " means that Z a - n.

1 86C g
31
The notation 2.3.1 means the 1-8-E reSult in the 3EE-Section of Chapter
2. The results are numbered consecutively in each section without

regard to the words lemma, proposition, and theorem.

vi

INTRODUCTION

The study of integral group rings Z(G)1 has received con-
siderable research attention for many years with much interest being
focused on the extent to which Z(G) determines G; this has been
called the group ring problem. In this connection, we mention a
recent paper of E.C. Dade, in [4], in which he exhibited 2 non-
isomorphic groups 61’ 62 such that F(Gl) 2:F(G2) for all fields
F. This is in contrast to a result of Whitcomb in [14] which shows
that Z(Gl) 1:2(62). One method of studying an object is to see how
it can be moved around; in particular, we are interested in the group
of all ring automorphisms of Z(G). Although the group ring prdblem
is not discussed in this thesis, it is possible that some of the
results will aid in solving a part of it; see Obayashi [8].

For the most part, our work is concerned with a slightly
specialized subgroup of 4, namely the subgroup 72d of all normalized
automorphisms of Z(G). First, we show that, when G' is abelian,
Aut(G) has a normal complement M = {f E nuqf(g) - g E A(G')A(G) for
all g E G} in flat In Chapter 1, several conditions are given which
guarantee that each element f of na’ is such that, for all g E G,
f(g) = uov(g)u.1 for some a 6 Aut(G) and some unit u in. Q(G);
both a and u depend on f. If f can be described in this way,

we say it has an elementary representation. In all of the results

 

 

1
See the previous pages for notation and terminology.
l

concerning elementary representations in Chapter 1, G' is abelian

and an elementary representation is obtained for all elements of 7K7
by showing that all elements of M have an elementary representation.
If G is nilpotent with nilpotence class s 2 or if G has at most
one non-linear irreducible representation over C, we show that M C I.
This implies that the elements of fld' have elementary representations.
When G has nilpotence class s 2, this is a result of Sehgal in [12].
If G = D4 is the dihedral group of order 8, then G is nilpotent of

class 2 so that every f E m is of the form f = Q o a where

l
o E Aut(D4) and Q1 6 I satisfies §1(g) - g E ACDl")A(D4) for all
g 6 D4. This is a recent result of Obayashi in [8]. Perhaps the
most satisfying result we have obtained is that if G has a cyclic
normal subgroup of prime index, then all elements of m have
elementary representations. A key lemma for this result is that if

f E 714 and f(C-Ig) ' 6g for g,g1 E G then |g‘ = ‘gfl; here 6'

is not necessarily abelian. It is also demonstrated that all elements
of m have elementary representations if G' has 1, 2 or 3 elements.
Chapter 2 contains preliminary results about a and 710.
If G' is abelian and M is as above, it is of interest when M s: I.
This question is not answered, in general, but it is shown that if M
is not trivial then M 0 I is not trivial. Then the assumption that
G' is abelian is drapped and we discuss when m = Aut (G). No
prOperties of Aut (G) enjoyed by m, in addition to those of Chapter
1, are also disclosed. f(A(l()) is analyzed if f E m has an

elementary representation and K 9 G. Finally, the center of 724 is

studied and shown to be trivial if Z(G) is trivial.

In Chapter 3, we describe a method for studying the question
of an elementary representation for elements of flfi' for arbitrary C.
Using this method, several examples are given to show that all elements
of m can have elementary representations even when G' is non-
abelian and simple. At present, we know of no group G for which
some element of m fails to have an elementary representation and

it is hoped that more research can be done on this topic.

CHAPTER 1

EIEMENTARY REPRESENTATIONS FOR NORMALIZED AUTOMORPHISMS

Section 1. Results from the Literature. We record here some results
which will be needed from the literature. The first 3 propositions
are listed in [9] and the proofs are included in the appendix.
Proposition 1.1.1 1 E A iff A sums to 0 on all conjugacy classes
of G.

Proposition 1.1.2 A 6 AP iff A sums to an integral multiple of
p on all conjugacy classes of G.

Proposition 1.1.3 If x,y E Z(G) are congruent modulo AP, then

xp and yp

are congruent modulo AP.

The next 4 results are taken from [14]. We emphasize that
H always denotes a group basis for Z(G).
Theorem 1.1.4 (Glauberman) If h E H, then 6% = a; for some
3 6 G. (E£ denotes the element of Z(G) obtained by adding all
conjugates of h in H.) This theorem shows that the center of H
and the center of G coincide. Also, it enables one to set up an
isomorphism Q from the lattice of normal subgroups of H onto the
lattice of normal subgroups of G. If M Q'H then Q(M) = K, where
K - {g 6 G‘C8 3 Ch for some h 6 M].

Theorem 1.1.5 If Q(M) = K then ‘M‘ = ‘K‘, Q(M') = K', and

 

A(M) = A(K)-

Theorem 1.1.6 If B is an abelian normal subgroup of H and
Q(B) = A, then there is an isomorphism. e of A onto B such that
if a E A and 9(a) = b then a and b are congruent modulo
A(A)A(G) .
Theorem 1.1.7 If G' is abelian then each h E H is congruent
to a unique g 6 G modulo A(G')A(G) and the mapping h a g is an
isomorphism of H onto G.

The next theorem is proved exactly like Theorem 1 of [12].
Theorem 1.1.8 Suppose f1,f2 E a' are such that f1(Cé) = f2(Cé)
for all g E G. Then there is a unit u in Q(G) such that

f2(g) I u f1(g)u-1 for all g E G.

Section 2. ANormal Complement for Aut(G) 111 m. In this section
we record the fact that M is a subgroup of finite index in a

and exhibit, when G' is abelian, a normal complement for Aut(G)

in nah Here we are identifying a E Aut(G) with the element of

NG' obtained by extending a linearly to all of Z(G).

Let f E a. For each g E C, set f(g) = {,(f(g))-f(g). One
easily checks that f extends linearly to an element of ¢7. Since
the only units of Z are i 1, it is clear that f E m. Also,

f(G) is a group basis for Z(G); this is true for any element of

flak If g E G, we have, by Theorem 1.1.4, f(Eé) I Eé(8) = Cgl for

some 31 E G. But if g,x2gx;1,...,xngx;1 is the set of all con-
. - -1 -1
jugates of g in G then f(cg) I f(g -l--x2gx2 +...+xngxn )

-1 -l -l -l
= “Hen-f(g) + L(f(x28x2 ))-f(ngxz ) +...+ L(f(xn8Xn ))f(xmsxn )
= L(f(g))f(Eg). Thus, if f e a and g e c then f(Eg) =- 1 cg

l
for some 31 E C. By Theorem 1.1.8, we see that the subgroup I

of all elements of d
a normal subgroup of a of finite
Lemma 1.2.1 7K7

It can be verified that an

is a subgroup of
m:
each g E G, f(Cé) I C? for some
g. I C man [6'3 I]

[a : mflm : I], m is of finite

1

Thus Since

which fix the center of Z(G)

pointwise is
index.
a of finite index.

element iff for

f E d’ is in mar
81 E G depending, of course, on
is finite and equal to

index in a.

