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RELATIONSHIPS BETWEEN THE RESTRICTED IDEALS
AND INDUCED MODULES OF THE GROUP RING 36

BY

Julie Rogers Kraay

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1981

I

7'

VII/5 // fj/l’

/ ,-
1". “7 / ,I

ABS TRACT

RELATIONSHIPS BETWEEN THE RESTRICTED IDEALS
AND INDIEED MODULES OF THE GROUP RING 3G

BY

Julie Rogers Kraay

Let H be a subgroup of the group G, 3 a field,
and I an ideal of 3G. We wish to determine when the
phenomenon I = (I n 33)$G occurs. Our first result,
an extension of a theorem published by D.S. Passman.
shows that there exists a unique normal subgroup, W, of
G such that I = (I n 3H)$G iff W'g_H. we also obtain

a second characterization which states that I = (I n $H)$G

31-!
man

 

iff I E AnnfliG NG, where N = and NG denotes the

tensor product N G33 3‘6 .

If we restrict our attention to ideals I

of the form I = [1 AnnaGM. where S is a non-empty set
MES

of irreducible $G-modules, then we obtain the following
additional partial characterizations. If Hi; G and
[G :H] < a, then I = (I n 3H)3G iff for each M E S
and for each irreducible $H—submodule W of M,

I EgAnnsGWG. If in addition to these hypotheses we assume

that [G:H] is a unit in $, we see that I = (I n 3H)$G

a ‘ G
iff I = PI Ann? M = (1 Ann (MH) , where MH denotes
MES G Mes 3G

M viewed as an FH-module by restricting the domain of
right multipliers to ER. Finally if we add to all
previous assumptions the additional one that G be
finite. we are able to conclude that I = (I n $H)?G
iff for each M 6 S and for each irreducible

)G

$G—submodule, L. of (MH it is true that L e S

(up to isomorphism).

‘we conclude the thesis with a brief chapter
concerning some relationships between the semisimplicity
of the group ring 3(G/H) and the phenomenon
Rad $G‘E (Rad $H)$G, where Rad denotes the Jacobson

radical.

ACKNOWLEDGMENTS

I want to express my deep gratitude to my mother.
my late father, and my brother and sister for their
constant encouragement during the course of my studies,
and for their courage and wisdom which far exceeds anything
I found in my books.

I am also deeply grateful to my husband who gave
me his enthusiastic support in spite of the inconveniences
this project presented.

I wish to thank all the faculty and staff of the
Math Department at Michigan State university for their
many kindnesses. They never cease to amaze me.

Special thanks go to my former teachers here.
The hours I spent in their classrooms were much enjoyed
and will be long remembered.

Finally, I want to thank Professor I. Sinha,
my thesis advisor, for all his patience, guidance. and

encouragement during the course of my research.

ii

Chapter

1.

2.

3.

4.

Introduction
Section 1.
Section 2.

Section 3.

On Ideals in
Section 1
Section 2

Section 3.

Section 4

Annihilators
Section 1.
Section 2.

Section 3.

Section 4.
Section 5.

Pr0perty p

TABLE OF CONTENTS

Group Rings and Their Medules

The Jacobson Radical

Relative Projectivity, PrOperty
p, and the Complete Reducibility

of Induced Modules .

3G and Their Restrictions to SR

Statement of the Problem

The Controller Subgroup .

Concerning Induced Modules

Concerning Chains of Subgroups

of Irreducible Modules

The General Case . .

Nbrmal Subgroupséof Finite Index

Finite Groups . . . .

The Annihilator of a Single

Irreducible $G-Modu1e .

Examples . . . . . .

and Semisimple Groups

iii

page

.‘10

10
13
16
21

24
24
29
36

42
48

54

CHAPTER 1 INTRODUCTION

§1. GroupiRings and Their Modules

Let G be a group and 3 a field. Then the group
ring 3G is the set of all finite sums of the form

2) a g, where a9 6 3. So 3G is a vector space over
966

3 with the members of G serving as a basis. If we
define addition componentwise and multiplication
distributively via the multiplication in the group, then

SG becomes an algebra over 3.

Let H be a subgroup of G and I an ideal of 3G.
Then 3H may be viewed as a subalgebra of TG, and
I n 3H is an ideal in 3H. Our main goal is to determine
those pairs (H.I) for which I = (I n 3H)$G. It turns
out that this phenomenon is intricately related to the
behavior of certain 3G- and $H+modules. Consequently,
a large portion of this thesis will be concerned with the
theory of modules. we begin by stating some elementary
results that will be crucial to our later work. Throughout
we will assume that all modules are right modules unless

otherwise specified.

Let M be an SG-module. By restricting the domain
of right multipliers to 3H, M may be viewed as an
3H-module. This 3H-module will be denoted MH‘

1

2
we will frequently exploit the following lemma, whose
proof involves an application of Zorn's lemma. (See

Passman [9], p. 224)

Lemma 1.1. Let H be a subgroup of G and let W
be an irreducible $H-module. Then there exists an
irreducible fiG-module. M, such that W is a submodule
of MH.
If [G:H] < a and if H‘A G, then the study of
3G, 3H, and their respective modules is greatly facilitated

by the following well-known theorem. (See Passman [9].

p. 281.)

Theorem 1.2. (Clifford's Theorem)

Let H be a normal subgroup of G of finite index
n, and let M be an irreducible $G-module. Then MH
has an irreducible SH-submodule W, and for suitable
xl,x2.....xm E G with m g,n. we have MH =
wxl 6 wxz G --- e wxm. an 3Hedirect sum of irreducible

3H—modules. In particular, :MH is completely reducible.

Just as to each SG-module .M there corresponds
the restricted SHsmodule "3' given an SH-module N
there correSponds the induced 3G-module N ®$H 3G, where

® denotes tensor product. This induced module will be

denoted NG. If {xi}i61 is a right transversal for H
in G, then NG = G N'® xi as vector spaces. It

is!
follows that if [G:H] < m and if dimgN is finite,
then dimyNG is also finite and in fact dimgNG =

[G:H] dimgN.

The following basic prOpositions concerning induced
modules will be used freely without further comment. (See

Curtis and Reiner [2].)

Progosition 1.3. If N is an SH-module such that

N = N1 e N2 as SH-direct sum, then NS = N?

fiG-direct sum.

e N? as

PrOpositionyl.4. If H and K are subgroups
G
of G such that H g K g G, then (N'K) a N6 as

$6-modules.

If M is an 3G-module, we denote by AnngGM the
annihilator of 3G in M. Thus AnnFGM. is an ideal in
3G. Such ideals will play an important role in our
later work, and we will use the following basic lemma

freely.

Lemma l.5. If M.a L as $G-modules, then
Ann$GM.= AnnyGL.

‘ggggg. Let T :ML4 L be an SG-isomorphism. Let
c e AnngsM, and let L e L. Then 1 = ¢(m) for some
m e M. and La = ¢(m)a = ¢(ma) = ¢(O) = 0. Thus
a e MSGL' and Annysn 5 Ann L. The Opposite inclusion

36
holds by symmetry.

§2. The Jacobson Radical

The Jacobson radical of the algebra 3G will be denoted
Rad 3G. Thus Rad 3G is the intersection of the annihilators
of the irreducible 3G-modules. (There are, of course,
other characterizations.) Many of our results will concern
a class of ideals of which Rad 3G is a minimal member.
Thus Rad ?G will be an important object of study for us,
both in its own right and as a specific example of an

ideal belonging to this more general class.

One of the most fundamental problems in group rings
is that of determining when Rad 36 = O. (we say that
36 is semisimple in this case.) For infinite groups,
conclusions exist for several important classes of groups.
such as solvable groups and linear groups, but both the
characteristic 0 and characteristic p cases remain
unsolved in general. However, the semisimplicity problem
has been solved for finite groups as indicated by the

following theorem:

Theorem légp (Maschke's theorem)

Let G be a finite group and 3 a field. If 3
has characteristh: 0, then 36 is semisimple. If
3 has characteristic p > 0, then 3G is semisimple

iff p does not divide \GI.

unfortunately, even for finite groups. it is not an
easy matter to describe Rad 3G once we know Rad 9G # 0.

One natural approach to this problem is to seek relationships

5
between Rad 36 and Rad 3H, where H is some fixed
subgroup of G. The next two theorems are basic results of
this sort. The first of these is an immediate consequence
of'Lemma 1.1. (See Passman [9], p. 273 for an alternate

proof.)

Theorem 1.7. Let H be a subgroup of G. Then

(Rad as) n an E Rad 3H.

‘gggggz Let a E (Rad $6) 0 3H. and let N be an
irreducible SH-module. By Lemma 1.1, there exists an
irreducible 3G-module, M, such that N _c_:_ “11' Since
a e Rad 3G, a annihilates M. Certainly, then, a
annihilates N. As N 'was an arbitrary irreducible

$H~module, a 6 Rad 3H.

If H is a normal subgroup of G with [G:H] < a.
then we have the following stronger result, whose proof
is a simple application of Clifford's Theorem. (See

Passman [9], p. 282).

Theorem 1.8. Let H.A G such that [G:H] < a.

Then (Rad 3G) 0 3H = Rad 3H.

Additional background results concerning the

Jacobson radical are given in the next section.

§3. Relative Projectivity, Property 29. and the Complete
Reducibility of Induced Modules.
Let H be a subgroup of G. An SG-module, M, is
said to be H-projective, or projective relative to 3H, if

every exact sequence of 36-modules O 4 L 4 N 4 M 4 O

6
which is split over 3H is also Split over 3G. Note
that if H = (1), then M is prrojective iff M is
projective in the usual sense, for O 4 L 4 N 4 M 4 0

always splits over 3 = 3H in this case.

D. G. Higman has characterized the H-projective
modules for those subgroups H having finite index in

G. (See Higman [3]).

Theorem 1.9 (Higman's Criteria)
Let [G:H] < a and let M. be an FG-module. Then

the following statements concerning M are equivalent:

(a) M is H-projective.

(b) M is isomorphic to a direct summand of (MH)G

(c) There exists an fiH—endomorphism n of M such

“ -1
that Ex. nxi=1

where {x.}?_ is a right
1:1 1 1 1-1

MI
transversal for H in G.

-1
1M

satisfies (c) above. So as an immediate consequence of

If [G:H] = n is a unit in 3, then n = n

Theorem 1.9 we have the following result:

Corollagy 1.10 (Higman)
If [G:H] = n is a unit in 3. then every fiG-module

is H-projective.

