AUTOMORPHISMS nxma sumomu. a I;
AND NORMAL Macon ..  *'

"mu “Viki-Dug". 0‘ p'I. D _____ H A:
MICHIGAN STATE ummsm _ _ _
Afiphonse H. Burtmns " ‘

1966

 

 

THESIS ’ "ififii‘filfifig:

LIBRAE k 31" k3
Michigan Mata: 4
University

 
 
    

    
 

This is to certifg that the
thesis entitled

Automorphisms Fixing Subnormal
and Normal Subgroups

presented by

Alphonse H. Baartmans

has been accepted towards fulfillment
of the requirements for

Mdegree inmtics

J. Adney

Major professor

 

Dale W

 

 

 

 

 

 

ABSTRACT

AUTOMORPHISMS FIXING SUBNORMAL
AND NORMAL SUBGROUPS

by Alphonse H. Baartmans

a

One of the main objects in the study of finite group
theory is the group of automorphisms of a finite group.
Our interest here centers around the set of all automorphisms
that fix chains of subgroups of a group. In particular, we
will consider the set of all automorphisms B0 that fix
every composition series, as well as the set of automor—
phiSms that fix all chief series of a group. If s : G =
Go > Gl > --- > Gn = l is a composition series of a solv—
able group, we define recursively the following:

80(5) = [e e A(G>/Gi = Gi for i = 0,1,2, ..., n]

Sk(s) = {e e A(G)/9/Gk_1/Gk = 1 for 1 i k i n}.

If we let C(G) denote the class of all composition

series of a finite solvable group, we may define B0

0 50(8) as well as B. = fl S.(s). In a similar
1 l
S€C(G) s€C(G)
manner, if we let D(G) denote the class of all chief
series of a group G, we may define A0 = 0 80(5) and
seD(G)
Ai(G) = fl Si(s).

seD(G)

In the course of study of a set of automorphisms E,
we are led to consider two special subgroups of the group

G; the group F(G;E) which consists of all X e G such

that x9 = x for all

9 e E, and the group M(G;E)
(XeX-l / x e G; 6 e E). If one can say something about

1

 

Alphonse H. Baartmans 2

the groups F(G;E) and M(G;E), then presumably one can
say something about how the automorphism acts on the group
G. In Chapter I, we will prove some elementary results
about the groups F(G;E) and M(G;E).

In Chapter II, we determine what conditions are imposed
on the groups B1 if we assume that G is abelian, nil-
potent, supersolvable and finally solvable. Some of the
results obtained are:

(1) If G is nilpotent v-group, then:

(1) B0 is abelian

(11) B a B = ... = B and is a v-group.

l 2 n
Our main interest is to determine what the structure
of the groups M(G;BO) and Bo must be if G is a solv-
able group. To this end we prove the following results for
M(G;Bo)x
(2) Let G be an arbitrary group.
(i) If H is a nilpotent subnormal subgroup
of G, then M(G;Bo) normalizes every
subgroup of H.
(ii) If H is an abelian normal subgroup of
G, then M(G;Bo) centralizes H.
By using result (2) effectively, we obtain the follow—
ing result for M(G;Bo) for a solvable group G.
(3) If G is a solvable group and F* denotes the
Fitting subgroup of G, then:
(i) M(G;Bo)' is an abelian group

(11) M(G;Bo)' _<_ Z(F*)

 

Alphonse H. Baartmans 3

(iii) M(G}Bo) is a normal subgroup of G, and is
nilpotent of class 3 2
(iv) M(H(G;Bo);Bo) 5 z(r*).
By using result (3), we are in a position to character-
ize the group BC by means of the following results:

(4) If G is a solvable v-group then:

(1) B0 is supersolvable
(ii) B; is an abelian v-group
(iii) Bo normalizes every subgroup of B8
(iv) Bo has a unique maximal v-subgroup B3,

which is the Hall‘W—subgroup of Bo'
(v) A v'-subgroup of B0 is abelian.

Upon completion of the above, we turn our attention
to the groups A1. Some of the results obtained for A1
are the following:

(5) If G is a p-group, then:

(i) The Sylow-p subgroup, Ag of A0, is nor—
mal in A(G)
(ii) A; is a p-group of class i n-1 and
Aég
(111) A; . An ,
(iv) A p'-subgroup of ,A0 is abelian
(v) A0 splits over A:
(vi) If H is a normal nilpotent subgroup of
A0, then H is a p-group
Next, we investigate AO(G) in case G is nilpotent

or supersolvable.

 

Alphonse H. Baartmans 4

We conclude the Chapter with a theorem relating A0,

B0 and I(G). We obtain:

(6) If G is a solvable group, then:

(i) Every normal subgroup of I(G) is normal
in A0
(ii) B0 belongs to the norm of I(G).

In Chapter III we try to determine what conditions are
imposed on the group G, if we assume that G admits an
automorphism that fixes all subnormal subgroups. In particu—
lar, we will investigate how the groups F(G;E) and M(G;E),
for E, a subgroup of B0, are imbedded in G. We obtain:

(7) If e E BO(G) and F = F(G;6) and M = M(G;e),
then the following are equivalent:

(i) F n M = 1
(ii) G is generated by F and M
(iii) G is a semi—direct product of F and M
(iv) M = M(M;6) = M(F*;6), where F* is the
Fitting subgroup of G.
(8) Let E be a subgroup of B0(G) and let F = F(G7E)
and M = M(G;E). If G is generated by F and M, then:
(i) F n M = 1
(ii) M = M(M,-E) = M(F*;E), where F* is the
Fitting subgroup of G
(iii) M is a Hall subgroup of F* and M : Z(F*)
(iv) F* is generated by M and F(F*7E)

(v) Every 9 e E is a power automorphism on F*.

Alphonse H. Baartmans 5

In Chapter II, we show that a group may admit an auto—
morphism 6 6 Bo such that (16],IGI) = 1. For these types
of automorphisms, as well as a more general class of auto—
morphisms, we obtain:

(9) If e e Bo, such that (|e[,|M(G;e)|) = 1, then:

(i) All conclusions of (7) hold.

(10) If E is a subgroup of B0, such that
([E],[M(G;E)() = 1, then:

(i) All conclusions of (8) hold.

(11) If (IBOI,IG|) = 1, then:

(i) G = M(G;E) >< F(G;E) and M(G;E) : Z(G)
and F(G;E) : G'.
(ii) F(G;E) is a normal Hall subgroup of G.
(iii) M(G;B0) is abelian and its p—Sylow sub—
groups are elementary abelian
(iv) Bo(F(GrBo)> = 1-

Next, we investigate the inner automorphisms of G
that fix all subnormal subgroups. In particular, we study
the group H , having the property that if g e N, then the
inner automorphism induced by g fixes all subnormal sub—
groups. We obtain:

(12) If G is solvable, then:

(i) E is supersolvable
(ii) If H is subnormal in N, then H is nor-
mal in N
(iii) E' is abelian and every subgroup of NI

is normal in N

 

 

 

 

of

Alphonse H. Baartmans 6

(iv) E : Z(F*).
We conclude the Chapter by investigating the structure

<

of G, if we assume that I(G) _.B0 or, equivalently, that

N=G.