We will show that Aut(G) has a normal complement in ‘ng

by exhibiting a surjective group homomorphism B: m -+ Aut(G)

and

noting that the exact sequence 1 —. ker B 131 'm E Aut(G) _. l is

split by the natural injection of Aut(G)

Let f e 720 so that f(G)

GI
if of f(G) onto G by setting,
if f(g)

the element of Aut(G) given by
each f E 71?.
need part of the next lemma.
Lemma 1.2.2 Suppose that G'
acteristic subgroup of G
f(A(K)) I A(K) for every f e 714.

Proof: We note that

f(A(1()) I A(K) for every f E 720,
Since A(K) is an ideal of Z(G),
for each R E K.

Suppose that K s G'

In order to show that B

is abelian and that

such that

f(A(K)) : MK)

and that

into m.

is a group basis for Z(G). If

is abelian, we can use Theorem 1.1.7 to exhibit an isomorphism

for each 3 E G, 1f(f(g)) ' 81

is congruent to g1 modulo A(G')A(G). Denote by of

o f|G and set 8(f) I of for

f

is a group homomorphism, we

K is a char-

K.S G' or G' s K. Then

for every f 6 72d implies that
since 724 is a subgroup of d.
f(M10) 5 £800

iff f(k-l) E A(K)

Q(fCK)) I A. Since Q pre-

serves intersections and Q(f(G')) = Q(f(G)') = G', we have

A = Q(f(K)) I Q(f(K) n f(G'))'I Q(f(K)) n Q(f(G')) = A-FIG'o Hence

A s 6'. Now let a e A. By Theorem 1.1.6, since f(x) is abelian,

III
93

we know that f(k) a a mod A(A)A(G) for some k E K. Thus f(k)
mod A(G')A(G) since A(A) c A(G'). This shows that a I of(k) E K,
since K is characteristic in G. Whence A.: K and, since Q
preserves order, A I K. Thus, if k E K, there is a k1 E K such that
f(k) - 1:1 E A(K)A(G) G (500- f(k) - k1 '3 f(k-l) - (kl-1) 6 A00
implies that f(k-l) 6 A(K) since (kl-l) 6 A(K). This completes
one part of the proof.

Suppose c' s K. If k e K then f(k) - of(k) e A(G')A(G)
c A(G') C A(K). Since K is characteristic in G, of(k) E K and
f(k-l) E A(K) as before. This completes the proof.
Lemma 1.2.3 B is a surjective homomorphism.

Proof: Let T E Aut(G). Then for each g 6 G, 1(g) ’ T(g) mod A(G')A(G)

and 3(T)(g) I 7(g). This shows that B is surjective. To show
that B is a homomorphism, let f1,f2 em and g E G. Then

f1 0 f2(s) = f1(f2(3)) - f1(of2(g) + v) for some v e A(G')A(G)-
Whence f, o f2(s) = £1(of2(s)) + film 2- of1<af2(g>> mod A(G')A(G).
by Lemma 1.2.2. Thus B(f1 o f2)(g) = B(f1) o 8(f2)(g) and B is

a homomorphism.

Theorem 1.2.4 If G' is abelian, then Aut(G) has a normal com-
plement M in m. In fact, M = {f E m‘flg) '="-. g mod A(G')A(G) for
each 3 E G}.

‘ggggfig We need only note that the natural injection of Aut(G) into
m splits the sequence 1 - ker B 131' mg Aut (G) -+ l and that

M = ker B.

Section 3. Some Sufficient Conditions for Elementary Representations.
The following definition is motivated by Theorem 2 of [12].
Definition. We say that f 6 a' has an elementary representation
(6.Eh) if f I Tu o a for some a E Aut(G) and some unit u in
Q(G). (Recall that Tu(x) I uxu.1 for all x E Z(G).)

The remainder of this chapter is devoted to giving some suf-
ficient conditions for all elements of M to have an elementary
representation. We remark that information about m quickly con-
verts to information about 0% see Section 2 for the construction of
an fem froman fea.

All of the sufficient conditions for the elements of ‘na' to
have an 6.&h include the assumption that G' is abelian. However,
this is not a necessary condition; in Chapter 3 we give examples to
show that all of the elements of 71a may have an 6.8. when G' is
non-abelian and simple. When G' is abelian, the decomposition of
m in Theorem 1.2.4 allows the statement that all elements of m
have an d.fih iff all elements of ker B have an elementary repre-
sentation.

Our method for showing that f 6 na' has an 6.Eh is to exhibit
a o E Aut(G) which does the same thing to class sums as f; that is,
f(Eé) I 0(Eé) for all g E C; an application of Theorem 1.1.8 says
that f I Tu o a for some unit u in. Q(G). In case G is abelian,
“G E Aut(G) for each f E 726]; this follows from G. Higman's result,
in [5], that the only units u of finite order in Z(G) with L(u) I l
are the elements of G. Later we will see that flu'I Aut(G) implies

that all subgroups of G are normal.

The next 2 lemmas will be needed in subsequent proofs; G' is
not necessarily abelian.
Lemma 1.3.1 If h E H, g E G are such that 6% I C; then b sums
to l on C8 and to 0 on all other conjugacy classes of G. (Since
for each h E H there is always a g 6 G with Eg I 6;, by Theorem
1.1.4, this shows that each element of a group basis sums to 1 on
some conjugacy class of G and to 0 on all other conjugacy classes
of G.)
‘Egggg: The result will follow from PrOposition 1.1.1 if we show that
h'8 E A. Applying L to [C

h
If u is a unit in Z(G) and x 6 Z(G) then u(xu-l) - x E A.

I 6?, we see that ‘Ch‘ I ‘Cg‘ I n, say.

Combining this comment with Eh I 6;, we see that n(h-g) 6 A. By
Proposition 1.1.1, h-g E A.

Lemma 1.3.2 If h 6 H, then h a g mod A(G')A(G) for some g1 E G.

1
In fact, if Eh I E; with g E G, then g1 may be chosen from gG'.
.ggggf; As in the proof of Lemma 1.3.1, E£ I 6% implies h-g 6 A.
Clearly A : A(G') so that h-g I z '(a-l)t(a) for some elements
t(a) E Z(G). Computing as Whitcombagid in proving Theorem 1.1.7, we

have that h-g s z (a-1)L(t(a)) a 2 (a‘(t(a)) - 1) a n a‘<t(a)) — 1

aEc' aec' aec'
mod A(G')A(G). Thus h E ( H 3L(t(a))-g) mod A(G')A(G). Whence
aec'
one value of g1 is H aL(t(a))°g E gG'.
366

The next lemma yields another proof for Theorem 2 of [12].
The manner in which it is used to prove this theorem (here, Theorem
1.1.4) makes the following question of interest. For what groups C

does the conclusion of Lemma 1.3.3 hold?

10

15mma.1.3.3 Suppose G is nilpotent of nilpotence class 5 2. If

h E H, g 6 G are such that h-g E A(G')A(G) then h sums to l on

C8 and to O on all other conjugacy classes of G.

Iggggf: The proof is by induction on ‘G‘. If G is abelian, then

G' I l, h I g and the result is trivial. Since G has class 5 2,

the set K I {[g,x] I g-1x-1gx\x E G} is a normal subgroup of G.