Subgroups H for which every fiG-module is H-projective
are of special interest. So, following Khatri and Sinha [6],

we state

7
Definition 1.11. Let H be a subgroup of G. Then
(3G,?H) is said to be a projective pairing iff every

$6-modu1e is Héprojective.

So Corollary 1.10 says that if [G:H] = n is a unit
in 3. then ($6,3H) is a projective pairing. In fact,
the converse holds as well. Khatri and Sinha [6] established
it for finite groups, and Gloria Potter [10] extended their

result to include infinite groups. In summary, we state

Theorem 1.12. Let H be a subgroup of G. Then

(SGJH) isa projective pairing iff [G:H] = n is a unit in 3.

It turns out that if H A_G. then the concept

of projective pairing is strongly related to the following

concept.

Definition 1.13. (Sinha [12])
Let H be a subgroup of G. Then the pair (3G,$H)

is said to have property p iff Rad 36 E (Rad 3H)$G.

The connection between prOperty p and projective
pairing for normal subgroups H is made evident by the

following theorem. (See Passman [9], p. 278).

Theorem 1.14. (Villamayor)
Let H be a normal subgroup of finite index such
that [G:H] is a unit in 3. Then Rad 3G = (Rad 3H)?G.

In particular,(36.3H) has preperty 9.

As an immediate corollary we have

8
Corollary 1.15. (Potter [10])
If H A_G and if (36.3H) is a projective pairing,

then (3G,?H) has property p.

For finite groups we have the following result
which further relates the concepts of project pairing

and prOperty p.

Theorem 1.16. (Motose and Ninomiya [7])
Let G be a finite group and H a subgroup of
G such that Rad 36 E (Rad 3H)?G. Then ($G,3H) is a

projective pairing.

In general, the concepts of prOperty p and projective
pairing are independent of each other. (That is, there
exist pairs ($G,3H) having proPerty p but not projective
pairing, and vice versa. See Potter [10].) However, both
of these are consequences of a third more stringent

condition, as described in the next theorem.

Theorem 1.17. Let H be a subgroup of G, and
consider the following statements:

(i) ($6,3H) is a projective pairing.
(ii) (3G,?H) has prOperty p.

(iii) For each irreducible 3H—module N. the
corresponding induced module, N6, is a completely reducible

3G-module.

Then (iii) = (i) (Potter [10]). and

(iii) a (ii) (Sinha and Srivastava [13])

If H.A G, then (iii) e (i). (Potter [10]).

If \G\ < a, then (iii) 6 (ii). (Motose and Ninomiya [7])

Because of the strong connections between conditions
(i), (ii) and (iii) of the previous theorem. Khatri [5]
and Potter [10] studied those groups G for which the
classes of subgroups satisfying (i), (ii) and (iii),
respectively, exactly coincide with each other. we make
no attempt to list all their results. Suffice it to say
that there are many non-trivial examples of such groups.
we do, however. mention one result along these lines. since

we will call upon it later.

Theorem l;l§; (Khatri [5])

Let p, q be distinct primes and let 3 be a field
of characteristic p. Suppose further that G is a
finite group of order p“, pq or pqz. Then for any
H g,G. H satisfies (i) (of Theorem 1.17) iff H satisfies
(ii) iff H satisfies (iii).

It is clear from the results described in this section
that the relationship between Rad 3G and Rad 3H depends
on such factors as [G:H]. the behavior of the induced
modules NG where N is an irreducible 3H-module, and
the behavior of modules of the form (MH)G, where M
is an irreducible. 3G-module. It is reasonable. therefore,
to examine these factors in our study of more general

ideals. This we do in the next two chapters.

CHAPTER 2 ON IDEALS IN 3G AND THEIR
RESTRICTIONS TO 3H

§l. Statement of the Problem
Let I be a fixed ideal in 3G. In this chapter
we seek conditions on H < G which are necessary and/or

sufficient for I = (I n 3H)3G to hold.

Let H g.G, and {gi}i6J = T be a right transversal
of H in G such that 1 e T. Then as = Z; (3H)gi
i6!
exhibits 3G as a free left 3H-module. Thus each

d 6 36 has a unique representation of the form a = 'Z) digi.

iGJ
where oi 6 3H and gi 6 T.
Given a = Z) a g 6 36, we may project or onto 3H
see 9
via the mapping 7TH : 3G 4 3H defined by 1TH(G) ==
1rfl( Z: a g) a Z) a g. If a has Z) Gigi as its unique

966 9 96H 9 161
representation with respect to the right transversal T

of H, it then follows that oi = nH(cg;1), Vi.

We begin with a basic lemma which restates the problem

in terms of 7H and the ai's.

Lemma 2.1. Let I be an ideal of 3G. Then. using

the above notation, the following are equivalent:

(1) I = (I n 3H)3G

] C1033

(2) Vc= Z digiEI, [c 16.0-

iGJ

i

10

11

(3) I 0 3H = WH(I)

n
Proof: (1) = (2). By assumption, a: Z) B

Y for
t=1 t t
some positive n, where Et 6 I n 3H and Yt 6 36.
write Yt = .Z) etigi' eti 6 3H. 9i 6 T. Thus
11 16.9 n
a= 2 3(2) €.g.)= 2(2) Be.)g.. Itfollows that
t=1 tier: “- 1 16.0 t=1 ttl l
n
vi 6 .a, “i = Z‘, steti. Now at e I A. as a steti e I. Also,
t=1 n
. I . = - a
Bteti 6 3H. So Vt, Bteti e n 3H 2 c1 £21 Btet1 e I n 3H

as required.

(2) = (3). If a 6 I 0 3H, then

9
ll

WH(G) e WH(I). Conversely. let a 6 I. Then

_ -1 _ -1 .
a - Z 1rH(cxgi )gi - 7TH(C1)+ Z IrH(dgi )g.. By assumption,

iEJ iEJ 1
l:\=gi
wH(ag;1) = ai belongs to I n 3H. Vi. In particular.

WH(G) e I n 3H. Since a e I was arbitrary, WH(I).E:I n 3H.

(3) = (1). Since I A_3G. it is clear
that (I n 3H)3G _C_ I. Conversely, let a 6 I. Then
1)

- ' - ‘1
a - ‘2) wH(agi gi. Since a e I, so does cgi . Thus

161
-l _ .
IrH(dgi ) e wH(I). But wH(I) — I 0 3H, by assumption.

Thus a e (I n 3H)3G.
we pause to give a couple examples of the phenomenon

I = (I n 3H)gG.

Theorem 2.2. Let H A.G such that [G:H] = n is

a unit in 3. Take I = Rad 36. Then I = (I n 3H)3G.

12
Proof: By Villamayor's Theorem (Thm 1.14),
Rad 3G = (Rad 3H)3G. But by Theorem 1.8 Rad 3H = (Rad 3G) n 3H.

Hence the result.

Uhder the assumptions of Theorem 2.2, we note that
from Lemma 2.1 it follows that Rad 3H = WH(Rad 3G) and,
equivalently, that those elements a belonging to Rad 36

are precisely the ones of the form a =‘Z2 digi, ci 6 Rad 3H.

Before turning to the next example, we mention

a piece of convenient notation. Let a = Z) agg 6 3G.
963

Then Supp a = {g 6 G\ag * 0]. Recall that 3G consists

of finite sums of the form )3 agg. Thus Va 6 3G,
96G

Supp a is a finite set.

New let G = (X) be an infinite cyclic group and
let H be a subgroup of G. Let 0 + I A 3G. In this
case 36 is known to be a principal ideal domain, so
I = a3G, some a 6 3G. By multiplication by xk, we

may choose a generator, a, for I of the form

' 2 n
* = 00.
() a ao+clx+a2x + +dnx , (101:0, dn+0 .
Theorem 2.3. Let G be infinite cyclic, H a

subgroup of G, and a3G = I A 36, where a is of the

form (*). Then I = (I n 3H)3G iff Supp a E:H-

Proof: If Supp 0 _c_ H, then I n 3H = o3H and so
(I n 3H)3G = (a3H)3G = a3G = 1. Conversely, suppose

I = (I n 3H)3G. By Lemma 2.1. 1TH(I) = I 0 3H. In

13
particular, wH(a) e I. SO WH(G) = my, some y 6 3G.
New if xt E Supp y, then t > 0 = xn+t e Supp ay = wH(d),
while t < 0 = x-t e Supp cy = wH(a). But each of these is

impossible since Supp wH(d) E {l,x,...,xn]. Thus t = 0

and Y E 3 =9 Supp IrH(cx) = Supp a. Hence Supp a g H.

§2. The Controller Subgroup
If H A.G and I = (I n 3H)3G, then H is said to
control I. The following lemma is taken from Passman's

text. (See Passman [9], p. 304).

Lemma 2.4. Let I be an ideal of 3G. Then there
exists a unique normal subgroup, W, called the controller
of I, with the property that H A_G controls I iff

Haw.

It is clear that the controller, W, of I described
in Lemma 2.4 is W = n H, where the intersection is over

all normal subgroups H for which I = (I n 3H)3G.

Lemma 2.4 extends to non-normal subgroups H as

indicated in the following theorem.

Theorem 2.5. Let I be an ideal of 36. Then there
exists a unique normal subgroup, W, of G such that

for H g_G, H not necessarily normal, I = (I n 3H)3G

iff Haw.

Proof: Let S = {ng.G\ I = (I n 3H)3G}. NOte that

G E S and so S + O. Set W = (1 H. We claim that
HES '

14

(1) W G

ID

(2) I (I n 3W)3G
Proof of (1): Let x e‘W, g E G. Given H E S we

1

have I = (I n 3H)3G = I =»g- 19 = [g-1(I n 3H)g][g-13Gg].

But g‘1(I n 3H)g = I n 339. Thus I = (I n 3H9)3G.

and Hg 6 S. Since x e‘w = (W H, x 6 H9 = gxg"1 e H.
HES

But H was an arbitrary member of S. So

gxg-le fl G=W, and WAG.

HES

Proof of (2): By Lemma 1.1, it suffices to show that
I n 3W = Irw(I). Clearly I n 3W = 1rw(I n 3W) c_:_ TTW(I).
we prove the converse by mimicking an argument given by
Passman in his proof of Lemma 2.4. Given a e I, we
need to show that WW(G) e I. we proceed by induction on
[Supp a]. If \Supp a] = 0, then a = 0 and certainly
nw(a) = 0 6 I. Suppose [Supp a] =11 > 0, and that the
result is true for all smaller support sizes. If a 6 3w,
then mw(d) = a 6 I, and we are done. 0n the other hand,
if a E aw; then, by the definition of ‘W, there exists
some H 6 S such that Supp a g H. Since H 6 S, we
have I = (I n 3H)3G and it follows from Lemma 2.1 that
WH(G) e I. Furthermore, \Supp nH(d)\ < [Supp 0], and
therefore, because H 2 W, induction yields

rw(a) = WW(WH(G)) e I.