AUTOMORPHISMS FIXING SUBNORMAL
AND NORMAL SUBGROUPS

BY

Alphonse H. Baartmans

A THESIS

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

DOCTOR OF PHILOSOPHY

Department of Mathematics

1966

 

encox
Dr. L
duril

was <

his

PRE FACE

The author wishes to express his appreciation for the
encouragement and advice given to him by his major professor,
Dr. J. E. Adney. This advice and suggestion, given to me
during the research and in the preparation of this thesis,
was of immeasurable value.

A note of thanks should go to Dr. W. E. Deskins, for

his assistance in the completion of the dissertation.

ii

 

 

INTR'

CHAP

CHAP

CHAP

INDE

BIBL

APPE

TABLE OF CONTENTS

Page
INTRODUCTION . . . . . . . . . . . . . . . . . . . 1 g
CHAPTER I . . . . . . . . . . . . . . . . . . . . 3
CHAPTER II . . . . . . . . . . . . . . . . . . . . 9
CHAPTER III . . . . . . . . . . . . . . . . . . . 43
INDEX OF NOTATION . . . . . . . . . . . . . . . . 71
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . 73

APPENDIX . . . . . . . . . . . . . . . . . . . . . 74

 

automc
subgrc
attem
compo:
of all
0.

we do
of a
to cc

grou;

<x°:

alarm

on t‘
abel
For
of .t
We 1
0&3:

the

DOB

INTRODUCTION

Our main object of investigation is the study of the
automorphisms of a finite group G that leave chains of
subgroups invariant. In particular, we will center our
attention on the set E5 of all automorphisms that fix all
composition series of a group, as well as on the set A0
of all automorphisms that fix all chief series of a group
G. _

In Chapter I, we define the groups E5 and A0 and
we define recursively the groups A; and Bi . In the study
of a set of automorphisms E of a group G, we are led
to consider two special subgroups of the group G; the
group F(G;E) , which consists of all x e G such that
Xe - x for all e e E, and the group M(G;E) -

(xox-l /x e G, e e E) . In Chapter I we will determine some
elementary properties of the groups F(G;E) and M(GsE)-

In Chapter II, we determine what aanditions are imposed
on the groups Bi if we assume that the group G is
abelian, nilpotent, supersolvable and finally solvable.

For a solvable group G, we shall determine the structure_
or the group Bo and the structure of the group M(GiBo)~
we will then focus our attention on the group A0 for the
case that G is a p-group and after that, for the case
that G is nilpotent. P

In Chapter III, we determine what conditions are im-
POBed on the group G, if we assume that G admits an

1

-—-—— ———— —- --—
a --_—.-——.-—u————

that

ticul
pr0p<
duce<
elude
solv.

all

2
automorphism that fixes all subnormal subgroups. In par—
ticular, we will try to see how the groups M(G;B0) and
F(G;BO) are imbedded in the group G. We shall see in
Chapter II that a group may admit an automorphism 6 6 B0,
such that (16),]Gl) = 1. These automorphisms, as well as
a more general class of automorphisms, impose strong condi—
tions on the group, as is shown in Chapter III.

We turn our attention next to the inner automorphisms
that fix all composition series of the group G. In par-
ticular, we will investigate the group N , having the
property that if x e N then the inner automorphism in-
duced by x fixes all subnormal subgroups. We will con-
clude the chapter by investigating what the structure of a
solvable group must be if every inner automorphism fixes

all subnormal subgroups.

 

 

If
automor;
automorl

we will

33
a or 1‘
fine
shall d
element

h°=h.

chairs
Bub81‘01

01‘ the
Which
. 30(5)
31(5)

CHAPTER I
BASIC PROPERTIES AND DEFINITIONS

If G is a group, we let A(G) denote the set of
automorphisms of the group G and I(G) the set of inner
automorphisms of G. If a is an automorphism of G ,

we will define invariance under a as follows:

Definition 1.1: A subgroup H is invariant under

 

c or fixed by a iff for every element h of H , ha
is an element of B. If H is invariant under a, we
shall denote this by H“ - H. A subgroup H is fixed
elementwise by c iff for every h , and element of H ,

ha-h.

We will be concerned with automorphisms that leave
chains of subgroups invariant. For an arbitrary chain of

Subgroups of a group G, we define the following:

Definition 1.2: Let s be a chain of subgroups

 

of the group G of length n) s: a e Go>G1> 0,2 -...>Gn .. l

which terminates in the identity. As in [43 we define

30(5) = [0 e A(G) / G: - G1 for i - l, 2, --~, n} and

, e .
31(3) - {0 e 81-1(s) / (Gix) - Gix for all x e Gi-l}

Definition 1.3;, If s is a chain of subgroups of a
group G: s: G£> Gl °°'>'Gn_1 > an - 1. We define as in
E1] the stability group A(s) of the chain 8 as follows:

 

0: we

I)
Helm

Eroup

.{9

u
A(s) - {e e A(a) / (01x)e - Gix for all x s a
and for i - 1, 2, ..., n).

1-1

We note that the stability group of a chain 3 of
length n is the group Sn(s) of Definition 1.2.

For an arbitrary chain 5 of length n, terminating
in the identity, Philip Hall in [1] has shown that A(s)
is nilpotent and of classsl/Z n(n—l).

As in [4] we will be mainly concerned with chains of
normal and subnormal subgroups of G; in particular, chief
series and composition series of G. If we let C(G) de-
note the class of all composition series of G, we may

define B°(G) thus:

Definition 1.4: 30(0) - B€Q(G) 30(5) and 31(a) a

 

n -
s£c(a) 81(3)} 1 l, 2, see, no
As in [4] we have that each B1 = Bi(G) is a normal
Blbgroup of the automorphisms group of G and that Bo -
9

30(0) - {9 s A(G) / H - H for all subnormal subgroups H
of a}.

‘ If we let D(G) denote the class of chief series of

G, we may similarly define AO(G)I as follows:

ngfinition 1.5: AO(G) - se9(a) 30(3) and A1(G) -
n
IED(G) 31(3)°
Again we have that each A1 - A1(G) is a normal sub-

group of the automorphism group of G and that A0 a AO(G)
‘ {0 e A(G) / Ha - H for all normal subgroups H of G}.

 

refe

ever

{6 E

grou
grou
subg

of

ther

(or
Sinc

Subx

and

136‘

fou.

'1.

5
In what follows, we will have frequent occasion to

refer to the following subgroup of the automorphism group:

Definition 1.6: The dilation group A(G) of a group
G is the set of all automorphisms 6 of G that leave

every subgroup of G invariant. In other words, A(G) —

{6 e A(G)/H6 = H for all subgroups H of G].

If 6 is an automorphism of G that fixes all sub—
groups of G , then 6 will surely fix all subnormal sub—
groups of G. Furthermore, if 6 fixes all subnormal
subgroups of G , then 6 must fix all normal subgroups
Of G. Therefore A(G) : B0(G) :.A0(G).

We will exhibit the subgroups B0, A0 , and A(G)
by means of two examples.

—1
Example 1: Let G = <a,b/a3 I b2 = 1, b ab = a2>,

then G is the symmetric group on three letters, A(G)
<O,B/os = 62 = 1; 3—105 = Q2> where a: a —> a ; b 4> ab
6: a —> a2; b —> b.
Since the alternating group on three letters is the only
subnormal subgroup of G , we must have A(G) = A0 = B0,
and A(G) = 1.
The next example shows that A0, B0, and A(G) may all

be distinct.

be the alternating group on

Example 2: Let G = A4
four letters. A4 = <a,b,c/a2 = b2

C-lbc = a> then A0 = A(G). B0(G) = ((1,6) where:

c3 = 1; c_ ac - ab;

 

A(G) o.
for all
element

11(9):)

g
A(G) c
Mun
Flier c

be done

then 1
1' 11W:

V110“ ;

“nub

”here!