Let n: Z(G) I’Z(G/K) be the ring epimorphism induced by the canonical
homomorphism of G onto G/K. Since R(H) is a group basis for
Z(G/K), by Lemma 3.1 of [2], and um - gK e A((G/K)')A(G/K), we have,
by induction, that R(h) sums to l on the conjugacy class of 3K

in G/K and to 0 on all other conjugacy classes of G/K. By Lemma
1.3.1, we know that h sums to l on some class, say C8 , of G

and to 0 on all others. Therefore, glK is a conjugate if gK so

that glK I xgx-IK for some x E G. That is, for some k I [g,x2] E K,

1

'1 - -1 -1 -1 -1
81 ' “3" k " x8" 8 *2 832= SESJ 1&3:le =s[s.x x2] = a conjugate

of g. Thus G8 I Cél and the proof is complete.

Theorem 1.3.4 (Sehgal [12]) If G is nilpotent of nilpotence class

S 2, then each f 6 M has an elementary representation.

'Ezggfi: Since G' s Z(G), we have the decomposition of Theorem 1.2.4, so
itsuffices to show that each f 6 ker B has an elementary representa-

tion. In fact, it will be shown that f(Cé) I C; for all g E G;

the result then follows by applying Theorem 1.1.8 with f I f and

1
f2 I the identity element of Aut(G). Suppose f(Cé) = C; for
1
some 6 G. f C I C I C so as in the roof of Lemma
31 ( s) f(g) 81 ’ "

1.3.1, f(g) . g1 E A. By Proposition 1.1.1, f(g) sums to l on C8
1
and to 0 on all other conjugacy classes of G. But f E ker 3

implies that f(g) - g e A(G')A(G) and so, by Lemma 1.3.3. f(g)

11

sums to 1 on Cg and to 0 on all other conjugacy classes of G.
Thus g1 is a conjugate of g and f(Cg) I Cg, as was claimed.

If G' is abelian and the conclusion of Lemma 1.3.3 is true
for G, then one can obtain an 6.8. for the elements of m by
repeating the proof of Theorem 1.3.4. The next lemma gives another
collection of groups for which the conclusion of Lemma 1.3.3 is valid.
Lemma 1.3.5 Suppose G has at most one non-linear irreducible re-
presentation over C. If h E H, g E G are such that h a g mod
A(G')A(G) then h sums to l on C8 and to 0 on all other conjugacy
classes of G.

.25222; By remarks in [13], G' is abelian. Suppose, by Theorem 1.1.4,
that Eh ' Eél for some 81 E C. By Lemma 1.3.1, we will be done

if g1 is a conjugate of g. Let p be any linear representation

of C over C and 1° its character. We can extend p and 19

to a ring homomorphism of Z(G) into C. Ch I Eél implies, as in
the proof of Lemma 1.3.1, that h-g1 E A C A(G'). Recalling the form
for elements of A(G'), we see that p(h) I p(g1) since G' is in
the kernel of p. By hypothesis, h-g 6 A(G')A(G) : A(G') so that
p(h) I p(g). Thus Ip(g) I 19(g1) for any linear representation

p of G. If g1 I 1 then h I l and so g I l, by Theorem 1.1.7.
If g I 1 then h I 1, by the same theorem, so that g1 I 1. Thus
we may assume that neither g nor g1 is 1. By an orthogonality
relation for complex characters, 0 I 2 Ip(g)'lp(l) I 2 1p(g1)1p(1)
where the sum is over all irreducibleprepresentationspof G. Since
Ip(8) ‘ 19(81) if p 18 linear and G has at most 1 non-linear

representation, we see that 1p(g) I xP(g1) for all irreducible

representations 9 of G. This certainly implies that g,g1 are

12

conjugate.

Corollary 1.3.6 If G satisfies the hypothesis of Lemma 1.3.5,
then each f E fid' has an elementary representation.

uggggf: See the remarks immediately preceding lemma 1.3.5.

We note that the alternating group on 4 symbols satisfies the

hypothesis of Lemma 1.3.5.

Section 4. Order of Elements is Preserved by_Normalized Automorphisms;

Ag_Application. The main result of this section is Corollary 1.4.3;
this corollary lends some motivation to the search for an 6.E§ for
elements of flan As a first application of this corollary, we prove
Theorem 1.4.5. Unless explicitly stated we do not assume G' abelian
in this section. The following lemma is proved much like Proposition 2
of [9].

Lemma 1.4.1 If ‘6 =46 for some h E H, g 6 G then C = E
h 3 hp 89

for any prime p.

Proof; As in the proof of Lemma 1.3.1, h-g E A C Ap. By Proposition

1.1.3, hp - gp E A . Theorem 1.1.4 says that C = E. for some
P hp 81

316 G and so hp - g1 6 AP, as was the case with h and g. Thus

gp - 31 E Ap and, by Proposition 1.1.2, g1 and gp are conjugate.
Whence 'C ='E =‘E .

hP 81 SP
Corollary 1.4.2 Under the hypothesis of Lemma 1.4.1, 6’“ I E
for any positive integer n. h 8

Proof: The proof is by induction on n; if n I l the result is
0'1 012 0’t
true by hypothesis. Let n I p1 p2 ... pt be the decomposition

of n into prime powers. By induction, we have that

13

I C and by Lemma 1.4.1 that

o oz ot-l o1 o2 ot-l
hpl p2 °"Pt p1 P2 “°Pt
8
‘E =‘E .
hn 8n

Corollary 1.4.3, Suppose f E na' and that f(Eé) I Cg for
l
$.31 e G. Then \s\ I ‘g1\ and ICg| = lcgl\.

Proof: ‘CS‘ I ‘Cg ] can be seen by applying L to the equality
1 ._
f CI I Ch . Since f G is a rou basis for Z G and C’ I C
( g) 31 ( ) g p ( ) f(g) g1.
the result ‘g‘ I ‘g1\ follows from Corollary 1.4.2 by noting that

lf<s>l = \sl-
From Corollary 1.4.3 we note that if f E a and f(Cg) I j; 681
for some g,g1 E G, then |g| I ‘gl‘. For if f is the element of
m derived from f as in Section 2, we have f(Eg) I L(f(g))f(C-g) I 631
Lemma 1.4.4 Suppose that G' is abelian and that f E ker B I M.
If f(Cé) I Cél for some g,g1 E G, then g1 E gG'. (Hence, an
f E ker B permutes the class sums corresponding to elements inside
the various cosets of G' in G.)
21222; By lemma 1.3.2, f(g) - g* 6 A(G')A(G) for some g* E g1 '.
The uniqueness statement in Theorem 1.1.7 and the fact that
f(g) ' 8 E A(G')A(G) implies that g* I g. Hence g1 € 86'.

As an application of the results in this section, we have the

following theorem.

16' = G',

gZG',...,ng' are the distinct cosets of G' in G. Further assume

Theorem 1.4.5 Suppose that G' is abelian and that g

that the following holds for each i I 1,2,...,m: If g,g* E giG'
and neither of g,g* is in, Z(G), then either ‘Cg‘ I ‘Cg ‘ or
*

‘8‘ I ‘g*‘ or g,g* are conjugate. Then each f E ker B I M fixes

14

each class sum corresponding to elements in G and hence all elements
of 714 have an elementary representation.