The theorem now follows easily. For if I = (I n 3H)3G,

then by definition of w; W’g H. Conversely, if W g,H,

15
then, by (2), I = (I n 3W)3G E (I n 3H)3G _I_:_ I, and so

I = (I n 3H)3G.

Corollary 2.6. If G is a simple group and 0 # I
is a preper ideal of 3G, then for all proper subgroups

H of G, (I n 3H)$G S I.

Proof: If I = (I n 3H)3G, then by the previous
theorem, I = (I n 3W)3G for some normal subgroup w
contained in H. As G is simple, the only such W is

<1), and in this case I = (I n 3W)3G is clearly impossible.

Example 2.7. Let G be a finite group, 3 a field
of characteristic p > 0, and I = Rad 36. By
Corollary 1.15, Theorem 1.16, and Theorem 1.12, if H.A G
then Rad 36 S (Rad 3H)3G iff H contains a Sylow-p-subgroup
of G. By Theorem 1.8, if H.A G then each of these is
equivalent to I = (I n 3H)3G. Hence if H ADG, then
I = (I n 3H)3G iff H contains a Sylowbp-subgroup of G.
In particular, it follows that w, the controller of I,
is the unique normal subgroup of G which is minimal

among all normal subgroups containing Sylowbp-subgroups.

Example 2.8. Let G = (X) be infinite cyclic, H
a subgroup of G, and I a non-zero ideal of 3G. we
have previously seen that I = a3G for some generator

2 n
d of the form d=c «H: x +---+anx , where

0 1 2
do # 0 and an # 0. Then it is clear from Theorem 2.3

X+C1

that ‘w, the controller of I, is the subgroup of G

which is generated by Supp 0.

16

§3. angerning Induced Modules

In this section we offer another characterization
of those subgroups H for which I = (I n 3H)3G. we

begin with a definition.

Definition 2.9. Let I be a fixed ideal of 36. Then

C(I) = H g G I 5 Ann N6 for each 3H—modu1e N such that

36
I n 3H EgAnngflN

Sinha [11] observed that a connection exists between
those subgroups H of G satisfying I = (I n 3H)3G

and the subgroups of G which belong to C(I). we quote

his result.

‘Lemma 2.10. (Sinha [11]) Let I A_3G. Then
(i) If H 6 C(I) then I = (I n 3H)3G.
(ii) If H A G, then H e C(I) iff I = (I n 3H)3G.

Unfortunately, we see from the definition of C(I)
that any direct application of Lemma 2.10 would involve
testing each 3H-module N, and the class of all 3Hemodules
is, at best, unwieldy. we now offer an improvement of
Lemma 2.10 which simplifies the criterion for membership
in C(I), and which shows that in fact C(I) consists

precisely of those subgroups H of G for which I = (I n 3H)3G.

Theorem 2:11- Let I be an ideal of 3G and H

a subgroup of G. Then the following are equivalent:

17

(i) I (I n 3H)3G

(ii) H e C(I)

3H

(iii) I g AnnyeNG, where N = In3H .

Proof: (i) =9 (ii) Suppose that I = (I n 3H)3G.
Then by Theorem 2.5, H 2 w, where w is the controller
of I. Let N be any 3H-modu1e such that I n 3H _C_ AnnaHN.

Certainly, then, I n 3W 5; AnnaHN.

Let [91] be a right transversal for H in G, and
let {kj] be a right transversal for W in G. Finally,
let a e I. Since W is the controller of I, I = (I n 3W)3G.

In particular, by Leanna 2.1, c: = 2 ajkj, where
cj 6 I n 3W, Vj. Note that since w A G,

-1
giajgi

G
a e AnngGN , and H 6 C(I).

6 I n 3W5 AnnZ‘HN, Vi,j. Thus (NG)d =

1 -
® gikj) - 0. Hence

(ii) a (iii) Suppose H e C(I). Then

since (I 0 3H) 5 AnngflN, it follows from the definition

of C(I) that I gAnnyGNG.

... . G
(iii) =9 (i) Suppose I g AnnyGN . Let
a = Z) digi e I, where {g1} is a right transversal for
H in G, and where oi 6 3H, Vi. By assumption, (NG)G = 0.
In particular, Vn e N we have (n ® l)d = (n ® 1)(Z‘, Gigi) =
=2“, (no. G g.) = 0. Thus no. = 0, Vi. Since n e N
i i i i
was arbitrary, we have Nci = 0, Vi. So each Ci 6 AnnyHN =

= I n 3H, and I = (I n 3H)3G.

18
Let H be a subgroup of G. New the class of
irreducible 3H-modu1es is always of special interest in
the study of 3H. Furthermore, in some cases this class
is completely determined: for example, in the case where
H is finite and 3 is algebraically closed of characteristic
0. So it is natural to wonder how C(I) compares to the

following class of subgroups of G.

Definition 2.12. Let I be an ideal of 3G. Then
we denote by A(I) the set of all H g,G such that for

G

each irreducible 3H-module N, I n 3H _c_:_ Ann N = I 5 Ann N .

3H 3G
From their respective definitions, it is clear that

C(I) E A(I). Equivalently, by Theorem 2.11, if H g G

such that I = (I n 3H)3G, then H 6 A(I). The converse

fails, as demonstrated by the next example.

Example lel. Let G = [1,a,b,ab], the Klein
4-group, and let H = (a). Let 3 be the field of two
elements. Since H is a p-group where p = char 3, it
is known that Rad 3H = w(3H), the augmentation ideal of 3H.

Since w(3H) is a maximal ideal in 3H of dimension 1,

it follows that ‘Egggifi’ is a field of dimension 1 over 3.
3H 9. . . . .
Thus Rad 3H ..3. Since each irreducible 3H-module is also

. . 3H .
an irreducible 'R53_3H" module, we see that there is but

one irreducible 3H+module, denote it N = (n), with the
action of H defined by n-l = n = n-a.

We compute AnnyGNG. Let c = a14-d2a4-a3b4-d4ab e AnnyeNG,

where each ai 6 3. Now an arbitrary member of NG has the

19

1)+(n52 O33
G

N , we have

form (n6 b), where each 6i E 3.

1 ®3H

Since a e Am3G

0 = [(n61 ® 1)-I-(n52 O b)](d14-c2a4-d3b4-d4ab)
= [n(5la1+6102+62a3+62014) ® 1] + [n(61d3+51d4+5201+62c2)
G b].
Thus V51 6 3, 51dl+51d2+52d3+52a4 = 0, and
51d3+5la4+52d1+52c2 = 0 .
If 61 = 1 and 52 = 0, we have c14-d2 = 0 = d34-d4.
If 51 = 0 and 52 = 1, 'we have “3"94 = 0 = ali-az.

If 61=l=62, we have c1+a2+a3+d4

The only other possibility is

ll
0

= 0 = 6 which imposes

61 20
no restrictions on the oi. Thus we conclude that

AnnSGNG = [c14-a2a4-c3b4-d4ab /a1 = c2 and c3 = d4].

Now let I = <1+a+b+ab>. That is,
I = {c(l+a+b+ab)/d E 3]. We claim that H 6 A(I),
but I # (I n 3H)3G. Recall that there is only the
trivial 3H-module, N, to consider. New I n 3H = 0,
so certainly I n 3H E AnngnN. So we must verify that

I _c_:_ AnnyGNG. But this is clear since

G _. ..
AnngGN - (c1+d2a+c3b+a4ab/c1 - (12, a3 -- a4] 2
{ala-czaI-c3b4-d4ab/cl = a2 = d3 = d4] = I.
Thus H e A(I). However, (I n 3H)3G = 0 # I.

Although C(I) and A(I) do not coincide in general,
there are special cases in which H E A(I) = H E C(I).
In fact, Theorem 2.11 gives rise to some of these, which we

now state as corollaries.

20
Corollary 2.;g, Let I be an ideal of 3G and
suppose that I n 3H is a maximal right ideal of 3H.

Then H E A(I) =9 I = (I n 3H)3G.

Proof: In this case N = IigH is an irreducible

 

 

3H—module such that I n 3H 2 AnnZ‘HN. Thus if H e A(I),

G
then I _c_ AnnSIGN

result follows from Theorem 2.11.

by the very definition of A(I). The

Corollary 2.15. Let I be an ideal of 3G and suppose

3H
Iflfr'H

H e A(I) a I = (I n 3H)3G.

that N =

 

is a completely reducible 3H—modu1e. Then

Proof: By assumption, N = N1 0 N2 Q ~-- 0 Nk'

where each Ni is an irreducible 3H-modu1e. New

I ('1 3H _C_ AnnyeN =9 I n 3H 5; AnnyHNi, Vi. If H e A(I),

then since each Ni is irreducible, it follows that
k

G .

I _c_ Ann3G(Ni) , Vi. Thus I E AnnyG(i<:1

Again, the result follows from Theorem 2.11.

G _ G
(Ni) ) - AnnsGN .

Corollary 2.16. Let I be an ideal of 3G, and
let H be a finite subgroup of G. Suppose that

I n 3H2 Rad 3H. Then H E A(I) =9 I = (I n 3H)3G.

 

Proof: In this case N = IfigH may be viewed as an
3H . 3H . .. ..
.Rad 3H - module. Since Rad 3H 15 semiSimple artinian,
all its right modules are completely reducible. In particular,

N is a completely reducible $63-17!- - module. It follows

that N is also completely reducible when viewed as an

3Hemodule. The result now follows from the previous corollary.

21
we conclude this section with a theorem which

describes another instance in which H E A(I) a I = (I n 3H)3G.

Theorem lel. Let I be an ideal of 3G and let
H be a finite subgroup of G. Suppose that (I n 3H)

is a prime ideal of 3H. Then H e A(I) a I = (I n 3H)3G.

Proof: Since H is finite, there exists an

3H
In3H'

*
L is of the form L = f%fi§ . ‘where L is a right ideal

 

irreducible right 3H-submodule L, say, of Furthermore,
*

of 3H containing I n 3H. Now I n 3H _C_ AnnszHL and
so since H E A(I) we have I g AnnyGLG. Let

c = Z) Gigi 6 I. where {91} is a right transversal for
H in G, and where oi 6 3H, Vi. Then (L§)c = 0.
In particular, V1 6 L, (L 8 1)a = (1 ® l)(Z) Gigi) =
Z: (Lei) ® gi = 0 a Adi = 0, Vi. Since I 6 L was

arbitrary, it follows that ci e AnnEHL, Vi.