A

6
c: a -—9 a: b -—9 b: c -—e as
B: a -> a: b -—9 b: c -> be
A(G) - 1, and we have that A(G)$ 130(0) 5,. AO(G).
In what follows we will have frequent occasion to use

the subgroups “(0:3) and F(G:2) which are defined below:

ggfinition 1.1: Let E be a complex of elements of
A(G) of a group G: an element ge G such that go - g
for all 0 c E is called an 3-fixed element. The E-fixed
elements form a group F(G:I), the E-fixed subgroup and
I(Gjl) - {g e G / g° - g for all 9 a I}.

nginition 1.8: Let E be a complex of elements of
A(G) of a group G. Then an element gag.l is called an
l-multiplier element. The group generated by the E-multi-
plier elements is called the B-multiplier group and will

be denoted by n(ms) - (3934/3 e o and e s s> .

Theorem 1.9: If E is a complex of elements of A(G),
than n(axs) is a normal subgroup of G. Moreover, M(G3£)
is invariant under E and is the smallest normal subgroup

whose factor group remains fixed, elementwise, by E.

- h
Proof: Let g e G. Consider (gag 1) where h is
0 -1 h -l'9 -lh
an arbitrary element of G. (g g ) - h g g

_ h-1h0(hO)-1808-lh _ h'1h°(h‘1g)°(h'1g)'1
- ((n‘l)°n)'1(n‘ls)°<n‘1s)‘l
therefore (Earl)h ¢ n(am). Since M(G_3E) -‘<s°s‘l/geh: can) .

we must

sequent]

Them

“(033).

Obnsequ

In

o/n it

13am

'for er]

(8°s'1,

or A(
963.

Them

(8 g 3
new
and '

7
we must have n(c;n)h - n(G3E); but h was arbitrary. Con-
sequently n(sz) is normal in G.

Let gag"l be a generator of n(GzE), and let a e E.

fhent (cos—l)“ - so“ (am)-1
- so“ (89)”1 8° (8'18) (30')-1
- [(ae)a (3°)-13 (sea-l) (sag-l)'l-

Since (go)a(g°)-l, gag"1 and gug-l all belong to
I(Gzn), we must have that (gag-l)“ is an element of M(G;E).
Consequently, n(ng) is invariant under E.

Suppose H is a normal subgroup of G such that
a/H is fixed elementwise by x. rhls implies that it fig
is a coset of H in G, then (Hg)e - ng - Hg or geg-1 e H
for arbitrary g e G and e c E. Hence M(G;E) -

(8.3'1/3 e e; o e 3) SH and the result follows.

Definition 1.10: E is said to be a normal complex
or 11(0) 1: c‘lec e s for all a e A(G) and for all

 

Theorem 1.11: Let E be a normal complex of A(G).

 

lhen: (1) I(Gsn) is characteristic in G.
(2) F(G}E) is characteristic in a.

2322;: Let g e G, 0 e E and a e A(G). Then
(898-1)“ - (go’s-10a(8a)-l, but c-19a 6.3 since E, is a
normal complex of ‘A(G); consequently (gag-1)“ e M(GJE)
and n(GxE) is characteristic in G.

 

subgr

then:

'C
gig

Ther:

M(G;]

<15

Cons

8

ml

Let g e F(G;E) and a e A(G). If e e E. Geo—1 =

is an element of E. Therefore as = 55. Hence (g0)e =

g _ _
(g )0, but 9 e E whence g6 = g. Hence (go>9 = (gG)o

= g01 or g0 E F(G;E) and F(G;E) is characteristic in G.
The next theorem shows that we only need to consider

subgroups of A(G).

Theorem 1.12: If E is a complex of elements of A(G),

then; (1) M(G;E) = M(G;<E>)

(2) F(G;E) - F(G;<E>).

Proof: If Q and 5 are elements of E, then

9‘19"B = (91‘ng) [(ga)'1<ga)61 e was).
Therefore M4G;<E>) Z M(G;E). Since E i (E), we have
M(G7E) : M(G7<E>) and therefore M(G;E) = M(G;<E>).

If g e F(G;E), then, for a and 6 elements of E,
9&6 = (90)‘3 = gQ = g and therefore F(G;E) : F(G;<E>).

Consequently F(G;E) = F(G:<E>)-

ties

exte

mine

and

to

If

If

CHAPTER II
STRUCTURE AND PROPERTIES OF THE GROUPS Bi AND Ai

Introduction:

The aim of this chapter is to investigate the proper—
ties of Bi and Ai' We will start with abelian groups,
extend our arguments to nilpotent groups and finally deter—
mine the structure and certain properties of the Bi and

Ai for supersolvable and solvable groups.

I. The Structure and Properties of the Groups Bi’ i = 1,2,

., n.

Theorem 2.1: If G is a direct product of groups H

and K, E a subgroup of Bo, E the restriction of E

H
to H, EK the restriction of E to K, then:
(1) M(G;E) = M(H7EH) x M(K7EK)
(2) F(G;E) = F(H;EH) x F(K7EK).
If ([H|,|K|) = l and E = B0, then:
(3) A(G) = A(H) x A(K)
(4) Bo(G) :.B0(H) x B0(K).
If G is nilpotent and (]H],lK]) = 1, then:
(5) B0 (G) = B0 (H) X B0 (K)
(6) A(G)/130(6) = A(H)/B001) >< A(K)/B0<K>

(7) If M(H;B0(H)) : z(H) and M(K;B0(K)) : Z(K),
then: M(G;B0(G)) : Z(G).

9

aunl'
kEI‘

conse

9
99

to

q E

and

norm

he

Cons

10
(8) If F(H;B0(H)) 3 H' and F(K;B0(K)) : K', then

F(G7B0(G)) : G'.

Proof: If G = H x K, then every element g e G has
a unique representation of the form g = hk with h e H,

k e K. The groups H and K are normal subgroups of G;

consequently if 6 e E then H6 = H and K9 = K. Let
969—1 be an E—multiplier element of G. Then
geg‘1 = (hk>9(hk)‘1
: hekek-lh—l
= heh-lkek—1 since the elements of

—1 —1
H and K permute. The elements heh and kek belong

—1
Consequently M(G;E) = <geg /

to M(H;EH) and M(K;EK).

' < O O U
g e G; e e E> - M(K,EK) x M(H,EH). Conversely M(H,EH)
and M(K;EK) are subgroups of M(G;E); moreover, they are

normal subgroups of G. For if g e G, then g = hk with

h E H, k e K. If u e H, then
- - — —1 —1 e
9 1(u 1u9)g = k 1h (u u )hk
= h—1(u—1ue)h
-1 6 —1 G p
= [(uh) (uh) ](h ) h e M(H;EH).

Consequently M(H;EH) = <u9u_1/u s H; 6 e E) is normal in G.

By a similar argument we obtain that M(K;EK) is normal in

) < H and M(K;E : K, we have that
H _

<
. . — 1.
M(H,EH) n M(K,EK) _ H n K

K)
Consequently the union of

G. Since M(H;E

 

Since

Cons

argu

Ther

are

morg

CD
1

ant:
is.