‘Egggf: Let f E ker B and g E G. If g E Z(G) then f(Cé) I 6?
since f(g) E Z(G) and f(g) - g E A(G')A(G) and we have the unique-
ness statement in Theorem 1.1.7. If g E Z(G), let f(Cé) I Cg* for
some g* E G. By Lemma 1.4.4, g* E gG'. By our hypothesis, g and

g* must be conjugate. Hence f(Cé) I Cé* I 6;.

Corollary 1.4.6 If ‘G" I 1,2, or 3 then each f E na' has an
elementary representation.

Proof: One can easily check that the hypothesis of Theorem 1.4.5

is satisfied.

Section 5. Groups with §_Cyclic Normal Subgroup g£_Prime Index.
In this section we exhibit a fairly large class of groups such that
every element of m has an elementary representation.
Theorem 1.5.1 Suppose G has a cyclic normal subgroup A of index
p, for some prime p. Then each f E flfl’ has an elementary repre-
sentation.
.2522E3 We first make some observations about how the conjugacy classes
of G are distributed among the cosets of G' in G and then produce
a o E Aut(G) which, when extended linearly to Z(G), does the same
thing to class sums of elements of G as an element f E ker 3;
clearly G' is cyclic.

Let A be generated by a and set [A] I |a] I m. A is a
maximal subgroup of G, so if b E A then G I A~<b> I <a>~<b>u
Since A 9 G, we have b-lab I ar for some positive integer r such

that (r,m) I greatest common divisor of r and m I 1. By 47.10

15

of [3], G' I <ar-1>. Since [G : A] I p, bp E A and bk E A for

any integer k such that 1 s k < p. Set d I (r-1,m). If %(G) I G,

then we are done, by Theorem 1.3.4. Thus we may assume that Z(G) I G.
m/d

Claim: %(G) I <a >. For we must have Z(G) s A since A is a

maximal subgroup of the non-abelian group G. ak E %(G) iff

b-lakb I ak. But b-1akb I akr so that ak E Z(G) iff ak I akr

iff ak(r'1) = iff m divides k(rnl) iff m/d divides k.

Thus Z(G) I (am/d> and hence has order d.
Next we claim that the following pd cosets of G' in G

2 d- 1C

are disjoint. The cosets are G',aG',a G',...,a d‘l

,bG', abG' ,...,a bG',
bzc',abzc',...,ad'lbzc',...,bp"lc',abp 1 H'.--- d Wlbp 1 . We verify

that the first d of these cosets are disjoint and leave the re-

mainder to systematic observation; recall that bk 4 A if k is an

integer such that 1 S k < p. Suppose alc' I 836' with

o s i,j s d-l and i > j. Thus a1“J = a‘Kr'l)

and (i-j) a
n(r-l) mod m for some integer n. Hence (r-l,m) I d divides
(i-j) and so i I j. Since [G : G'] I g I ;%§ I pd, these pd
cosets are the distinct cosets of G' in G.
Finally we claim that aibjG' is the conjugacy class of G
containing aibj for any i,j such that O s i S (d-l) and
1 s j s (p-l). That is, all except the first d cosets of 6'
listed above are complete conjugacy classes in G. The argument is
as follows. If g E A then g E-%(G) or g has exactly p distinct

conjugates in G, since the centralizer of g in G is A or G.

We are working under the assumption that %(G) C A, so that there

are \Z(c)| +l pJZ(G)l

Id +-—;§ conjugacy classes of G in-

side A. By Corollary 47.15 of [3], there are pd +_E%Q irreducible

16

representations of G and hence this many conjugacy classes of G.

If x is the number of conjugacy classes of G outside A, we must

have x + d +% I pd +121? , so that x I d(p-l). Each coset of

6' must contain at least one conjugacy class of G and there are

d(p-l) cosets of G' outside of A. Hence albjG' I C i j as
8 b

claimed.
If f E ker a, we produce a o E Aut(G) such that f(Cg) I

0(Eé) for all g E C. By Lemma 1.4.4, f(Cg) I 5'3 for some
a
aS 6 ac'. Set 0(a) . as and 0(b) - b and extend this to a11

of c by defining 0(albj) = aisbj. By Corollary 1.4.3,

‘a‘ = ‘as‘ so that o E Aut(G). By Corollary 1.4.2, f(E i) =
a
C I C I C . I 0(C i) for any positive integer i. Hence

f(a)i ais 0(a1) a

0(Cé) a f(Eé) if g e A. By lemma 1.4.4, f(E

.19) ‘ Eaibj

integers i,j such that O s i s (d-l), 1 s j s (p-l) since

for any

C ibj is the only conjugacy class of G inside aibjG'. The proof
8 o 0

will be complete if d(aibj) E aleG' for these values of i,j.

i+k (r-l)

We have aibjG' I G'aibj I {a 1 bj‘k1 is a non-negative integer]

and s

[l + k(r-l)] mod m for some non-negative integer k -since
as E aG'. Therefore 0(aibj) I aisbj I a1<1+k(r-l))bj E aibjG', as
was to be shown.

We conclude this Chapter with some remarks about the connection
between group bases and elementary representations of elements of flat
Suppose that all elements of flU' have elementary representations
and that H 2:6 for all group bases H of Z(G). Let H be a
fixed group basis for Z(G) and suppose a is an isomorphism of

G onto H. a can be extended linearly to an element of flu? which

17

we also denote by a. Let a I Tu o o be an elementary representa-
tion for do Then H I Q(G) I Tu 0 0(6) I Tu(G) I uGu-1 I G“. That

is, H can be obtained from G by conjugation by some unit in Q(G).

CHAPTER 2

PRELIMINARY RESULTS ABOUT 724

Section 1. Properties 9£_ M_£Ll, An interesting problem, directly
related to obtaining an elementary representation for the elements
of flU' when G' is abelian, is to find when ker B c I. We have
seen that ker B c I for the groups in Theorem 1.3.4, Lemma 1.3.5

and Theorem 1.4.5. Set N I ker B n I. Since L7 : I] is finite

ker B = ker B aII-ker B
N ker a n 1" I

finite. In connection with this problem, we also have the follow-

 

and , we see that [ker B: N] is

ing proposition.

Proposition 2.1.1 If G' is abelian and N = ker a n I is trivial
then ker B is trivial.

Proof: LBt f E ker B and g E G. We need to show that f(g) I g.

T _1 defined by T _1 (x) I g-1f(g)xf(g-1)g for all
g f(g) s f(g)
x E Z(G) is an element of m. We claim that T _ is in
g f(g)
ker B. If g* E G then, since f E ker a, f(g*) I g* +-Ag for
*

some Ag 6 A(G')A(G). If x E Z(G) then T _1 (x) I

1 * 1 1 s f(g) 1
- - -1 - -
g f(s)xf(s )s = g (s +-Ag)X(g +*A _1)s I (1 +'8 Ag)x(1 +'A -13)
s g
a x +~xA -18 + g-lAgx + g-lAng _1g 5 x mod A(G')A(G) since
3 s
A(G')A(G) is an ideal of Z(G). Thus T _1 E ker a n I, so
a f(g)

g-1f(g) E-%(Z(G)), by hypothesis. Set f(g) I gz for some 2 E Z(Z(G)).
Since ‘f(g)‘ I |g‘, z is a unit of finite order in Z(Z(C)). By

18

19

Theorem 2.1 of [2], z is an element of -%(G). But this implies,
by Theorem 1.1.7, that f(g) I g since f E ker B. This completes
the proof.