Let 0 # L 6 L. Then I = L*4-I n 3H, where
'k * 'k
1. 6 L 5.3 3H, but I, ,6 I n 3H. Since (Ii 6 AnnyflL, Vi,
* *
we have 1 di 6 I 0 3H. Since I Z'I 0 3H and since”
I 0:3H is prime, it follows that oi e I n 3H, Vi.
Thus a E (I n 3H)3G, and as a E I was arbitrary, we

have I = (I n 3H)3G.

§4. Concerning Chains of Subgrogpg

We conclude this chapter with a couple of results
dealing with the situation where K < H < G, and where
'one of the subgroups K, H belongs to A(I), I being

an ideal of 3G.

22
Theorem 2.18. Let K < H < G and let I be an ideal
of 3G. Suppose H 6 A(I), K 6 A(I 0 3H), and that for
each irreducible 3Kpmodule, N, -the corresponding induced

module NH is completely reducible. Then K e A(I).

‘ggggg: Let N be an irreducible 3Krmodule such
that N(I n 3K) = 0. Since K E A(I n 3H), it follows that
(NH)(I n 3H) = 0. New by assumption, NH is completely
reducible; say NH = L1 S ... e Lt' where the Li are
irreducible 3H-modules. Since (NH)(I n 3H) = 0, it is

certainly true that Li(I n 3H) = 0, Vi. But since
L

H E A(I), this implies that I _c:_ 0 Ann (L.)G =
i=1 36 i
G G G _ G__ G
Ann?G(L1 6 L2 G ~-- @ Lt) - Ann36(NH) - AnnFGN , and so K E A(I).

Theorem 2.20. Suppose K A_H < G 'with [H:K] = n
a unit in 3, and let I be an ideal of 36. Then if

K belongs to A(I), so does H.

Proof: Suppose N is an irreducible 3H-module such
that N(I n 3H) = 0. By Clifford's Theorem,
NR = N1 6 N: e --- e Nt, where each Ni is an irreducible
3K+module. Since N(I n 3H) = 0, we certainly have

N(I n 3K) = 0. Thus Ni(I [I 3K) = 0, Vi. Since

K e A(I), it follows that I _c._' Ann96(Ni)G, Vi.
G G .. c _ G
Therefore, I g Am3G(Nl 0 N2 9 e Nt) - Ann$G(NK)

Now since [H:K] is a unit in 3, (3H,3K) is a
projective pairing by Theorem 1.12. It follows from

Theorem 1.9 that N is isomorphic to a direct summand

23

of (NK)H, say (N'K)H a N'@ L, for some 3H-module L.

So we have the following: (N'K)G a [(NK)H]G 2 [N G L]G a NG a LG.

Since I annihilates (NK)G, it follows that I also

annihilates NG. Thus H e A(I).

CHAPTER 3 ANNIHILATORS OF IRREDUCIBLE MODULES

§1. The General Case

Let G be a group and let Irr(3G) denote the class
of all irreducible 3G-modules. In this chapter we
restrict our attention to ideals I of the form

I = (1 AnngGM. ‘where S is a non-empty subset of
M68

Irr(3G). we note that Rad 3G = F) AnnFGM is an
MEIrr(3G)

ideal of this form, as is AnngGM for any M 6 Irr(3G).
Of course, in some cases Rad 3G = 0. However, the
augmentation ideal of 3G, m(3G), is the annihilator
of the trivial irreducible 3G-module and w(3G) # 0
provided [G] > 1. So the class of ideals under

consideration is never trivial.

we begin with a definitionr

Definition 3.1. Let G be a group, let 3 be fixed, let
(I a! S E Irr(3G), and let I = fl AnnngM' Then we say
mes
H g,G has property p with respect to I and S, or
(1
M E (WirrechH Ann3HW)3G.
MES

H e p(Is), if I = In Ann

mes ’6

We note that our definition depends on the choice of

S as well as I, since it is not clear that I = PI Ann M
MES 36
has a unique representation of this form.

24

25

We now pause to consider an important example.

Example 3.2. Take S = Irr(3G) and

I = F) AnnyGM = Rad 3G. Then H 6 p(Is) iff
MEIrr(3G)
I = Rad 3G E_(Rad 3H)3G. This is so since by Lemma 1.1,

given any irreducible 3H-module, W, there exists some

M E Irr(3G) such that W E MH' Thus Rad 3H = fl AnnyHW.
wirredEMH
MES

In this special case, therefore, we see that
H e p(Is) iff (3G,3H) has prOperty p in the sense of
Definition 1.13. So Definition 3.1 may be viewed as a

generalization of the concept of prOperty p.

Our main interest is still the study of those pairs
(I,H), I A ac, H g G, for which I = (I n 3H)3G. If I

is of the form I = fl Anna‘GM for some S E Irr(3G), then

MES
it is clear that I n 3H = (1 AnnyfiM. Thus I = (I n 3H)3G
MES
iff I = ( (I A )3G. Actually, I E ( fl Ann )3G is
M68 “nan“ MES 3H“

sufficient for I = (I n 3H)3G since the Opposite inclusion

[ fl AnnaHM13G = (I n 3H)3G E I automatically holds.
Hes

For future reference, we summarize these comments in a lemma.

Lemma 3.3. Let G be a group, [3 79’ S E Irr(3G), and

I = F] Ann M. Then I = (I n 3H)3G iff

mes 9‘6
I=flAn ME(flAnn )3G.
Mes n36 mes ”M
New let I be of the form I = (1 Ann M, where

M65 36

S E Irr(3G). Then as we'd expect from the similarities

26
between Lemma 3.3 and Definition 3.1, the class consisting
of those subgroups H of G for which I = (I n 3H)3G is
strongly related.to the class p(IS). Indeed, as we shall
later see, these classes coincide in several important

special cases. In general we have

Theorem 3.4. Let G be a group, (J 7! S E Irr(3G),

 

and let I = (1 Ann AM. If H is a subgroup of G such
.mes 36

that I = (I n 3H)3G, then H e p(IS).

Proof: Suppose I = (I n 3H)3G. Then by Lemma 3.3

we have I = 0 Ann ME ( fl AnnyHM)3G. But

MES ’6 MES
( fl AnngHMWG E ( fl Anns‘HW)3G, since any element
MES Wirred

EEMH

MES
of 3H which annihilates some M e S must certainly

annihilate all its irreducible 3H—submodules.

We note that in the case I = Rad 3G, Theorem 3.4
simply says that if Rad 3G = [(Rad 36) n 3H]3G, then
Rad 36 E (Rad 3H)3G. Otherwise put, if H contains the
controller of Rad 3G, then (3G,3H) has property p.

This was also an immediate consequence of Theorem 1.7.

The following example shows that in general H E p(Is)

is not sufficient for I = (I n 3H)3G.

Example 3.5. Let G = 33, char 3 = 2, S = Irr(3G),

and I = Rad 3G. Let H be a subgroup of G of order 2.

Then since [G:H] is a unit in 3, it follows from

27
Theorem 1.12 that;(3G,3H) is a projective pairing. Further-
more, since |G] = 2-3, we conclude from Theorem 1.18

that (3G,3H) has property p. Equivalently, H E p(IS).

However, I f (I n 3H)3G. For suppose equality holds. Then
H contains the controller subgroup, W, of I = Rad 36. But
the only normal subgroup of G contained in H is <1).
This means that W = <1) and so consequently
(Rad 3G) n 3W = (Rad 3G) n 3 = 0. Since ‘W controls I,
we have I = Rad 3G = [(Rad 3G) n 3W]3G = 0. But by
Maschke's Theorem (Thm 1.6), Rad 3G # 0. Thus

I = (I n 3H)3G is impossible.

Let I = M25 AnngsM for some S E Irr 3G, and let
G, denote the class of all subgroups H of G such that
I = (I n 3H)3G. Since H 6 c. iff H contains the
controller of I, it is evident that the class c. has
the prOperty that HgKgG, H€G=KEG. It is
not clear that the same property holds for the class
9(13). but the next theorem provides a result somewhat
along these lines.

Theorem 3.6. Let I = (W Ann M for some
mes 36

¢ 7" S E Irr(3G). Suppose that H 6 p(IS) and K is a
subgroup of G such that H'l K;g G and [K:H] < n.

Then K e p(IS).

Proof: Let L be any irreducible 3K~module. Then
by Clifford's Theorem LH is completely reducible. It
is therefore clear that Ann3HL = r) AnngHw. In

W irred c:

—LH

28

particular, this is true for any L E MK' where M e S.

Thus, since H E p(I ), we have I = 0 Ann
5 mes

C
so“ —

[ n AnnyflW] 3c; 5 [ fl ( 0 Ann Hw) ] so

a ' ' 3
Wirred irred Wirred
5 ”H L E MK 5 LB

M68 M68

-.- [ n Ann3HI,]3G El fl Annstlt’r'G .

Lirred E MK Lirred S “K

M68 M68

So by definition, K 6 p(Is).

Corollary 3.7. Let I = fl Ann M for some
mes 9‘6

(J 7’ S E Irr(3G). Suppose that H 6 p(IS) and that K
is a subgroup of G such that H < K < G, [K:H] < a,

and H is subnormal in K. Then K E p(IS).

Proof: Repeated applications of Theorem 3.6.
The following theorem gives a sufficient condition
for H e P(IS).

_'l'heo:_:em3.§_. Let I = 0 Ann M for some
MES 36

(Z 7! S E Irr(3G) . Suppose H g G has the property that
for each M e S and for each irreducible 3H-module

G
WE MH' it is true that AnniGW 2 I. Then H E p(IS).

Proof: Let Z aixi E I, where [xi] is a right

transversal for H in G and where ai 6 3H, Vi. Let
M e S and let W be any irreducible 3H—submodule of MH.

By assumption 2 aixi annihilates W6 = Z W ® xi. In-
i

29

particular, for each w E W we have 0 = (W 8 l)(Z) aixi) =
i

Z) (wa. ® x.). Thus we. = 0, Vi. Since w e W ‘was
i i i i ‘ ,
arbitrary, a1 6 Ann3HW, Vi. Furthermore, since M. was

an arbitrary member of S and W‘ an arbitrary irreducible

3H—submodule of ME, it follows that for each i,

ai e irrgl Annyaflk Thus Z} aixi 6 ( irrgl Ann$EW) 3G,
w SEMH W EEMH
M655 M653

and H e p(IS).