If

ll
M(H;EH, and M(K;EK) is a direct product and M(H;EH) X
M(K;EK) : M(G;E). Therefore M(G;E) = M(H;EH) x M(K-E )
I K .
If g e F(G;E), then g = hk with h e H and k e K.

Since _ 6 -1
1 - 9 g

_ 6 —l

— (hk) (hk)

_ 6 —1 —

— (h h ) (kek 1), the last equality
implies that heh‘1 = 1 and kek‘1 = 1. Hence h e F(H;EH) i
and k e F(K;EK), and therefore F(G;E) : F(H7EH) x F(K;EK).

Conversely F(H;EH} and F(K;E are normal subgroups of

K)

F(G;E); for if h e F(G;E g = hlkl e F(G;E), then

H),
g’lhg = k;lh;1hh1k1 = hIlhhl.

Consequently F(G;E is normal in F(G;E). By a similar

H)

argument we obtain that F(K;EK) is normal in F(G;E).

Therefore F(G;EH) x F(K;EK) : F(G;E) and (2) follows.
Let G = H x K with (|H],|K|) = 1, then H and K

are characteristic subgroups of G. If 9H is an auto-

morphism of H and 6K is an automorphism of K, then

9 = 9H ' 9K is an automorphism of G. The product
9 = 6H ' 6K is to be interpreted as follows: 6 acts on
H as 9H does and on K as 6K does. Then 9H is an

automorphism of H. Similarly 6 restricted to K, 9K,
is an automorphism of K. Therefore A(G):.A(H) x A(K).

If 6 is an automorphism of G then 9 restricted to

H, 9H, is an automorphism of H. Similarly 9 restricted
<

to K, 6K, is an automorphism of K. Therefore A(G) _

A(H) X A(K) and therefore A(G) = A(H) x A(K).

 

 

fc

dc

DC

h:

h

12
Every automorphism 6 6 Bo may now be written in the

form 9 = 9H - 9K where 6H = G/H and 6K = e/K. If 9

H
does not fix all subnormal subgroups of H, then 6 does
not fix all subnormal subgroups G, similarly for 9K
Therefore BO(G) : B0(H) x B0(K).

If 9H e 30(H), 9K e B0(K), where 9K induces the

identity on H and 6 induces the identity on K, then

H
6 = 6H ' 9K is an automorphism of G. If G is nilpotent,
by Theorem [T—l], every subgroup of G is subnormal in G7

hence if 9 e B0(G), then 9 must fix all subgroups of G7

hence if y e G, then ya = ys<Y7e>, where (s(y;9). [Yl)

= 1. If g = hk is an element of G, then

9
96 = (hk) -
e e
If e = 9H - 9K then g9 = h Hk K
= hs(h;6) ks(k7e), where

(S(h;e),|h]) = 1 and (s(k;e),jk|) = 1. If t is an

integer, then gt = (hk)t = htkt. Consider the system:
t E s(h;6) mod 1h]
t E s(k;6) mod [k].
This has a unique solution, t E to mod lg] = lhllkl.
t t to
e e e _ s(h;6) s(k,—e)= o o: _
Consequently g6 = (hk) = h k — h k h k 9

Hence 9 fixes every cyclic subgroup of G and 6 15 a

dilation of G. Consequently B0(G) = B0(H) X B0(K), and
(5) follows.

From (3) and (5) we obtain (6).

 

 

 

wit]

of

If

If

Th:

Wi'

gr

Of

St

l3
z(G) = z(H) x z(K) and G' = H' x K', together

with (1) and (2) give (7) and (8).

Corollary 2.2: If a group G is the direct product

of subgroups H1, H2, ---, Hn' E a subgroup of B0 and

Bi the restriction of E to Hi’ i = 1,2,°--,n, then:
(1) M(G7E) = M(H17E1) X M(H27E2) --- X M(Hn7En>' \
(2) F(G;E) = F(H1;E1) x F(H2;E2) ... x F(Hn;En).

If the orders of the Hi are relatively prime in pairs and

E = 30(6), then:
(3) A(G) = A(Hl) x A(Hz) ... x A(Hn)-
(4) B0(G) : Bo(H1) X B0(H2) --- X 30(Hn)'

If the orders of the Hi are relatively prime in pairs and
G is nilpotent, then:

(5) B0(G) = B0(H1) X B(H2) --- X B(Hn)-

(6) A(G)/50(G) = A(H1)/BO(H1) X A(H2)/B0(H2) --- X

A(Hn)/B0(Hn).
(7) If M(Hi:Bo(Hi)) : z(Hi), then M(G;B0) : Z(G).
(8) If F(Hi;B0(Hi)) 1 Hi, then F(G;B0) : G'.
The proof of the Corollary is the same as that of Theorem 2.1.
With the aid of Theorem 2.1 and Corollary 2.2 we are now in

a position to discuss the Bi for abelian and nilpotent

groups.

(A) B0, B1, B2, ---, Bn for abelian groups.

If G is an abelian group, then G is a direct product

of its Sylow-subgroups. Since the orders of these Sylow—

SUbgroups are relatively prime in pairs, we may apply

 

 

Cor

shc

thi

l4
Corollary 2.2, and it suffices to consider Bo for abelian
p-groups. If G is an abelian group, then B0(G) = A(G),
for if U is a subgroup of G, then U is normal in G7
consequently U6 = U, for 9 e B0(G) and 9 e B0(G) fixes
all subgroups of G; therefore B0(G) : A(G). But A(G) :
B0(G), therefore BO(G) = A(G). R. H. Jaschke in [3] has

shown:

Theorem 2.3: If G is an abelian group, and A(G) is
the set of dilations of G, then
(1) A(G) is an abelian normal subgroup of A(G).
(2) If 9 e A(G) then 6 has the form g6 = gs(6)
where 5(9) is an integer relatively prime to
the exponent of G.

(3) A(G) is isomorphic to the prime residue classes

modulo the exponent of G.

B0:

Corollary 2.4: If G is an abelian group, then
A(G), the set of dilations of G, and B0(G) satisfies all
conclusions of Theorem 2.3.

Proof: This follows immediately from the discussion
preceding Theorem 2.3.

Definition 2.5: If G is a group, the subgroup of G
is called the

generated by all the non-generators of G,

Frattini subgroup G.
It can be easily shown that the Frattini subgroup of

G is the intersection of all maximal subgroups of G. The

Frattini subgroup of G will be denoted by ¢(G).

15
We will next state two Lemmas that will be needed in

what follows.

Lemma 2.6: If G = H1 x H2 x H3 x -'- x Hn, then
¢(G) = ¢(H1) x ¢(H2) x --. x ¢(Hn).

Proof: The proof of this may be found in [6, page 165].

Lemma 2.7: Let G be a group and let U be a subgroup of
A(G) such that [U] = pk where k is an integer and k i 1.
Let M1 and M2 be subgroups of G with M2 4 M1. If
M1 = M1 and M: = M2 for all 9 e U and le/le = p,

then every 9 e U must induce the identity on Ml/Mz-

Proof: Let 6 e U; let sz be a coset of M2 in

M1; then (sz)9 = sze. Since [Ml/M2) = p, we must have

sze = szk where (k,p) = 1 and k < p. Consider the

n
action of 6p on Ml/M2=

n
P . n
(M2X)e : M2X Since 19] = p ‘
On the other hand:
1'] 1'1
P 9 ep‘1
(sz)9 = (MZX )
n
p—1
k 6
= (M2X >
P
= M2 Xk

but by Fermat's Theorem

n
kp E k mod p.