By Theorem 1.1.8, any f E I can be written as f I Tu for
some unit u in Q(G). The next preposition shows that conjugation
by certain units cannot be an element of ker 3.

Proposition 2.1.2 Suppose G is nilpotent and G' is abelian.

Let u E Z(G) be a unit of finite order with Tu E N I ker B n I.
Then u I i;g1 for some g1 E %(G) so that Tu is the identity of
d.

‘ggggf: Tu E ker B says that ugu“1 - g E A(G')A(G) for all g E C.

By Lemma 8 of [6], u E i:g* mod A(G')A(G) for some g* E G. Thus
1

' -l -1
8*88* 8 mod A(G')A(G) for all g E G and g g*gg* a 1

mod A(G')A(G). By the results of [11], g-1g*gg;1 I 1 so that

3* E Z(G). Hence u(:;g;1) 1 mod A(G')A(G) and u(i;g;1) is a
unit of finite order in Z(G). Again by [11], u(:;g;1) I 1 so that

u I j;g*. This completes the proof.

Section 2. When is flKII Aut(G)? G' is not necessarily abelian

 

in this section. First, we consider when na’I Aut(G). The next
proof follows the ideas of Theorems 9 and 10 in [5].

Proposition 2.2.1 If nU'I Aut(G) then every subgroup of G is
normal.

.ggggf: Let 1 I g1 be any element of G. It suffices to show that
gzglg;1 is a power of g1 for any g2 E C. Set P I g2(1-g1) and
Q I 1 +g1 +'gi +...+'g:-1, the sum of the distinct powers of g1.

Since PQ I 0, (1-32P)(1 + 3QP) I 1 so that 1-3QP is a unit in

20

Z(G) with inverse 1 + 32F. Thus conjugation by 1-3QP is an auto-
morphism of Z(G) which is clearly in 724. By the hypothesis,

(1-3QP)gl(1 + 3QP) is an element of G. But

fl
0

(1’32P)81(1 + 3QP) (1'3QP)(81 + 3QP) since ng

ll
0

g1 + IQP - 3QPg1 since HQ

81 + 3(QP - Q1381)-
[g1 + 3(QP - QPg1)] E C says that QP - QPg1 I 0 since we are
working in Z(G). QP I 32 + 8182 + Sigz +---+ 831-182 ' 8281 '
313281 - gzlgzg1 -...- gri-1g2g1. We may suppose that no summand of
QP with a minus sign is g2 since g2 I gzg1 implies g1 I l and
g2 I gigzg1 implies gii I gzglggl. Expanding QPg1 we have

2 n-l 2 2
QPEI = 8281 + 813231 + glgzgl +.-.+ 31 8281 - 8281 - 818281 --~-
gq-lgzgzl. The equality of QP and QPg1 implies that g2 I gjlgzg1
for some j I l,...,n-l. Whence gij I gzglg;1 and we are done.

AS a partial converse, we have the following result.
Proposition 2.2.2 If G is abelian or is a 2-group in which every
subgroup is normal, then 72d I Aut(G).

E222: We use the following reSult from [1]. If G is abelian or
is a 2-group in which every subgroup is normal and u E Z(G) is a
unit of finite order, then u I : g1 for some g1 E C. Now if

f E 724 and g E G, we have, by this result, that f(g) E G. Thus

if f E 724 then f restricted to G, denoted f\G’ is such that

“G e Aut(G). Hence m = Aut(G).

Section 3. What i_§_ f(ggKn? Here we consider f(A(K)) where

f E a has an elementary representation and K 9 G.

21

Proposition 2.3.1 Suppose f E a' has an elementary representation,
say f I Tu o O. with o E Aut(G) and u E Q(G). Then f(A(K)) I
A(K) iff 0(K) IK where K SC.

329$: One easily checks that Tu E a. It is clear also that

Tu E 724. Thus Tu(G) is a group basis for Z(G).

If 6 is the isomorphism between the lattice of normal sub-
groups of Tu(G) and the lattice of normal subgroups of G described
in Section 1 of Chapter 1, we claim that Q(TU(K)) I K. Suppose
Q(Tu(K)) I L 9 G. If L1,L2,...,Lt are representatives of the con-
jugacy classes of G inside L and k1,k2,...,kt are representatives

of the conjugacy classes of G inside K, then C +-C- +...+E I

‘1 L2 Lt
- - - — -l - -l - -1
C +C +...+C Inc 11 +uC u +...+uC u I
-l -l -l k k k
ukli uk.u uk u 1 2 t
1 2 t
‘E +-E' +. +-E' . But 0 is a basis for Z(G), so L = K. By
R1 R2 kt

Theorem 1.1.5, A(uxu'1) = A(Tu(K)) - A(K).

Now we can show that 0(K) I K implies f(A(K)) I A(K). To
show that f(A(K)) C A(K), it suffices to show that f(k-l) E A(K)
for each R E K since A(K) is an ideal of Z(G). f(k-l) I f(k) - l I
u¢3(k)u-1 - l I uklu-1 - l for some k1 E K I o(K). Thus
f(k-l) E A(uKu-1) I A(K). To show A(K) C f(A(K)), it suffices to
show that uku-1 - l E f(A(K)) for each k E K since f(A(K)) is
an ideal of 2(a) and d(uRu'l) = A(K). But f(a'1(k)) = Tuo(o'1(k))
I uku-1 implies that f(a-1(k)-l) I uknu-1 - 1, so A(uKu-1) I A(K) c
f(A(K)) and we have f(A(K)) I A(K).

For the reverse implication, we suppose f(A(K)) I A(K) and

show d(x) - K. It suffices, since ‘K‘ is finite, to show a(k) e K

for all k e K. f(k-l) = uq(1t)u’1 - l is in A(K) = A(uKu-1), by

I.’.I\.l.l .[ ulli [ [Al I l I i lull ll ill I .l

22

hypothesis. Thus, uo~(k)u.1 - l I z (uklum1 - 1)t(k1) with
k16K
t(k ) e Z(G) for all k e K. So a(k) - 1 = z u'1(uk u“1 - 1)t(k )u
l 1 1 1
klEK
= z u'1(uk1u'1 - l)uu'1t(k1)u = z (kl-l)t'(k1) with t'(k1) =
klEK klEK

u'1t(k1)u e Z(G) since u'12(c)u = u'1(u2(c)u'1)u = Z(G). Thus
o(k) - 1 E A(K). If n denotes the canonical ring homomorphism of
Z(G) onto Z(G/K), n(o(k)) - I'= a(k)K - 1': 6' since A(K) is

the kernel of n. Thus o(k) E K and the proof is complete.

Section 4. How _i_s_ fig like Aut(G)? Some of the properties of

 

Aut(G) enjoyed by d and 72d have been detailed previously. The
next two propositions give additional information along these lines.