Corollary 3.9. Let I = (1 Ann M for some
-—-——' .M68 36

¢ 79’ S E Irr(3G). Then A(I) E p(IS), where A(I) is as

in definition 2.12.

{22922: Let H 6 A(I) and let W be an irreducible
3H—submodule of MH, where M e S. By its definition,
I annihilates M. Since W E MH' it is therefore clear
that I n 3H E Anngfiw. But since H E A(I), this implies
that I EgAnnyGwG. So the hypotheses of Theorem 3.8 are

satisfied and H 6 p(IS).

§2. Nbrmal Subgroups of Finite Index

If the subgroup H of G is normal in G and has
finite index in G, then Clifford's Theorem may be applied
to extend the results of the previous section. Indeed,
in this case the converses of Theorem 3.4 and Theorem 3.8

both hold. we begin with the converse of Theorem 3.4.

30

Theorem 3.10. Let I = F] AnnyGM for some
M68

0’ 7! S E Irr(3G). If H _A_ G and [G:H] < co, then

I = (I n 3H)3G iff H e p(IS) iff H e A(I).

Proof: By Clifford's Theorem, each M.e S is
completely reducible as an 3H-modu1e. So an element of
3H annihilates MH iff it annihilates each of its

irreducible 3H-submodules. Thus Fl Ann M =

MES 36
r) Ann W. It is therefore clear from Definition 3.1
irr d 3H
w e c:MH
M658

and Lemma 3.3 that I = (I n 3H)3G iff H e p(IS).

New I = (I n 3H)3G iff H E C(I) (Thm 2.11) and
cII) E A(I) by their respective definitions. Furthermore,

A(I) c p(IS) by Corollary 3.9. The result follows.

we note that if I is not of the form I = (1 Ann M
.Mes 3G

for some S E Irr(3G), then it is possible that I # (I n 3H)3G
even though H A,G, [G:H] < a and H e A(I). Such was the
case in Example 2.13 of the previous chapter where we took

I to be an ideal properly contained in Rad 36.

we now turn out attention to the coverse of Theorem 3.8
in the case H A,G and [G:H] < a. In view of our last
result, we employ a slightly different wording than that used
originally in the statement of Theorem 3.8.

Theorem 3.11. Let I = (3 Ann M for some
MES 36

¢ 7! S E Irr(3G). Suppose H _A_ G and [G:H] < on. Then

31

I = (I n 3H)3G iff for each M 6 S and for each

irreducible 3H-module W E MH it is true that I E Annyng-

The following simple lemma is used in the proof of

Theorem 3.11.

Lemma 3.12. Let I = (1 Ann M for some

(I 5-! S E Irr(3G), and suppose H A G. Then

I 8 3H = Fl AnnyHM is a G-invariant ideal of 3G.
MES

Proof: Let a e I n 3H and x e G. Then for each

x l

Mes, Ma =M(x-c1x)EMax. But IEAnnch by its

definition. Hence Mo = 0 which forces Max = 0. So

ax e AnngeM for each M e S, and consequently ax E I.
x

Also, since H‘l G and c 6 3H, we have a 6 3H. Thus

dx 6 I n 3H and I n 3H is G-invariant.

Proof of Theorem 3.11: In view of Theorems 3.8 and
3.10 we need only show that if I = (I n 3H)3G, then
I E AnnyGWG for each irreducible 3H-module W such that

WENh, some Mes.

n
So suppose that I = (I n 3H)3G. Let a = ‘2) aixi E I,
i=1

)1}

i=1 is a right transversal for H in G and

where {xi

where each at 6 3H. Then by Lemma 2.1, each
Oi e I n 3H = M28 AnnS‘HM. Now I n 3H is a G-invariant

ideal by the previous lemma. So in particularllfor each i
x.
and j and for each M e S we have that oil e AnnaHM.

32
New let W be an irreducible 3H-submodule of MH'

x 0

where M e S. Then since a.3 6 Ann HM for each i
_1 i 3
x.
and j, certainly oiJ e AnngfiW. Hence (WG)a =

wGIZaiin=IZ wexjHE aux); 2(23 W®x.ii=ax)
. 3= -1 i=1 j: -1 i=1 3
n n x"1 n n x. '1
Z (Z W®G:j x.xi) = 2(2‘3 Waij ®x.xi) =0.
j=l i=1 3 j= i=1 3

Since a e I was arbitrary, I E Ann we, as required.

we pause to interpret Theorems 3.10 and 3.11 in
the special case I = Rad 3G.

Corgllary 3.13. Let I = Rad 3G, H.A G, and
[G:H] < a. Then the following are equivalent.

(i) Rad 3G E (Rad 3H)3G

(ii) Rad 3G [(Rad 36) fl 3H]3G

(iii) For each irreducible 3H-modu1e W, Rad 3G‘E.AnnyGWG.
If in addition we assume that [G:H] = n is a unit
in 3, we obtain the following result.

Theorem 3,;3, Let I = Fl AnngsM for some
M68

¢¥SEIrr(3G). If HAG and [G:H] =n isaunit in 3,

then I=(In3H)3G iff 1: 0 Ann M= ('1 Ann ()6
MES ’6 mes ’6 Mk

Proof: First suppose that I = (I n 3H)3G. Let
M 6 S. Since [G:H] = n is a unit in 3, (3G,3H) is a
projective pairing by Theorem 1.12. It follows from Theorem 1.9
G

that M is a component of (MH) Consequently,

G .
AnnZ‘GM2 Ann:‘G(MH) . Letting M range over S we have

33

G
I = n Ann M2 0 Ann ( ) .
mes 3G Mes 3G MI-I

Clifford's Theorem guarantees that for each M e S,

0n the other hand,

MH is of the form MH = Wi e w: 9 -~- e Wt, where the

W1 are irreducible 3H-modules. Since I = (I n 3H)3G,

we have by Theorem 3.11 that I E AnngG(Wi)G, Vi. Thus

t
G _ G .
I E 121 AnngG(Wi) - Ann3G(M‘I-l) . Letting M range over
. G G
Swehave IEflAn ().SoI=flAnn(),
mes “ac MH mes 36 MH

as required.

_ G
Conversely, suppose I - I? Ann3G(MH) Let M E S

MES
and W be an irreducible 3H-submodu1e of ME. Then

wG _c_ (MH)G and so AnnyGWG 2 Ann$G(MH)G 2 n AnnyG(MH)G = I.

M68
By Theorem 3.11, I = (I n 3H)3G.
Theorem 3.14 suggests a condition sufficient for
I = (I n 3H)3G, as seen in our next result.

Theorem 3.15. Let I = F] AnnsGM for some
MES

¢ 72’ S E Irr(3G), and let H A G such that [G:H] = n
is a unit in 3. Suppose that for each M 6 S and for
)6

each irreducible 3G-submodule, L, of (ME it is true

that L 6 8 (up to isomorphism). Then I = (I n 3H)3G.

.ggggg: Since H‘l G and [G:H] = n is a unit in 3,
it follows from Theorems 1.12 and 1.17 that for each
irreducible 3H-module, W, the corresponding induced
module, W6, is a completely reducible 3G-module. New

by Clifford's Theorem and Proposition 1.3, for each

34

G. G__ G G
M E S, (MB) is of the form (MH) - W1 @ e Wt' where
the wi are irreducible 3H-modu1es. Since each (W‘i)G is

completely reducible, so is (MH)G. It follows that

PI Ann3G(MH)G is precisely the intersection of the
M68

annihilators of the irreducible 3G-components appearing

in the (MH)G.

Since by assumption the only such components

are among the members of S, we certainly have

G
PI Ann ( ) 2, PI Ann? M.
mes 3‘6 MH mes G
On the other hand, since by Theorem 1.12 (3G,3H) is
a projective pairing, we have by Theorem 1.9 that M is

a component of (MH)G, V M.e 5. Consequently,

G
(I AnnSGM2 PI AnngG(MH)

M68 M68
_ G
Thus (W AnnaGM - (1 Ann G(MH) , and by Theorem 3.14,

mes mes 3
I = (I n 3H)3G.

Theorem 3.15 gives rise to an interesting corollary,

but before stating it we need a definition.

Definition 3.16. An 3G-module, L, is said to be
homogeneous if it is a direct sum of, say, n c0pies of

an irreducible 3G-modu1e M.

Corollagy 3.17. Let I = AnngGM for some M e Irr(3G),
and let H'A G such that [G:H] = n is a unit in 3.

If (MH)G is a homogeneous 3G-modu1e, then I = (I n 3H)3G.

Proof: Since (3G,3H) is a projective pairing, by

Theorem 1.9 M is a component of (MH)G. Since (MH)G is

- b

35

homogeneous, M is the only irreducible component. Thus

the hypotheses of Theorem 3.15 are satisfied, and I = (I n 3H)3G.

we now take time out to consider an example which

illustrates some of the concepts discussed so far.

Example 3.18. Let G = D4 = <a,b/a2 = b4 = l, ab = b3a)

and let H be the central subgroup of G generated by
2. Let 3 be an algebraically closed field of characteristic
0. Then, in particular, the hypotheses of Theorem 3.15

are satisfied.

New H and G have the following character tables:

 

 

 

H 1 b2 G 1 h2 (b,b3} [a,ab2] {ab,ab3}
I1 1 1 x1 1 1 1 1 1
I2 1 -1 x2 1 1 -1 1 -1
x3 1 1 -1 -1 1
x4 1 1 1 -1 -1
x5 2 -2 0 0 0

 

Let M be the irreducible 3G-module corresponding
to x5, let I = AnngGM, and let W be the irreducible

3G-module corresponding to $2. Than an easy computation shows
that (X5)H = ZIZ, and (I2)G = 2X5.

In terms of the corresponding irreducible modules, we

G a M 6 M G M G M. Thus (Ma)G is homogeneous,

have (NH)
and so I = (I n 3H)3G by Corollary 3.17. Since H is a

minimal non-identity normal subgroup of G, it follows that

36
H is the controller of I. Consequently, if K‘g G,

then I = (I n 3K)3G iff K 2 H.

we haven't been able to prove the converse of
Theorem 3.15 as stated, but the converse does hold under
the added assumption that IGI < a. The proof of this fact

is the main result of the next section.

§3. Finite Groups

If G is finite, then 36 is a finite dimensional
algebra. In this case, the structure of the ideals of
3G is more easily ascertained. Since we are only

interested in ideals I of the form I = (1 Ann M,

.MES 36
and since each such ideal corresponds uniquely to some
. 3G . . . . .
ideal of Rad 3G , much inSight can be gained by examining
3G . . . . . .
Rad 3G , which is a semisimple, artinian algebra. The

following facts concerning Ragg3a' (which are also true

for semisimple artinian algebras in general) will prove
useful. See Isaacs [4] and Curtis and Reiner [2] for

more details.