Consequently:

P
k ) _ M2Xk
Therefore:
sz = sz or k = l and therefore

(sz)e = sz and 6 induces the

identity on Ml/Mz and the lemma is proven.

We are now in a position to apply our results to
abelian p—groups and abelian groups in general. We will
first focus our attention on abelian p—groups and then

generalize it to arbitrary groups.

Theorem 2.8: Let G be an abelian p—group, then:

(1) B1 is an abelian p-group

Proof: Let G be an abelian p—group, then B1(G) is
a subgroup of B0(G). By Theorem 2.3, we have that B0 is
an abelian group, hence B1 is an abelian group. Let

6 e B1(G), then 6 induces the identity automorphism on

G/M for every maximal normal subgroup M of G. Let MX

be a coset of M in G, where x is an arbitrary element

9 - .
of G. Then (Mx)e = Mxe = MX, hence x X 1 e M. Since M

9-1_ .
and X are arbitrary, we must have x x e n{M/M max1mal

normal}. Since all subgroups of G are normal, we have

6 -1 _
fl{M/M is maximal normal} = ¢(G). Hence x x e ¢(G) and

therefore M(G;B1) : ¢(G). By Theorem 1.9, B1 must induce

the identity on G/¢(G). By Theorem [T—2] we have that B1

is a p-group. Consider now two subgroups M1 and M2 of

17
G, where M2 is a maximal normal subgroup of M1‘

If 9 6 B1 , M? = M1 and Me = M Therefore 9 e B

2 2' 1

must induce an automorphism on Ml/Me. Since 'Ml/Mel = p
and [GI = pt for some positive integer t, we must have by
Lemma 2.7 that G induces the identity on Ml/Mg' Since
M1 and M2 were arbitrary, 9 must induce the identity
on all composition factors of G. Therefore 9 e Bn and
consequently Bl'é Bn. Since Bn S Bl by definition, we
must have B1 = B2

Although Bl £ B0 in Theorem 2.8, we may not conclude

«or = Bn’ and the Theorem follows.

that Bo - Bl’ as may be seen from the following example.

Example 3: Let G be the elementary abelian group
of order 9. Then G = <a,b/[a,b] = l, a3 = b3 = 1). Then
B0 = <q/a2 = 1), where aa a a2 5 ha a be. On the other
hand, if e e 131(0), then 9 must induce the identity on
Q/¢(G). Since 4(0) = l, we must have that 0 must induce
the identity on G. Consequently 9 = 1 and therefore

Bl = 1. We have therefore that Bl(G) i BO(G).

Definition 2.2: Let 7 be a non—empty set of primes.

A v-number is a natural number each of whose prime factors
13 in w: a w-element of a group G is an element whose
order is a w-number; a w-group is a group each of whose

elements is a n-element.

For any set of primes ‘w, the number 1 is a w-number;
the element 1 is a s-element; and the subgroup 1 is a

v-subgroup.

 

 

by

oh:
the
is

18
The set of all the primes not in y will be denoted
by 7'.

Theorem 2.10: Let G be a v-group. If s is a

 

chain G - Go:> G1 > °°°> Gn = l of subgroups of G, then
the stability group A(s) is a v-group. Furthermore, if s
is a chain of length 2, say 8 : G > 61 > G2 = 1, then

A(s) is a w-group if G1 is a r-group.

';rro_of: Let [cl = p, e e A(s) and (p,]G|) = 1.

Apply induction on the length of the chain. Since 9
fixes the chains 8' : G1 > 02 > G3 °°°> Gn = l, we have
‘9 e A(s’). Since (|G|,]G2I) = l, e restricted to G1
is the identity.

Let g e G, then geg-l e 01. Let g6 = gx with
x 6 G1: then gep - g; hence gOp = gxp, and g = gxp
and xp = 1. But (|p|,lGl|) = 1, hence xp f 1 unless x = 1.
If x a 1, then g9 = gx = g and 6 is the identity. The
first part of the theorem follows by induction.

Let s be a chain of length 2, say s: G > G1 > G2 = 1.
Let he 6 A(s) such that [9] = p and (p.|cl|) = 1. If
8's 0, then geg'l a G1 and $9 = 318 whgri 31 e Gl-
Since '9] = p,_ gep a g; but geyemis)?~ = 858..
Therefore g - gfg and gf'n 1. [Sinbe (PalGll) = 1:
sf ,4 1 unless 81 =- 1. JIf (:1 =1. then 89
and 9 is the identity. Therefore ifGl is a w-group,

then A(s) is a w-group.

19
Theorem 2.11: If G is an abelian w—group,then:
(1) Bo is an abelian group
(2) BI is a v—group

(3) B1=Bz=...=B.

Proof: The group Bo is a subgroup of A(G) so by

I
Theorem 2.3 we have that A(G) is an abelian group and
consequently Bo is an abelian group.

By Theorem 2.3 every' 6 e B0 has the form g9 = g5(6)
where 5(6) is an integer relatively prime to the exponent
of G. We will show that if 6 6 B1, then 5(6) E 1 mod pi

for each prime divisor pi of the order of G.

Let M be a maximal subgroup of G of index pi°

Let Mx be a coset of M in G. If 6 6 B1, then (Mx)e =
MXQ = MX. Since Bl : A(G), we must have that x9 = X5(6)
where (5(6). exp G) = 1. Consequently (Mx)9 = MXe = st(9).
Therefore st(6) = Mx ; hence X5(6)—1 e M. Since the

index of M in G is equal to pi , we must have 5(6)—1

= 0 mod pi and therefore 5(6) 5 1 mod pi“ Since G

is an abelian group, then G must have a maximal subgroup
of index pi for each prime divisor pi of G. Consequ-

ently 5(6) =-1 mod pi.for each prime divisor pi of the

order of G.

Let G./G.+1 be a composition factor of G of order
1 i
' . nd let G E B .
pi. Let Gi+1X be a coset of Gi+1 in G1 a 1
Therefore (G. x)6 = G. Xs(e). From the previous para~
1+1 i+1

graph we have that 5(6) 5 1 mod pi. Therefore 5(6) = 1+kpi

20
where ‘k_ is an inte er. C . 6 = 6
k : onsequently (Gl+1x) Gi+1X
1+ p. p- p- k
= i = i 1 _ i i
Gi+1x Gi+1x x - Gi+1(x ) x, but x e Gi+1
i k 6 k
d th f . - = p
an ere ore (x ) e Gl+1, hence (Gi+1x) Gi+1(x ) X

= G . We have,therefore,that each 6 6 B1 must induce

i+1X
the identity on each composition factor. Hence B1 i'Bn7
but,by definition, Br1 : Bl. and therefore B1 = Bn' Since

Bl 1 B2 1 ... Z Em and B1 = Bn’ we must have that
Bl-B2= ... ‘Bn.

The group B1 induces the identity on all composition
factors of G; hence Bl must belong to the stability group
of every composition series of G. By Theorem 2.10 this

implies that B1 is a w—group.

(B) B0, B1, ..., Br1 in case G is Nilpotent;

As was done for abelian groups, we will first discuss
the case for which G. is a non—abelian p-group, and then

generalize to nilpotent groups.

Lemma 2.12: If G is a nilpotent group,then

Bo(G) = A(G), the set of dilations of G.