Proposition 2.4.1 If f E d and f(A(K)) C A(K) for K 9G, then
f permutes the class sums of elements of G inside K; hence

f(K) = E' where E'= z k.
kEK
Proof: Since f(k) - l E A(K) for all k E K, L(f(k)) I l for

all k E K. If k E K, we know that f(Ck) I i;Cé for some g E G.
We must choose the plus sign by the preceding remark. We need to

show that g E K. Since f(k) - k I f(k-l) - (k-l) E A(K), by

hypothesis, f(Ck) "E I Cg - Ck E A(K). Let n again denote the

k
canonical ring homomorphism of Z(G) onto Z(G/K). Applying n

to C - C we obtain gK +x2gx2 K +u..+"xngxn1K - ‘0

g k OK=O,

1.‘

where { x x.1 x -1] is the set of all conju ates of g
8: 282 an” ngxn 8

in G. Whence gK I K and g E K.

Corollary 2.4.2 If G' is abelian and K is a characteristic

subgroup of G with K S G' or G' S K, then any f E flfl' permutes

the class sums of elements of G inside K.

23

Proof: By Lemma 1.2.2, f(A(K)) I A(K) and the reSult follows from
Proposition 2.4.1.

Proposition 2.4.3 If f E a' and f(g) I g1 for g,g1 E G then

f'E =‘E .

( 8) 31

Proof: We know that f(Cé) I 1:6? for some g2 E G. For any
2

x E G, L(f(xgx-1)) L(f(g)) I L(g1) I 1 so we must choose the plus

Sign. Thus f(Cg) C8 and f(g) - g2 I g1 - g2 E A as in the

2
proof of Lemma 1.3.1. By Proposition 1.1.1, g1 and g2 are con-

'u ate so C. I C. I f‘C .
J g g1 82 ( 8)

Section 5. The Center 25‘ flgh This section gives some information

about the center of d.

Proposition 2.5.1 Let G be arbitrary. If f E‘nj' commutes with

 

T8 for all g E G then f restricted to G, denoted f takes

is
G onto G. Moreover, f\G is a central automorphism of G.
.2E22E3 We recall that o E Aut(G) is central if o(g) E g %(G)

for all g E C. By hypothesis f 0 T8 I T8 0 f for all g E G. If
s1 6 c then f o Tg(g1) = f(s)f(s1)£<s'1) = Ts o f(sl) = s f(sps'1
so s'1f<g)f<s1> =f<sl)s‘1£<s>. Hence 8-113(8) is in mm).
Let f(g) = gz for some 2 e-z(2(c)). Since \f(g)| = |g\, z is

a unit of finite order with L(z) I 1. By Theorem 2.1 of [2],

z E Z(G). Thus f(g) I gz E G. This shows that f]G is a central

automorphism of G and completes the proof.

Corollary 2.5.2 If G is arbitrary with Z(G) I 1 then na' con-
tains no non-trivial elements which centralize the group {Tg‘g E G}.

As a result, fla' contains no non-trivial elements in the center of

24

72a is trivial.
Proof: This follows from Proposition 2.5.1 and the fact that if

%(G) I 1 then there are no non-trivial central automorphisms of

G.

CHAPTER 3

A METHOD FOR STUDYING M AND SOME EXAMPLES

Section 1. Introduction. In this chapter, we examine a method for

 

dealing with the question of when an elementary representation exists
which is particularly useful in case the character table of the group
is available, or partially available, and one suspects that I I‘fldl
The method is to extend f E m linearly to an automorphism of C(G)
and see what f does to the identities of the simple components of
C(G). This method is then used to study 724 when G I Sn’ the
symmetric group on n symbols, for small values of up We find that
if n s 10 and G I Sn then any f E 714 has an elementary repre-
sentation. Thus it is possible that all elements of fld' can have

elementary representations without G' being abelian.

Section 2. The Method. Let G be an arbitrary finite group. If

 

f E flag we can extend f linearly to a ring homomorphism of C(G)
into C(G) which we also denote by f. Actually f is an auto-
morphism of C(G). For, let 2 agg be an arbitrary element of
C(G) with as E C. For each gEGE G, g I f(xg) for some xg E Z(G)
so that 2 agg I 2 a8f(xg) I Z f(agxg) I f(z agxg). This shows that
f maps C(G) onto C(G) and since C(G) is a finite dimensional
vector Space over C, f is a monomorphism of C(G) onto C(G).
Whence f is an automorphism of C(G). Let {Si]2=1 denote the

finite set of simple components of C(G) Such that C(G) I 81(3..43 Sn

25

26

and let ei be the multiplicative identity of Si for i I l,2,...,n.

Since f effects a permutation of the set {81]231, f also gives

rise to a permutation of the set {ei]i=1. By Theorem 33.8 of [3],

j 1 “ J j
E I (g ) C where I is the irreducible
jg G i=1 1 8i

 

we have that e

character of G afforded by a minimal left ideal of 8j and

g1 I l,g2,...,gn is a set of conjugacy class representatives for G.
We record a series of results framed in this setting; the usefulness
of these results will be demonstrated in the next section. Through-
out the remainder Of this chapter, f will denote the linear extension
of an element of m to C(G) as described above.

Proposition 3.2.1 Suppose f(E' ) I‘C and f(e.) I e . Then
83 82 1 j

 

1%,) - 11(33).

Proof: f(e)If(L‘i-L‘uk1 2: 1(gk)Cg 8.117% 2: I.—_(gk)f(cgko)
kIl

Since f(ei) I e , we have the equation

J

<*) Tifi-kz: 1(g)f(Cg)= ‘31 z Ij(s)C
k 3k G kIl 1‘ 3k

Recall that {Eg }r1:=1 forms a c-basis for %(C(G)) and that
k

f(Cé ) I Cgl since glI 1. Thus, from (*), we obtain

 

l
the equality 4-1‘1-1— 1 (l) = 4%)— so that r (1) = 1,3 (1).
Comparing the coefficients of G3 on the left and right side of (*),

it is clear that -TéTl’Ii (33) I g-TéTl-I;(g2). Therefore, 11(33)

13(32) as was to be shown.
‘529255. The above proof shows that if f(ei) = ej then 11(1) 3 13(1).

Corollary 3.2.2 If f(E ) = E and f(e ) = e then 11(33) ...
1 s3 83 i J
1 (s3)-

27

Proof: Take g2 I g3 in Proposition 3.2.1.

._ ._ 1 _

Corollary 3.2.3 If f(Cgs) = cg2 and f(ei) - ei then 1 (g2) —
i

I (83) :

Proof: Take 1 I j in Proposition 3.2.1.

 

Corollary 3.2.6 -Suppose G has only one conjugacy class Cg with

‘Cg‘ elements. If 19(g) I 11(g) then f(ei) I ej.
Proof: By Corollary 1.4.3, f(Cé) I 6;. Since Ij(g) I 11(g), f(ei) I ej

by Corollary 3.2.2.
Corollary 3.2.5 Suppose G has only one irreducible complex char-

acter 11 of degree Ii(l). If 1}(g ) I Ié(g ) then f(C' ) $46
2 3 g3 g2
Proof: By the Remark preceding Corollary 3.2.2, f(ei) I ei. Since

11(32) I Ii(g3), f(C' ) I Cl by Corollary 3.2.3.
g3 82
Lemma 3.2.6 Suppose G has exactly 2 irreducible complex characters

of degree 1. Denote the non-trivial character of degree 1 by 12.