PrOposition 3,l2. Let G be a finite group.

a. Each irreducible 3G-module may be viewed as an

irreducible 3G

‘§§a_35'- module, and Vice versa. Furthermore,
there are only finitely many of these (up to isomorphism).
3G __
Say Irr(3G) - Irr(Rad 3G) - {M1,M2,...,Mt}.
b. Each (right) ‘figggga - module is completely

36

m - module IS

reducible. Equivalently, each (right)

37

. . . 3G
the sum of its irreduc1ble Rad 3G - modules.
$6 - — .... . . . .—

i. Each ‘Ii is a simple ideal

ii. Each 'Ii is the sum of all the right ideals of

3G )
Rad 3G

3G
Rad 36

was defined in part a.

which are isomorphic to Mi' where Mi 6 Irr(

I
9
HI

iii. Ann M. -

3G 3 . . i
Rad 3G 1#3

d. For each i, l g_i g t, let Ii be that unique

right ideal containing Rad 3G which corresponds to 'Ii

36

Rad 3G ‘ The“

under the canonical homomorphism 3G'4

AnngGMj = figj Ii'

The considerations listed in prOposition 3.19 lead

us to the following theorem:

Theorem 3.20. Let G be a finite group. Then the

ideals I of the form I = (1 Ann
_ MES

are precisely those prOper ideals of '36 which contain

3GM' (Z 7! S E Irr(3G),

Rad 36.

Proof: (we use the notation of preposition 3.19)

 

Clearly, if I = (1 AnanM: then I 2 Rad 36.
M68

I .
Conversely. suppose 1‘2,Rad 36. Then Rad $6 15 an
36 . .
Rad 33 ' mOdU1e- Consequently. by Pr0p081tion 3.19 (b),
I - . . . 3G _
Rad 3G is the sum of its irreduc1ble 'figa-ga' modules,
3G

namely, those minimal right ideals of which are

Rad 3G

38

___£__
Rad 3G '

3.19 (c) are simple, Vi we have that

contained in New since the Iii of prOposition

I -— .
Rad 3G n Ii 13

either empty or equal to ‘Ii itself. These considerations

. I . .
lead to the concluSion that REE—35’ is the direct sum of some
of the iii; wolog, it is the sum of the first k of
I _. _. ._
them. Thus Rad 3G = I1 o 12 G -- o Ik, where 1 g_k g t.
New by Proposition 3.19 (c), Ann M. = o 'I.. It
3G 1 1%. i
Rad 3G 3

. I _ - _
follows eaSIly that Rad 3G Il e 12 o o Ik -
ijk Ann 36 Mi'
1> Rad 36

We claim that I = (W Ann3GMi° For let a 6 I. Then

i)k
I
(ad-Rad 3G) 6 Rad 3G - in; Ann 3G Mi, and so for
> Rad 3G

each i ) k, (Mi)c = Mi(d4-Rad 36) = 0. Thus

a 6 PI AnnyeM. 0n the other hand, suppose a 6 PI Ann M.

i)k i)k ’6
Then Vi ) k, Mi(ot+Rad 3G) = (Mi)a = 0. So
_ I
(ad-Rad 3G) 6 31k Ann 3G Mi - REE—35' It follows that
> Rad 3G

dd-Rad 3G = BI-Rad 3G for some B e I. Consequently
a = 34-y, some Y E Rad 36. Since I 2_Rad 36, y E I.

Thus a E I, and I = PI Ann

Mi' as claimed.
i)k

3G

Corollary 3.21. If char 3 = 0, or if
char 3 = p does not divide [G], then every prOper ideal

of 3G 'is of the form I = (1 AnnyGM.
MES

Proof: By Maschke's Theorem, Rad 36 = 0 in this

case .

39
In the finite group case, therefore, we see that
our hypothesis that I be of the form I = (1 Ann M

mes 3‘3
is not terribly restrictive.

It is also clear from Pr0position 3.19 that if G is

finite, then each prOper ideal I containing Rad 36

has a unique representation of the form I = PI Ann M,
36
M68

G 7! S E Irr(3G). For from (I AnnyGM = (I AnngGM it

mes mes'
follows that (W Ann 3G M.= (1 Ann 36 M and,

M68 -——-—- .MES’ -————-—
Rad 36 Rad 36

due to the direct sum decomposition described in Proposition
3.19 (c), that S = 8'. Since, therefore, I has a unique
representation, Definition 3.1 no longer depends on S

and we may simply say that H e p(I) if H is a subgroup
of G satisfying Definition 3.1. With these comments

in mind, we now state our next theorem.

Theorem 3.22. Let G be a finite group, I a
prOper ideal of 3G, and suppose that char 3 = 0 or
char 3 = p does not divide \GI. ,Then V H g_G, '

I = (I n 3H)3G e H e p(I).

Proof: By Corollary 3.21 and our comments above,

I = PI Ann M for some unique S c Irr 3G, S # ¢. Let
mes 36 _

HZEDG. Then V M.e S , MH is completely reducible

since 3H is semisimple and artinian. Thus

AnngflM = irreg] Annsfiw. By comparing Definition 3.1

‘W 55”};

and Lemma 3.3, we now see that H e p(I) iff I = (I n 3H)3G.

40
we now prove the converse of Theorem 3.15 for

finite groups.

Theorem 3.23. Let G be a finite group and H a
normal subgroup such that [G:H] is a unit in 3. Let

I = PI Ann
MES

I = (I n 3H)3G iff for each M,e S and for each irreducible

3GM' for some 0' 7! S E Irr(3G). Then

submodule L of (M’H)G it is true that L e S (up to

isomorphism).

Proof: <‘.= Theorem 3.15

= Let M 6 S and let L be an irreducible

3G-submodule of (MH)G. Suppose L f S. Now since

G G G
L E (MI-I) , we have AnniGL 2 m3G(MI-I) 2 M28 Ann36(MI-I) .
G
But by Theorem 3.14, 0 Ann ( ) - (I An m = I. It
mes 3’6 MB mes n36
follows that I = ((1 Ann3GM) n Annch. But this violates

M68

the fact that I has a unique representation of the form

I = PI Ann GM, as discussed before the statement of
mes 5‘

Theorem 3.22. Thus L e S, and the theorem is proved.

Paraphrasing Theorem 3.23, we see that (I n 3H)3G
is smaller than I precisely when for some M e S, (M'H)G

contains an irreducible submodule outside of S.

Of special interest is the case I = AnnyGM, some

M e Irr(3G). we have

Corollary 3.24. Let G be a finite group and H

a normal subgroup such that [G:H] is a unit in 3. Let

41
I = AnnyGM for some m e Irr 3G. Then I = (I n 3H)3G

iff (ME)G

is homogeneous.

Proof: We recall that as discussed in the proof
of Theorem 3.15, (MH)G is completely reducible. Since
[G:H] is a unit in 3, we have from Theorems 1.12 and

1.9 that M is a component of (MH)G. Thus (M'H)G is
homogeneous if M is its only irreducible 3G-submodule, and

the result now follows from the theorem.

we remark that if [G:H] = n is not a unit in 3,
then Theorem 3.23 fails. For instance, let G be a
finite abelian p-group and 3 an algebraically closed field
of characteristic p. Then 3G has only one irreducible
module, namely the trivial one - call it M. Take
I = AnnyeM = w(3G), the augmentation ideal of 36.
Then for H $_G, we have I n 3H = w(3H), and it is
well known (see, for example, Connell [1]) that
w(3G) 2 w(3H)3G. Thus I = (I n 3H)3G fails. 0n the
other hand, since M is the only irreducible 3G-module,
certainly .M is the only irreducible 3G-submodule of (MH)G.
(Of course, it would not be accurate to say that (M'H)G

is homogeneous in this case, since it fails to be completely

reducible.)

We conclude this section by remarking that if G
is finite and I is an ideal containing Rad 36, then
Theorem 3.23 provides a complete solution to the problem
of determining those subgroups H of G for which

I = (I n 3H)3G. The procedure is to use Theorem 3.23 to test

42
the normal subgroups of G, and to eventually locate the
controller of I. (Recall that the controller is always a
normal subgroup.) Admittedly, this procedure will be
formidable in some cases. But if, for example, 3 is
an algebraically closed field of characteristic 0, then
the computations may easily be performed in terms of
characters. In section 5 of this chapter, we will illustrate
these comments with some examples. First, however, we turn
out attention to the special case I = AnnyGM, where

M e Irr(3G).

§4. The Annihilatopof a Single Irreducible 3G-Module
Throughout this section we will assume that 3 is
an algebraically closed field of characteristic 0. Our
goal is to study the phenomenon I = (I n 3H)3G in the
case that G is finite and I = AnnycM for some
M e Irr(3G). Many of the results in this section will
depend heavily on character theory. we will sometimes
use characters and modules interchangeably. For example,
Irr(3G) will denote both the set of distinct (up to

isomorphism) irreducible 3G-modules and the set of

irreducible characters of G, depending on the context.

The following version of Clifford's Theorem will
facilitate a later result. Although it is stated in
terms of characters, it may be interpreted in terms of

modules as well. (See Isaacs [4], p. 79.)

43
Theorem 3.25. (Clifford) Let G be a finite group,
H‘A G, and x e Irr(3G). Let 9 be an irreducible
constituent of XH and suppose e = 91,92, . . . .91: are the
distinct conjugates of e in G. Then

XH = e(el+-92+....+-et), where e is the multiplicity

of e in XH'

Note, in particular, that if e is an irreducible
constituent of XH' then all the other irreducible
constituents of XH are conjugates of e and, conversely,
that any conjugate of e is a constituent of XH'

Furthermore, each conjugate occurs with the same multiplicity.

The conjugates in G of irreducible 3H-modules (or
characters) have a bearing on our problem. The following
theorem is crucial. (See Curtis and Reiner [2], p. 329

for the proof.)

Theorem 3.26. Let G be a finite group, let
H A,G, and let T be an irreducible representation of H.
Then the induced representation TG is irreducible iff
V x g'H the representations T and Tx :h 4 T(x-lhx) are
disjoint.