Proof: If G is nilpotent,we have by Theorem [T—l]

that every subgroup of G is subnormal in G. Since B0

fixes all subnormal subgroups of G, we must have that B0

fixes all subgroups of G. Hence 6 e B implies 6 e A(G),
and B0(G) :.A(G). If an automorphism fixes all subgroups

Of a group G, then it surely fixes all subnormal subgroups

21
of G; therefore for any group G, A(G) i B0(G). Therefore

we must have that BO(G) = A(G), the set of dilations of G.

Next we will state a Theorem by Huppert [2] which will

be needed in what follows.

Theorem 2.13: If G is a non abelian p—groug then

 

BO(G) is a p—group.

Theorem 2.14: If G is a non abelian p—group,then
the following hold.
(1) B0 is a p—group

(2) B0=B1 ... =Bn.

Proof: By Theorem 2.13 we obtain (1).
Let 60 e B0. Let M1 and M2 be two subgroups of G
such that M2 :M1 and [Ml/M2] = p. Now M2 and M1

. 9
are subnormal in G. Therefore if 6 e B0(G), then M2 -

M2 , M? = M1 and consequently 6 must induce an auto-
_ t
morphism on Ml/Mz. But [Ml/le = p and [6[ — p for some

positive integer t. By Lemma 2.7 this implies that 6

must induce the identity on the factor group Ml/Mz. Since

9, M1, and M2 are arbitrarx,we must have that BO induces

- <
the identity on all composition factors.I{ence Bo _ Bn’ but

Bn i B0; hence B0 = Bn'

Definition 2.15: If G is a group, the intersection
of all the normalizers of all cyclic subgroups of G is

called the norm of G and will be denoted by N(G).

 

22
We will state some results which will be needed in

later development, which have already been shown by H. J.

Jaschke in [3].

Theorem 2.16: If G is a group then the following

 

hold:
(1) nus-A(G)) i N(G)
(2) G/F(G;A(G)) is nilpotent of class 2.
(3) A(G) n I(G) : Z(I(G)), hence A(G) n I(G) consist
of central automorphisms.

(4) M(N(G):L(G)) : Z(G).

If G is nilpotent,then BO = A(G) and the above
Theorem holds if we replace A(G) by BO. With the aid
of the above Theorem we are now in a position to discuss

Bo for nilpotent groups.

Theorem 2.17: If G is a nilpotent w—grouplthen if
noneof its Sylow subgroups are abelian:

(1) B0 is an abelian Tr—group°

(2) B0 = B1 ... = Bn.

If G has an abelian Sylow subgroup then:

(3) B0 is an abelian group

(4) B1 = B2 ... = Bn and is a w—group.
- = S X S "° X S be a decomposi-
Proof. Let G p1 p2 pn

Of G into a direct product of its Sylow subgroups. By

Corollary 2.3, we have that B0(G) = B0(Sp1) x B0(Sp2) °°°

). Since B1(G) i.Bo(G), every 6 s B1(G) may be
n

XB
0(Sp

23
written in the form 6 = 61 ‘ 62 "' am where 6. e B0(S ).
l p.
1
Moreover 6i acts trivially on Sp. for j #i .
3

We will show that 6i E 31(Sp ) for all i. Let Mi
1

be a maximal subgroup of S . Then M. = S X S X ...
pl 1 p1 p2

X Sp X Mi X S '-- X S is a maximal subgroup of G.
i—1 i+1 pn
Since 6 e 81(G), we must have that for every x e Sp ,
i

- Mixe = Mix. Consequently x9);1 5 M. for every

I
m
l

X e S , and therefore M(S :6) < M.. Since S is a
i pi ‘ 1 pi

characteristic subgroup of G, we have M(S 76).: S .

-Therefore M(S ;6) i M. 0 S = M.. Since M. was
pi 1 pi l i

arbitrary, M(S ;6) i M. for all maximal subgroups M.
pi l l
Of S . Since 6 restricted to S is equal to 6i
) j.Mi for all maxi—

we must have that M(Sp :6) - M(Spi7 i

mal subgroups M. of S . Consequently 6i induces the

1 pi
identity on all factor groups sp./Mi where Mi is a
l
maximal subgroup of Sp and therefore 6i 6 B1(Sp.>°
i , 1
By Theorems 2.8 and 2.14, we must have 6i e Bn(Sp ).
. i

Let Ml/Mz be a composition factor of G, of order pi.

We may choose xi 6 S

Let M2X be a coset of M2 in M1. pi

as a coset representative of sz. For if x e G, then

_ . ... h X. E S and
x — x1 ... Xi Xi+1 xn where eac ] pj
1’1
' - -~- . hen
lle = pjj . Let Y _ X1X2 ... Xi—l Xi+1 Xn T

24
mi pni- ni pni pni
x = xiy7 but xp = (xiy) l = xil y l = y l . Since
ni ni
pi p-
[Mi/M2] = pi’ we must have x 6 M2 and y l 6 M2.
n. n.
Since jyi] =‘v. pjjl we have (pil,]y]) = l and
n.
i

P- p.
(y l) = (y). However y l G M2; consequently y 6 M2.

he = . = ..
T refore M2x szxl szl

Let 6 e B1(G); then (sz)e = (szi)6 = sze. Since

1
9 6i

6 = 61 62 -°° 6. ... 6 and x. e S , we have x. = x. ,

i n 1 pi l i
9 91
and (sz) = szi. But 61 € B1(Sp ) and therefore
i
l6i] = p: for some integer t. Furthermore, [Ml/le = pi.

Therefore by Lemma 2.7, 6i must induce the identity of Ml/M2~

6.
6 l _ _
Consequently (sz)e = sze = szi = szi - MZXi ~ M2X,

and 6 induces the identity on Ml/MZ. The composition
factor Ml/M2 was arbitrary; consequently 6 must induce

the identity on all composition factors. Therefore

<

B1(G) :.Bn(G), but by definition Bn(G) _.Bl(G)/ hence

B1(G) = Bn(G).

If none of the Sylow subgroups of G are abelian, then

B0 = Bl. For if 6 6 B0! then 9 = 61 . 62 ... 6n where

‘ ' ’ - ' .- rou ,
9i 6 Bo(Spi). Since each Spi is a non abelian pl g p

we must have by Theorem 2.14 that 6i 6 Bn<sp.) for all i.
i

If M is a maximal subgroup of G of index pi’ we may

Choose x. e S as a coset representative of M in G.
l

<t>

 

<Szt>

25

9.
6 _ 6 _ i . _ t
Consequently (Mxi) - Mxi — Mxi . Since ]6i[ — pi for
some integer t and jG/M] = pi, we must have by Lemma 2.7
9 _ 9 _ i _
that Mxi — Mxi. Therefore (MXi) — Mxi - Mxi and 6

induces the identity on G/M. Since M was an arbitrary
maximal subgroup of G, 6 must induce the identity on
G/M for all maximal subgroups M of G. Consequently

6 6 B1(G). The element 6 was chosen arbitrarily and
therefore we must have BO(G) = B1(G). This establishes

the Theorem.

We shall now give an example to illustrate some of

these concepts and properties that were derived above.

Example 4: G a p—group.
Let G = <slt/ s3 - t2 = 1; t'ls t = 53>-
The lattice of subgroups (subnormal subgroups)

is the following:

<Sz,t> <S> <5215t>

 

\ .
<54,t> <54,szt> <s2> <s3t> gst>

<S4t> <s6t>

 

   
 

z(G> = (84>

B0 is

26

generated by the following automorphism:

a : t ——> t s ——> s3
V : t ~> t S —>. s 7 .