2 2 - -
1f 1 (32) i 1 (33) then f(C ) s c .
g3 82
Proof: If e1 is the idempotent of C(G) associated with the

trivial character of G, then f(el) I e1 since f permutes the
class sums of elements of G. Hence f(ez) I e2 by the Remark pre»

ceding Corollary 3.2.2, where e is the idempotent of C(G) assoc-

2
iated with I2. By Corollary 3.2.3, f(Cé ) I C; as was to be shown.

3 2
Corollary 3.2.7 Let G I Sn' Any f E 71a trust take a class sum of
elements of G inside Ah, the alternating group on n symbols, to
a class sum of elements of G inside An.
Proof: Since An I 3; and [sn : Ah] I 2, Sn has exactly 2 irreducible
complex characters of degree 1. If 12 denotes the non-trivial

linear character Of sn then I?(g) I 1 if g E An and -1 if

g E An. Let g3 E An be arbitrary and g2 be any element of Sn

 

Ill III I I
I I'll! III. II

28

not in An. Clearly 12(32) # 12(g3). Whence, by Lemma 3.2.6,

f(CgB) # cgz.

Section 3. Ihg_Examples. This section is devoted to showing that

if G I Sn’ n I 2,3,...,10, then each f E fluV has an elementary
representation. The methods utilize Corollary 1.4.3 and the results
of Section 2, Chapter 3. All information about the character tables
of these groups is in [7].

Theorem 3.3.1 If G = Sn for any n = 2,3,...,10 then each f E 710
has an elementary representation.

m: (1) If G =82 then MIAutm) by Proposition 2.2.2.

(2) If G I 33 then each f E fla' fixes each class sum by Corollary
1.4.3 and the fact that no 2 conjugacy classes of 53 have the same
number of elements. Hence, each f E m is of the form '1'u for
some u 6 Q(S3)-

(3) Suppose G = 34' Conjugacy class representatives for S4 may
be listed as follows: g1 = l, g2 I (12), g3 I (123), g4 I (1234),

gS I (12)(34). The number of elements in each class is 1,6,8,6,3
respectively and the order of the elements in each class is l,2,3,4,2
respectively. Since there is exactly one class containing 1,8 or 3

elements, f fixes C. , C. , C , by Corollary 1.4.3. Since the
81 g3 85
elements of C and C8 are not of the same order, f also fixes
2 4
C and C , by Corollaryll.4.3. Hence f I T for some
82 34 “
ueqmp.

(4) Suppose G I 35. As in the case of 84’ one can examine the
number and order of the elements in the conjugacy classes of S5

and use Corollary 1.4.3 to conclude that f I Tu for some u E Q(SS)°

29

(5) Suppose G I 36. The character table for G is given in Table 1.
By Corollary 1.4.3, f fixes ‘C , C. , C. . By Corollary 3.2.7, f

81 85 87
also fixed 039 and C84 since g9 E A6 and g4 ( A6. We now con-
sider two cases.

Case I: f(C. ) i C. . By Corollary 1.4.3, f(C ) =IC and
_ _ g3 311' g3 g3
f(C ) I C . It will be shown that f fixes all other class
811 811 _ _
sums of elements of G. Since f(Cg ) I Cg , Corollary 3.2.2 says
4 4__ _
. Since f(C ) I C , Corollary 3.2.2
7 g3 83
requires that f(ez) I e2 or f(ez) I e10. Thus f(ez) I e

that f(ez) I e2 or f(e2) I e

2. Since
12(32) # 12(8 ) and f(ez) I e , f(C, ) I E. by Corollary 3.2.3.
10 2 32 g2
Hence f(C, ) I'E . Since 12(g ) I 12(g ) and f(e ) I e ,
810 g10 6 8 2 2
f(E )=E by Corollary 3.2.3. Hence f(E )=E . Thus f
36 36 88 88
fixes all class sums of elements of G and hence f I Tu for some

u E Q(SE).

Case II. f(C’ ) I'C . By 11.4.3 of [10], there is an element
83 g11 _ _

a E Aut(s ) such that d(C ) I C . We emphasize the fact that

6 g3 811
f E m is arbitrary with f(Cg ) I 6g so that f may in fact
3 11

be a. It is claimed that f interchanges C and C and
_ _ g2 810

interchanges C8 and Cg . Once this claim is established, it

6 8
will be clear that f(Cg) I 0(C8) for all g E 36 so that

f I Tu o a for some u E Q(S6)'
Why is the claim true? We know f(C. ) I'C . If f(ez) I e

34 34
then, by Corollary 3.2.2, I5(g4) I I?(g4). Since this is not the

5

case, f(e2) i e5. If f(ez) I e10 then, by Corollary 3.2.2,

I}o(g4) I 12(34). Since this is not the case, f(ez) # e If

10'
2 2 - ‘-

f(e2) I e2, then I (83) ‘ I (311) Since f(Cg3) I C311, by

Corollary 3.2.3. Since 12(g3) I 12(g11), f(ez) # e2. Thus

30

H H- H H- H H- H H- H H- H HHH
H- H H- H o o H H- N m- n oHH
o m- H o H- o H H o N- a as
H N o H- o H N- o H N- oH we
N m H- o o H- H H H- H- m NH
N- o o o H o o o N- o 0H 0H
N m- H- o o H H H- H- H n ma
H N- o H o H- N- o H N oH as
o m H o H- o H H- o m a ma
H- H- H- H- o o H H N m m NH
H H H H H H H H H H H Ha
cs mH om oNH ssH oNH ms om os mH H .mu.
AemqVHmNHV AomVAsmVANHV AonanVANHv AomsmNHv AnsmNHv AmsNVANHV AsmVANHV AsMNHV ANNHV ANHV H m m>Huauaom
HHm on am mm Nw ow mm aw mm Nu Hm -muamm mmmHu

mummsficoo

em you wanna umuomumso "a mam<H

 

I I'll |Ir I.- ‘II. I I! III 1| 'II I I

31

f(ez) I e by the Remark preceding Corollary 3.2.2. Since f(ez) I e

7 7
and fig ) I 12(3 ), f(E ) I E by Corollary 3.2.2. Thus f(E ) =
2 2 g2 S2 2 82
IE by Corollary 1.4.3. Since f(e ) I e and I7(8 ) I I (g ),
810 2 7 6 6
f(C, ) I E, by Corollary 3.2.2. Hence f(C. ) I E. by Corollary
86 86 36 88
1.4.3. This establishes the claim.

(6) Suppose G I S7. We will show that any f E fin? fixes all class
sums and hence has an elementary representation. Exactly the same

statement can be made if G I 38’ 39 or S and the method of proof

10

is so much like the case G I 37 that these 3 cases are left to the
reader. Now to proceed with G I 57- One notices that the class

of transpositions in 87 has 21 elements and no other class has
exactly 21 elements. We will apply Corollary 3.2.4 to show that

. 15
f(ei) I e for each idempotent ei, i I 132a°°°315' Since {ei}1=1

i
is a C-basis for Z(C(S7)), it is clear that f fixes all class sums

of elements in 37. We reproduce a part of the character table for

S from [7] in Table 2. Clearly f(el) I e and by the Remark

7 1

preceding Corollary 3.2.2, f(eIS) I e This Remark also shows

15'

_ 14
that f(e8) e8 and that f(ez) — e2 or e14. But I (g2) I

2
I (82) 80 that f(ez) I e2, by Corollary 3.2.4. Hence f(e14) I e14.

Similarly f(ea) I e4, f(eIZ) I e12 and f(e6) I f( I e

eo’ 610) 10
and f(e7) I e7, f(e9) I e9. By Corollary 3.2.4, f(e3) cannot be

as, e11, or e so f(e3) I e3. Similarly f(es) I e5, f(ell) I e

13
and f(e13) I e1

11

3. Thus f fixes all of the ei's and we are done.

32

TABIE 2: Partial Character Table for s

7
32 = (12)
81 = 1 Transpositions
11 1 1-
12 6 4
I3 14 6
16 15 5
I? 14 7 4
I6 35 5
I7 21 1
I? 20 0
19 21 -1
110 35 -5
111 14 -4
112 15 -5
I13 14 -6
I14 6 _4
I15 1 _1

LIST OF REFERENCES

10

10.
11.