Theorem 3.26 actually says that the representation
TG being irreducible is equivalent to T having [G:H]
distinct conjugate representations. Or, in terms of modules,
if W’ is an irreducible 3H-module, then WG is irreducible

iff there are [G:H] distinct isomorphism classes of

3H-modules conjugate to W. These considerations give rise

to the following theorem:

44
Theorem 3.27. Let G be a finite group, let
H l_G, and let I = AnngGM for some M 6 Irr(3G).

Suppose MH = W1 9 ... @ Wt' where the ‘Wi are
distinct (up to isomorphism) conjugates and where t = [G:H].

Then I = (I n 3H)3G.

G is irreducible.

.gggggz By Theorem 3.26, each (Wi)
Furthermore, by the Frobenius Reciprocity Theorem, for
each i the multiplicity of M in (Wi)G is the same
as the multiplicity of Wi in NH: namely, the multiplicity
)G

is one. Thus (IIH a- (wl)G e (W2)G e o (thG a:

G

M.@ M.@ --- o M, and so (ME) is homogeneous. It follows

t times
from Corollary 3.24 that I = (I n 3H)3G.
The situation described in Theorem 3.27 can indeed

occur. Consider, for example, this next result.

Theopem 3.28. (Isaacs [4], p. 86) Let G be a finite
group and let H.l G such that [G:H] = p is prime. Suppose

M e Irr(3G). Then either
(a) MH is irreducible, or

P
(b) ME = e ‘w., where the W. are irreducible,
i=1 1 1

distinct (up to isomorphism), and conjugate.
Rewording Theorem 3.28 to better suit the context

of this paper, we have

Corollary 3.29. Let G lbe a finite group, and let
H'l G such that [G:H] = p is prime. Let M E Irr(3G)

and let I = AnnyGM. If MH is reducible, then I = (I n 3H)3G.

45
Our next result is actually a restatement of
Theorem 3.27. we include it because it lends itself to

an interesting application.

Theorem 3.30. Let G be a finite group, H.l G,
and suppose there exists an irreducible character, 9,
of H such that 9G 6 Irr(3G). Then I = (I n 3H)3G,
where M is the irreducible 3G-module associated with
G

x = e and I = AnngGM.

Proof: Since x = 96, it follows from the

Frobenius Reciprocity Theorem that e is a constituent
of multiplicity one in XH' So if e = 91'62""'et
are the distinct conjugates of e in G, we have by
Clifford's Theorem (Theorem 3.25) that XH = el+-92+----+-et.
Furthermore, since as is irreducible, by Theorem 3.26

9 has [G:H] distinct conjugates. Thus t = [G:H] and

I = (I n 3H)3G by Theorem 3.27.

Before proceeding to the promised application of

Theorem 3.30, we will need the following definition.

Definition 3.3l. A finite group G is called
Frobenius with kernel N and complement H if G = NH,

NAG, HnN=1, and Hon=1 forall xeG-H.

we now quote a result from Isaac's text.

Theorem 3.32. (Isaac's [4], p. 94) Let G be
a finite Frobenius group with kernel N‘l G. Then for

each character X e Irr(3G) with N E ker x we have

X = o6 for some m E Irr(3N).

46
As an immediate consequence of Theorems 3.30 and

3.32, we can now state

Corollary 3.33. Let G be a finite Frobenius group

with kernel N l_G. Let x e Irr(3G) such that
N E ker x. Finally, let I = AnnsGM where M is the
irreducible 3G-module corresponding to x. Then

I = (I n 3N)3G.

The proceeding results provide some examples of
the phenomenon I = (I n 3H)3G, where I = AnngGM
for some M e Irr(3G). we now turn out attention to some

situations in which this phenomenon cannot occur.

Theorem 3.34. Let G be a finite group, let H g,G,
and let M. be an irreducible 3G-modu1e with associated

irreducible character x. Finally, let I = AnanM.

If )((1)2 < [G:H], then I 7! (I f) 3H)3G.

Proof: Let n = [G:H]. New, by assumption,
dim(Hom3(M,M)) ='x(l)2 < n. Thus any n linear trans-

formations on M are linearly dependent. In particular,
n
)

if {x i=1

i is a right transversal for H in G, then

the linear transformations corresponding to the xi

via the'representation associated with M are linearly
dependent. So there exist a1 6 3, not all ai = 0,
such that a1x14-a2x24----4-anxn corresponds to the zero
linear transformation. In other words,

alxld-a2x24----4-anxn E AnngeM = I.

47
Suppose alxli-a2x24----4-anxn e (I n 3H)3G = (AnnaHM)3G.
Then by Lemma 2.1 it follows that ai E AnnFHM, Vi. But
the ai are members of 3 and so this forces ai = 0 Vi,
a contradiction. Thus Z} a.x. z (I n 3H)3G, and so

i i
I a (I C) 3H)3G.

Corollary 3.35. Let G be a finite group, let H
be a prOper normal subgroup, and let M be an irreducible
3G-module with associated irreducible character x. Finally,

let I = AnngGM. If H E ker x, then I 7! (I n 3H)3G.

Proof: Suppose H E ker x. Then x may be viewed

as an irreducible character of G/H. Now [G:H] = Z) ¢(l)2.
IGIrr(G/H)

Since x e Irr(G/H) and since IIrr(G/H)I ) 1, it follows
that x(l)2 < [G:H].

Corollary 3.36. Let G be a finite group, let M
be an irreducible 3G-module with associated irreducible
character x, and let I = AnngGM. Suppose x(1) = 1.
Then for H g_G, I = (I n 3H)3G e H = G.

Proof: Let ‘w be the controller of I. Then
I = (I n 3W)3G and so by Theorem 3.34 we must have
x(1) = l 2_[G:W]. This forces ‘W = G, and the result

follows.

Corollary 3.37. Let G be a finite abelian group.

Let M e Irr(3G) and I = AnnaGM. Then for H.S.Gv

I=(In3H)3GeH=G.

48
Proof: In this case x(1) = l for all

irreducible characters X of G.

§5. Examples

we conclude this chapter with a couple of examples
which illustrate the use of Theorem 3.23 and the results
of the previous section. Once again, we assume throughout
that 3 is an algebraically closed field of characteristic

0.

Example 3.38. Let G = S3 = <(12).(123)), and
let H = <(123)) A G. Then the reSpective character tables

of G and H are

 

 

G 1 [(12),(13),(23)} {(123,(132)} H l (123) (132)
x1 1 1 1 I1 1 1 A 1
x2 1 —l 1 I2 1 w w2
x3 2 0 -1 I3 1 w2 w

 

 

For each i, let Mi be the irreducible 3G-module
corresponding to xi. Recall that by Corollary 3.21, each

proper ideal of 3G is of the form I = (W AnniGM,
M68

where (Z 7! S E Irr(3G). For each ideal of I, we now

compute its controller subgroup.

If I = Annme1 or AnniGMz, then, by Corollary 3.36,

G is the controller of I.

49

Suppose I = Ann M . NOw (M3) is reducible since

3G 3 H
H is abelian and so all its irreducible modules must be
one-dimensional. It follows from Corollary 3.29 that
I = (I n 3H)3G. NOw H is the unique non-trivial normal
subgroup of G, and the controller of I must be a normal
subgroup other than one. We conclude, therefore, that
H is the controller of I.

Suppose I = (AnngGMl) n (AnnyeM2)° Now

- - G—
(x1)H - (x2)H - I1, and (I1) - XII-x2. It follows from
Theorem 3.23 that I = (I n 3H)3G. Once again we conclude

that H is the controller of I.

Suppose I = (Ann3GMl) fl (AnngGM3). Since x2 is
)G

. G _ _ .

a constituent of [(X1)H] - (I1 - XII-x2, it follows
from Theorem 3.23 that I # (I n 3H)3G. Since H is
the only non-trivial candidate for the controller of I,

we are forced to the conclusion that G is the controller.

An analagous computation shows that G is the

controller of I = (Ann3GM2) n (Ann3GM3)°

The only remaining ideals of 36 are 0 and 3G
itself. Clearly <1) is the controller in each of these

cases .

Example 3.39. We now do a more exhaustive study of

3

the group G a = <a,b/'a2 = b2 = l, ab = b a) of example

. D4 . . . .
3.18. In the previous example, we listed the ideals of
3G and computed the controller subgroup of each. This

time we will list the normal subgroups H of G, and

50
in each case determine which ideals of 3G are controlled

by H. The normal subgroups of G are the following:

<1),<b2),<b),Hl = <l,a,b2,ab2),H2 = <1,b2,ab,ab3),G

Before proceding with our computations, we repeat

the character table of G for convenience.

 

G 1 b2 [b,b3] [a,ab2} {ab,ab3}
x1 1 1 1 1 1
x2 1 l -l 1 —1
x3 1 l -l -l 1
x4 1 1 1 -1 -1
x5 2 -2 0 0 0

 

For each i, l g_i g_5, let Mi be the irreducible
3G-modu1e associated with xi. We now consider the normal

subgroups of G one by one.

H = <1): It is clear that H is the controller of

0 and 3G, and that H controls no other ideal of 3G.

H = <b2): Then H has character table
1 b2
#1 1 1

I2 1 -1

51

Furthermore, (x1)H = (x2)H = (X3)H = (X4)H = )1

)G

(I1 = xl+x2+x3+x4

G
(W2) - 2X5

It follows from these computations and Theorem 3.23 that
H controls the non-trivial ideals AnnyGMS and

(Ann3GMl) n (Ann ) n (Ann M ) n (AnnyGMa). Since H

3GM2 3G 3
is a minimal non—identity normal subgroup of G, we see

that H is the controller of these ideals.

H = (b): Then H has character table

 

 

H 1 b b2 b3

V1 1 l l 1

I2 1 -1 l -1

I3 1 1 -1 -1

I4 1 -i -1 1
Furthermore, (x1)H = I1 ()1)G = Xll'X4

(X2)H ‘ *2 (4'2)G = x24-x3

(x3)H = 11’2 ()3)G = x5

”4’11 2 “’1 ”’4’G = x5

(XS)H = I3-tI4

It follows from these computations and Theorem 3.23

that in addition to those ideals controlled by <b2),

52
H = <b) controls the non-triVIal ideals (AnnyGMl) n (Ann3GM4)'
(““3sz
and (Ann3GM2) n (Ann3GM3) n (AnnycMB). Furthermore,

) n (Ann3GM3)' (Ann3GMl) n (AnngGM4) n (AnngGMS),

<b) is the controller of each of these additional ideals.