Therefore:

Bo = «haw/[mm = [my] = [Buy] = 1; a2 = r32 = y2 = 1).
Hence B0 is an elementary abelian group of order 80

M(GiBo) — <S4>

F(G;Bo) ‘ (s4,t>.

Example 5: G nilpotent. G = D8 x C9 where D8 is

the dihedral group of order 8

and C9 = <a/a9 = 1> is

a cyclic group of order 9. Then A(G) is generated by:
<6 : g6 — g5 for all g e G>,

1.8” A(G) = 30(5) = (9/96 = 1).
M(GiBo> _ C3 = <a3>
F(G;Bo) - D8.

All automorphisms of B0

The

(C)

In order
we shall need
in order that

development.

are power automorphisms.

group B0(G) for a solvable group G.

to discuss the case B0 for solvable groups

several Lemmas. We shall present those first,

they may be used for reference in the later

If G is a solvable group,then G pos—

Lemma 2.18:

sesses a non—trivial,normal,nilpotent subgroup.

 

27
Proof: Since G is solvable, G has a derived series

 

a - e°g c' a 0': ...z0“ - 1. Nov 01/01“ is abelian
and since every term in the derived series is characteristic,
is have Gn-l is normal in G. Hence Gn'l / 1 is a
normal nilpotent (abelian) subgroup of G.

Definition 2.12: The maximum, normal nilpotent sub—
group of a group G is called the Fitting subgroup of G.

Lemma 2.20: If G is a solvable group, then G has

a nontrivial Fitting subgroup.

3223;: By Theorem [T-i] the product of two normal
nilpotent groups is a normal nilpotent group of G. Since
G is finite, G has only a finite number of normal nil-
potent groups. The product of all the normal nilpotent
groups is the Fitting subgroup of G.

Lemma 2.21: Let G be an arbitrary group and let
M(GIBO) be the Bo-multiplier group of G. If H is a
subnormal nilpotent subgroup of G, then M(GsBo) normalizes H.

Proof: If H is a subnormal nilpotent subgroup of G,
then every subgroup of H is subnormal in H. Since H is

subnormal in G, we have by the transitivity of subnormality

that every subgroup of H is subnormal in G. Therefore if

9 ¢ 30(0), then a must fix all subgroups of H.
Let x e a and let h e K, then x'lhx e 3“. Since

H; is isomorphic to H, H3‘ is nilpotent. Furthermore,

28
since any automorphism maps a composition series onto a com-
position series, Hx must occur in a composition series for
G, and H; is subnormal. Therefore, Ex 13 a subnormal
nilpotent subgroup of G. Consequently, if 9 e B°(G), then
0 must fix all subgroups of 3". Then:
(x-lhx)o - (x-lhx)'(x-lhx’e) where (s(x-1hx,9),|x-1hx|) - 1.
0n the other hand (x.1hx)o - x‘ehexe
- x-ehs(h3°)xG where (s1h39),|h|) - 1.
Therefore (x-1hx)s(x-lhx’°) - x'°h°(h’°)x0
lhx36)x _ x-Ghs(h30)xe.
Hence xcx'lh'(x-lh13°) (x°x‘1)'1 - h°(h3°)

_ I '
x lhs\x

Since (s(x_1hx39),|x'lhx|) - 1, this implies that l
s(h30) and hs(x hx39)

(I(x‘lhxze),|h|) - 1. Consequently h 1

are generators of < h >. Therefore x x— normalizes every
subgroup of H. Since xox-l was an arbitrary generator
or M(Gxno), we have that M(Gjno) normalizes every subgroup

of H.

Lemma 2.22: Let G be an arbitrary group and let M
be the Bo-multiplier subgroup of G. If A is a normal
abelian subgroup of G, then M centralizes A.

Proof: If A is a normal abelian subgroup of G,

then eVery subgroup of A is subnormal in G. Therefore

every 0 e BO(G) must fix all subgroups of A. By
Theorem 2.3, e a 30(G) restricted to A has the form:
a. - a'(9) for all a e A, where (s(0), exp A) - 1. Let

x G G. 0 e B (G) and a e A, then x‘lax c A. Therefore

29
(x'lax). - (x'lax)’(°) where (5(0), exp A) - 1
- x-1 a'(°) x - x"1 a0 x
an the other hand: (x-lax)o - (x°)"l ao x9.
x-l) - a9.

Since ‘9 restricted to A is an automorphism of A, 191-1

Therefore (x°)'1 3.0 1° x'1 aox . or (x‘x-l)”1 a°(xO
must centralize all elements of A. Consequently,
n(cgso) -<x'1x°/ x e a, e e B°(c)> must centralize A.

Zflgggg§_g;gfix Let G be a solvable group, n = M(GzBo),
and let F* denote the Fitting subgroup of G, then:

(1) -M' is abelian

(2) M' - M(G;B°)' g z(pe)

(3) M is nilpotent of class 5 2

(4) 14013130) s 2(11*).

2222;: By Lemma 2.21, M normalizes every subgroup
of the F*. Every element of M induces an automorphism
0n F*, and since it also fixes every subgroup of F*,
it must induce a dilation on F*. Since the dilations of
a group form an abelian group, we must have that the inner
automorphisms induced by M' fix all elements of F*.
Therefore N' must centralize F*. By Theorem T~4 , the
centralizer of F9 is contained in F*3 hence M'.§ z(ra).
Since z(re) is an abelian group, we have M' is abelian;
hence (1)-and (2) follow. The center of) F* is a normal

abelian subgroup; hence, by Lemma 2.22, M must centralize

30
the Z(F*). Since M' : Z(F*), then M centralizes M'.
Hence M' is contained in the center of M. Hence M is
nilpotent of class 2 or abelian.

Let x e M, a e F*, and 6 6 Bo, then:

If x 6 M, then x induces a dilation on F*, and since
the dilations of a nilpotent group form an abelian group,

we must have:

_1 _
(x ax)e = x 1 x
Consequently
6 -1 6 6 _ -1 6
(x ) a x - x a x
or ((Xe)x_1)—1a6((xe)x_1) = a6 or xex—1 centralizes
F*. By Theorem [T—4L we must have xex_1 : Z(F*). There—
fore M(M;Bo) = <x9x‘y x c M, e eBo> : Z(F*).

Theorem 2.24: If G is solvablelthen the following

hold;

(1) B0 is solvable

(2) B6 is abelian.

Proof: If G is solvable,then G has a composition
series whose factors are of prime order. Let G =
Go > G1 > 62 > -~- > Gn = 1 be such a composition series
su h th t G /G I = . By definition G9 = G. for all

C a li 1+1 pi 1 1

6 6 B0. Hence Bo must induce an automorphism on each of

. ' hese rou 5 are of
the factor groups Gi/Gi+1 Since t g p

order pi’ their automorphism group has order pi—l and

31
is cyclic. Hence 86 must induce the identity on all such
factbrs. Hence 85 is nilpotent of class n-l by the
Theorem of Philip Hall [T-5]. But this implies that B0
is solvable. We can,however, obtain a better condition on
the nilpotent class of B5. Since BO fixes G/M.@;Bo),
and the restriction of B0 to M(G;B0) is abelian, we
have that 85 fixes G/M(G;Bo) and M(G;B0). Hence B5
fixes a chain of length 2. By Philip Hall's Theorem [T—5],

we must have that B6 is abelian.