12.

13.

14.

LIST OF REFERENCES

S. Berman, 92 Certain Properties 2f Integral Grouerings,
Dokl. Akad. Nauk. SSSR, 91(1953), 7-9.

 

J. Cohn and D. Livingstone, On the Structure Lf Group Algebras,
l, Canadian Journal of Mathematics, 17(1965), 583-593.

 

C. Curtis and I. Reiner, Representation Theory pf Finite Groups
and Associative Algebras, (New York, 1962).

 

E. C. Dade, Deux gr wpes finis distincts ayant 1a meme algebra
de groupggsur tout corps, Mathematische Zeitschrift, 119(1971),
345-348.

 

G. Higman, The Units Lf Group-Rings, Proceedings of London
Mathematical Society, 46(1940), 231- 248.

D.A. Jackson, The Groups g£_Units pf the Integral Group Rings
g£_Finite Metabelian and Finite Nilpotent Groups, Quarterly
Journal of Mathematics, Oxford (2), 20(1969), 319-331.

D. E. Littlewood, Group,Characters and the Structure Lf Grou
Proceedings of London Mathematical Society, 39(1935), _150-199.

T. Obayashi, gntegral Group Rings g£_Finite Groups, Osaka Journal
of Mathematics, 7(1970), 253-266.

D. Passman, Isomorphic Groups and Grouerings, Pacific Journal
of Mathematics, 15(1965), 561-583.

W.R. Scott, Group Theory, (New Jersey, 1964).

S.K. Sehgal, ALRemark gn_g Paper pf D.A. Jackson, unpublished.

 

, Qg_the Isomorphism pf lptegral Grouerings, L,
Canadian Journal of Mathematics, 21(1969), 410-413.

 

G. Seitz, Finite Groups having_on1y One Irreducible Representation
Lf Degree Greater than One, Proceedings of American Mathematical
Society, 19(1968), 459- 461.

AH Whitcomb, The Group_Ring Problem, Ph.D. Thesis, University of
Chicago.

33

APPENDIX

APPENDIX

The purpose of this appendix is to prove Propositions 1.1.1,
1.1.2, 1.1.3. The source of these proofs is primarily page 596 of
[3]; this material is therein credited to Brauer.
Proposition 1.1.1 k E A iff A sums to 0 on all conjugacy classes
of G.
22222; (I) If k E A. then 1 is a finite Z-linear combination of
elements of the form ab-ba with a,b E Z(G). It thus suffices to
show that each element ab-ba sums to 0 on all conjugacy classes of
G. But this is clear because 8182 and 3281 are conjugate in G
for any g1,g2 E G.
(c) Suppose a E Z(G) sums to 0 on all conjugacy classes of G.
If g1 I 1,32,...,gn is a set of conjugacy class representatives

for G then a I X +-X +...+Xn where Xi I 2 a g for

l 2
gECg
-1 i -1
i l,2,...,n and a8 6 2. Let gi,ngix2 ,,,,,xmigixmi be the
conjugates of gi in G. Since 3 a I 0, we can write
SEC
31
xi = a131 ' a181 + azngixz ' azgi +"°+ amixmigixmi ' amigi

-1 -1 . .
32(ngix2 - gi) +...+-ami(xmigixmi gi), which 15 clearly

an element of A. Thus a E A-
Proposition 1.1.2 1 E AP iff K sums to elements of oz on the

conjugacy classes of G.

34

35

2532:: CI) If A E AP then X I x +-y with x E A, y E PZ(G).

The result now follows from Proposition 1.1.1 since elements of

pZ(G) sum to elements of pZ on the conjugacy classes of G.

(I9 Lat 31 I 1,g2,...,gt be a set of conjugacy class representatives
for G. If A sums to elements of pz on the conjugacy classes of

G, then X = X1 +-X2 +u..+'Xt Where Xi I Z a g with 2 a I pn

gEC 8 gEC 1
31 31
for some ni E Z. Thus, 1 I X1 - pn1 +-X2 - pn2 +...+Xt - pnt +
p(n1 +-n2 +u..+'nt) which is an element of AP Since Xi - pni E A

by Proposition 1.1.1.

Proposition 1.1.3 If x1,x2 E Z(G) are Such that x1 - x2 E AP,

P
1

Proof: First, we discuss statements (A) and (B) and then complete

then x - x5 E AP.
the proof.
(A) If. x,y e z<c> then (x + y)? a x? + y? mod AP. Why? The
details of this calculation are left to the reader.
(B) If X 6 A then xp E AP. Why? We first show that
(ore-ea)" e "p for any cuB 6 us). By (A). (aB'Bo')p _=_
(amp-r (-l)9(ga)" mod Ap' If p -= 2, then (801)" a
-(ea)p mod AP and if p is odd, then (-l)p = -1. Hence
(dB'Ba)p I (GB)p ' (Ba)? mod Ap- If Y IB(oB)p-1 then
GY'YG ‘ 08(a8)p-1 - B(oB)p-1a I (QB)p - (Ba)p so that
(as-8oz)" a (as)p - (so)? = cry-ya :-= 0 mod AP. Thus
(dB-Ba)p E AP, as was to be shown. The proof of (B) can be
completed by induction on the number of summands in 1 E A.
For suppose ni E Z and ai’BiE Z(G) for i I l,2,...,k.

Then

36

[n1(Blal-a181) + “2(8202-0282) +. . .+ nkmkok-okakn"

[“l(51"1'°'15 1) +° ° ’+ nk-1(Bk-10’l<-l-Q’k-lek-lnp

+ n:(Bkok-ak8k)p mod AP, by (A). But we have
n:(3wyqukak)p E AP and by the induction hypothesis
[“l(51°‘l‘°'131) +°”+ “k-l(Bk-lak-l‘ak-lBk-lnp 6 Ap' Thus
[almlal-alal) +. . .+ nkwkak-okekn" e AP.

To complete the proof of the Proposition, let x1,x2 be

as in the hypothesis. Then x1 - x2 I 1 +pr for some 1 E A,

1p E pZ(G). Hence

_ P: p p p d A
(x1 x2) (1+1p) 1 “Fl-p mo Ap- by ()

P
d b
0+1}, II!) AP: YCB)

0 mod AP, since A: E pZ(G).

)P _ P P

- _ - P _ P
But (x1 x2 _ x1 x2 mod AP’ so that x1 x2 E AP, as was to

be shown.