 

 

H = H1: Then H has character table
H l a ab2 b2
I1 1 1 l 1
I2 1 -1 -1 1
I3 1 —1 1 -1
I4 1 l -1 -1
NG‘” “(13’s = *1 W1)G z x1")‘2
(1(2):: = )1 (‘12)G = X3+X4
(x3)H = I2 (I3IG = x5
(x4)H = )2 ”W6 = X5

(XS)H = 1‘3 + “’4
It follows from these computations and Theorem 3.23

that in addition to the ideals controlled by <b2), H = H1

controls the non-trivial ideals (AnanMl) n (AnngGMz),

(Ann M3) n (Ann M ), (Ann3GMl) n (AnanMz) 0 (Ann3GM5)' and

3G 3G 4

(Ann3GM3) n (AnnSGMA) n (Ann3GM5)’ Furthermore, H = H1

is the controller of each of these additional ideals.

53

H = Hz: This case is completely analagous to the

case H = H so we omit the details. It turns out that

1!
H = H2 is the controller of the ideals (Ann3GMl) n (AnnyGMB),
(AnngGMz) n (AnngeM4), (Annych) fl (Ann3GM3) n (Ann3GM5)'

and (Annyst) n (Ann3GM4) n (Ann3GMS)'

This now exhausts all normal subgroups except
G itself. So G is the controller of all ideals not
explicitly mentioned above. These are all of the form

(I Anna‘GM, where (Z 7‘ S E Irr 3G, and we will not
MES .

bother listing them. We remark, however, that from
the start we knew that G is the controller of AnanMl,

Ann3GM2’ AnnyeM3' and AnnFGM4 by Corollary 3.36.

CHAPTER 4 PROPERTY 9 AND SEMISIMPLE GROUPS

wallace [14] introduced the concept of JK-groups.

Their definition is as follows:

Definition 4.1. Let K be a field and G a group.
Then G is a JK-group if for all groups T and normal
subgroups S with T/S a G the pair (3T,3S) has property

p.

It is clear that if G is a JK-group then G is
semisimple. For G/l a G = (KG,K) has prOperty p = Rad KG = 0.
Examples of JK-groups include locally finite or abelian
groups having no elements of order p where p = char K.

(See Passman [9], p. 293).

We now generalize this concept.

Definition 4.2. Let 3 be a field and let X be a
class of groups which is closed under homomorphic images.
Then a group G e X is a J(X)-group if VT 6 X and

VS A_T with T/S 2 G, the pair (3T,3S) has property p.

Once again, it is clear that any J(X)-group must
be semisimple. Furthermore, any Jngroup belonging to X

is a J(X)-group.

54

55
Of particular interest to us are certain classes, X,
of groups which we will call p-classes. we will use the

notation PJE H. to indicate that the subgroup H of G
G

contains a conjugate of the subgroup P.

Definition 4.3. Let 3 be a fixed field, and let
X be a fixed class of groups closed under homomorphic
images such that for each G e X there exists a collection

p(G) of subgroups of G satisfying the following proPerties:
(a) V P E p(G), (3G,3P) has property p.

(b) If H g_G such that (3G,3H) has prOperty P.

then for each P E p(G), P g_H.
G

(c) If w l_G, then p(G/W) = {<1)} iff for each

P e p(G), P g,w.
G

(d) If G1, G2 6 X and m :61 4 G2 is an isomorphism,

then V P e 9(61). p(P) e p(Gz).
Then X is called a p-class and for each G 6 X, the

members of p(G) [are called p-subgroups of G.

we pause to mention some examples of p-classes.

Example 4.4. Let 3 have characteristic p and
let X be the class of finite groups having normal p-Sylow-
subgroups. For each G 6 X, let p(G) be the p—Sylow-
subgroup of G. We verify (a), (b) and (c) of the previous

definition. The remainder is clear.

Part (a) holds by Theorem 1.12 and Corollary 1.15.

56
Suppose H g,G such that (3G,3H) is a projective
pairing. So by Theorem 1.12 [G:H] is a unit in 3 and

H contains P. Thus part (b) of Definition 4.3 holds.

New suppose w A.G 9 p(G/W) = {<1)J. Then 3(G/W)
is semisimple. By Maschke's Theorem, it follows that G/W
has no elements of order p and W contains P. Conversely.
if G/W contains no elements of order p , then <1) is the
EPSylow-subgroup of G/W’ and so p(G/W) = {<1)}. Thus

part (c) of Definition 4.3 holds.

Before proceeding to our next example, we need some

preliminary definitions and results.

Definition 4.5. Let G be a locally finite group.
Then a subgroup A of G is said to be locally subnormal
in G if A is finite and is subnormal in all finite

subgroups of G which contain it.

P
The characteristic subgroup I (G) of G was defined

by Passman in [8].

Definition 4.6. (Passman) Let G be a locally finite

group. Then

(p(G) _ Q/A is locally subnormal in G and c>
I A is generated by elements of order p

P
Passman proved the following result concerning ‘f (G).

Theorem 4.7. (Passman [8])

Let 3 be a field of characteristic p ) 0 and G a
locally finite group. If H l_G such that Rad 3G‘E (Rad 3H)3G,

then H 2 (p(G).

57

Finally we quote a result from Passman's text [9].

Theorem 4.8. Let H A_G ‘with G/H locally finite.
If either char 3 = 0 or char 3 = p and G/H is a p'-group,

then Rad 36 = (Rad 3H)3G.

we now make the following observations:

Theorem 4.9. Let 3 have characteristic p ) 0. Let
X be a class of locally finite groups closed under homomorphic
images such that V G E X, the p-Sylowesubgroups of G
are all conjugate to each other. Further suppose that for
each G e X, (p(G) 2_P' where P is a p-Sylow-subgroup
of G. Then X is a p-class where for each G e X,

p(G) is the set of Sylow-p-subgroups of G.

Proof: Definition 4.3 (a) follows from Theorem 4.8.

For part (b), if H g.G such that (3G,3H)
P
has property p, then by Theorem 4.7 H 2 f (G) 2 P, where
P is a p-Slethubgroup of G. Since by assumption all

p-Sylow-subgroups of G are conjugate, (b) holds.

For part (c), if W’l G such that
p(G/W) = {<1)}, then G/W is a p’-group and w contains
a p-Sylow-subgroup of G. Since all p-Sylow—subgroups of G
are conjugate, P g W‘V P e p(G). Conversely, if P g,w,
G G
then G/w is a p’-group and so p(G/W) = [<1)].
Part (d) is clear.

Example 4.10. As an example of a p-class satisfying

the hypotheses of Theorem 4.9, let X be the class of

58
locally finite nilpotent groups. Then each G E X is the
direct sum of its Sylow subgroups. (See wehrfritz and Kegel[15],
p. 63). In particular, G has a unique normal p-Sylow-subgroup,
P, and p(G) = {P}. So we need only verify that for each
G e X and corresponding Sylow subgroup P, P E (p(G) .
From the definition of (p(G), it clearly suffices to
show that V x 6 P, <x) is locally subnormal in G. So
let H be a finite subgroup of G containing <x). Then
P n H A_H, and so it only remains to show that (X) is
subnormal in P n H. But P n H is a finite nilpotent
group and so each of its subgroups is subnormal. In particular,
<x) is subnormal in P n H, as required. [See Huppert:

Chap. 3].

The following lemma further describes p-classes, X,

as defined in Definition 4.3.

Lemma 4.11. Let X be a p-class. Then

(i) Rad 3P = 0 for some P 6 p(G) iff p(G) = {<1)}.

(ii) Any two members of p(G) are conjugate and,

conversely, if P belongs to p(G) so do all its conjugates.

(iii) If p(G) consists of a single normal subgroup P,

then p(G/p) = {<1>)-

(iv) If in addition Rad 3G = (Rad SPYJG V P E p(G),

then P g WgGeRad 3PERad 3w.
N¢(W)

Proof: (i) If p(G) = {<1)}, then certainly
Rad 3P = 0 -for each P E p(G). Conversely, suppose Rad 3P = 0

for each P e p(G). Then by Definition 4.3 (a),

59
Rad 3G E (Rad 3P)3G = 0. Thus Rad 3G = 0 and (36,3)

has prOperty p. By Definition 4.3 (b) P g_l V P e p(G).
G
This forces p(G) = {<1)}.

(ii) Let P1,P2

the pair (3G,3P1) has property p. So by Definition 4.3 (b),

6 p(G). By Definition 4.3 (a),

P2'é'Pl' Analagously, Pl é P2“ Thus P

conjugate. 0n the other hand, let P e p(G) and x 6 G.

1 and P2 are

NGw the map m :G 4 G given by p(g) = 9x is an automorphism
of G. By Definition 4.3 (d), o(P) = 9‘ e p(G). Thus

all conjugates of P belong to p(G).

(iii) This is immediate from Definition

4.3 (c).

(iv) Suppose Px g.W, where P E p(G) and
x e N¢(W). Then (Rad 31:)x = Rad as“ = Rad 39* n 3w =
(Rad 3P n szx g [(Rad 3P)3G n 3w1x = [(Rad 36) n 3w1x
E (Rad 3W)x. (The last inclusion follows from Theorem 1.7).

Thus Rad 3P E Rad 3w.

We cloSe this chapter with a theorem which relates the
semisimplicity of 3(G/H) to property p in the case that

G belongs to a p-class X.

Theorem 4ll2. Let X be a p-class of groups such
that ‘Rad 3G = (Rad 3P)3G V G e X and V P e p(G). Then for
each G e X

(i) 3(G/H) is semisimple iff G/H is a J(X)-group.

(ii) If H A.G, then (3G,3H) has prOperty p iff

G/H is a J(X)-group.

60
Proof: (i) It is clear that if G/H is a J(X)-group
then 3(G/H) is semisimple. Conversely, suppose 3(G/H)

is semisimple. Let T e X and S A.T such that T/S

H?

G/H.
Thus 3(T/S) is semisimple. Since X is closed under
homomorphic images, T/S e X. Let P e p(T/S). Then

0 = Rad 3(T/S) = (Rad 3P)3(T/S) = Rad 3P = 0. By Lemma 4.11
(i), p(T/S) = <1). It follows from Definition 4.3 (c)

that for each Q 6 p(T), Q g_S. Lemma 4.11 (iv) now yields
T

Rad 30 E Rad 3S. Thus Rad 3T = (Rad 3Q)3T E (Rad 3S)3T
and (3T,3S) has property p. By definition, G/H is

a J(X)—group.

(ii) Suppose H l.G such that (3G,3H)

has property p. By Definition 4.3 (b), P g_H for each
G

P 6 p(G). It follows from Definition 4.3 (c) that
p(G/H) = {<1)}. By Lemma 4.11 (i), 3(G/H) is semisimple.

By part (i), G/H is a J(X)-group. The converse is immediate.

BIBLIOGRAPHY

9.

10.

11.

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62

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