Lemma 2.25: Let H be a normal subgroup of G. Let
S be the subgroup of A(G) consisting of all automorphisms
of G such that M(G;S) i H and such that H i F(G;S),
then:

(1) S is abelian

(2) M(G7S) : z(H).

Proof: (1) follows immediately from the Theorem by
Philip Hall [T—5], since S is contained in the stability
subgroup of the chain G > H > 1.

6 .
Let h 6 H, g e G then [h,g] = [h,g] Since H 4 G.

On the other hand: _1 _1 9
[h,gJG = [h g hg]

Since gegn1 e H, let g9 = kg with k e H, then:
—1 6 —1 6
[he]6 = h (g) h g
— —1
= h 1(kg) h (kg)

= [h,kg]

32

but [h.kg] = [h,gnh,k19
hence [h,gle = [h,gnh,k19
= [h,g]-
Therefore [k,h] = 1 hence k = gag-16 z(H).
Theorem 2.26: If G is solvable then B0 normalizes

every subgroup of B5.

Proof: Let M = M(G;B0), and let F* denote the
Fitting subgroup of G. We have that M.: F*; consequently
Bo induces the identity on G/F*. The group Bo restricted
to F* induces dilation on F*, and since the dilations
of a group form an abelian group, we must have that B5
centralizes F*. The group 85 therefore induces the
identity on G/F* and on F* and consequently belongs to
the stability group of the chain G > F* > 1. By Lemma 2.25,
we have that M(G7B6) : Z(F*), and by Lemma 2.21 we have

that M i N(F*). Let Q 6 Bo, g e G, then:

ga = nag with n0 6 N(F*)
o_1 . *
g = n _1g With n _1 E buF ).
o o
-1
Since (gO)Q - g, we have that
o o o
(g ) — (nag)
-1
= n n _lg 7
o
—1
consequently n: n _1 = 1-

 

33

6

Let 9 E BI e If - '
0 g e G, g - N69 With ne 6 Z(F*),
then:
090-1 6 ‘1
g = (n g) Q
o
—1
= 6 6 a
(mag >
—1
= < 6P -
nag Since 6 centralizes F*
0-1
_ (naneg)
-1 -1
- n n6 n _1g
0
but n6 E Z(F*), which is an abelian normal subgroup; hence
0—1 5(0-1> 1
n9 = n where (5(Q ), exp Z(F*)) = 1.
-1 I —1
Moreover n3 6 Z(F*) and therefore permutes with n0
a
and n _1.
o
6 —1 -1
Therefore g0 o = n n; n g
a l
_ a o
- n9 (nQ n _1)g
d
—1
: mg g
—1
-1
S(o ) -1 _
but g9 = n;(0 %. Therefore we have that a6a 1 ‘

and therefore B0 normalizes every subgroup of B0.

 

##-
r- W

 

 

 

 

 

34
Theorem 2.31: If G is solvable, then no is super-
solvable.

I
Proof: Since B0 is solvable, 3; must occur in a

chief series of 30‘ We may refine this chief series to
a composition series for Bo: hence we have
30-H°>Hl> °°° “1'36 K1+1>---)Ig‘-l. Now every
subgroup of I; is normal in Bo: hence the groups Hi,
Hi+1. "°, 35 are normal subgroups of Bo. Since all sub-
groups of no that contain Bé are normal in So, the above
composition series is a chief series for Bo. Since the
factors of a composition series for a solvable group are
of prime order, the above series is a chief series whose
chief factors have prime order.

we might be inclined to show that B0 is nilpotent
1! G is solvable. This is not possible, as may be seen
from Example 1. In this example G a 83, the.symmetric
Iroup on three letters, and consequently G is solvable,
but so - A(G) - 83 is not nilpotent.

At this point it might be worthwhile to illustrate some
or the problems that may occur in what follows by means of

an example.

Example 6: Let G be the semi-direct product of a.
°Y¢110 group of order 7 and a cyclic group of order .2;
Then G -‘<a,b/a7 - l, b2 - l, bab - a6). The chief series

or G is its composition series: G >,<a>>|l. Therefore,

I

35
Ab - so - A(G). n(c) -.<a,s/u7 - 1, 56 . 1 and s‘las - o5 >
where a : a -—> a B : a -—¢ a5
: b -—+ a b : b -—> b.

lots, however, that 3 divides [Bo] but 3 does not divide
the order of G. In other words: prime divisors of [Bo]
are not necessarily prime divisors of the order of G.

The previous example shows that although G is a
r-group, no need not be a r-group. If G is a r-group,
and Bo possesses r-elemcnts, we let B: denote a maximal

r-subgroup of Bo. If Bo has no r—element, we let B3 = 1.

Definition 2.28: A Hall w-subgroup of a group G is

 

a subgroup whose index is a 1'-number.

Theorem 2.22: If G is a solvable r-group, then:

(1) Bo has a unique maximal r-subgroup, B; , which
contains every r-element and every r—subgroup of 30°

(2) B: is the Hall w-subgroup of G.

(3) so splits over 33.

(4) 36 is a r-group and a r'-subgroup of B0 is abelian.

2222;: If Bo has no r-elements, then B; a l and
(1): (2), and (3) hold. Without loss of generality, we
may assume that Bo has r—elements. 35 fixes each
3ub8roup in a composition series of G and induces the
identity on every composition factor of G. Hence Bé
belongs to the stability group of every composition chain.
thce, by Theorem 2.10, B; is a r-group. Let B; be a

36
maximal w-subgroup of Bo that contains B6 , then B: is

normal in Bo. If x is a r-element of Bo’ and x A B3 ,

let X denote the product of B3 and (at) . Then B3 is
IBgl l<x>l
properly contained in K. On the other hand, [K] =- ——-—-—

W; n <X>l .
Since |Bg|, |<x> l, [133mm l are all 1r-numbers, we must
have that [KI is a r-number and therefore K is a
w-subgroup of Bo.
Since B3 is a maximal w-subgroup of G and B3 is

properly contained in K, we contradict the maximality of B3;

 

consequently x 5 BS; or K - B3.

If p is a prime divisor of [Bo/53': then 30/33
has an element x of order p. By the Homomorphism Theorem
Bo must contain a subgroup H such that B; 5 H.5 Bo and
[stg] - p. If p is a r-number, then [HI = [11:33] lsgl

 

is a w-number; consequently H is a w-subgroup which
contradicts the maximality of B3. Therefore p is
a v'-number and anngj is a w'-number. Therefore B3
is a Hall w-subgroup of 30' If H is a Hall w-subgroup
Of Bo’ then H is a w-subgroup of Bo‘ By the first part
of the theorem, H's Bg. Consider the index of H in Bo:
[30:11] - [Bong] [ng H]. Since [BozH] is a 1r'-number
EBgzflfl is a w'-number. Since B; is a w-group, fiBgzflfl is a
v'-number iff [Egafll - 1 or H a B3. Hence 33 is the
Hall w-subgroup of Bo.

Since B3' is a normal Hall w-subgroup of Bo’ we may
aPply'the Schur-Zassenhaus Theorem [T-6] from which we obtain
that B° - B: H and 83(\ H = 1. Consequently Bo splits

 

37

over B; .
If H is a 1'-subgroup of B0’ then H's B6 , but
B5 is a r-subgroup of Bo. Since H is a w'-subgroup of

30’ we must have H' - l or H is abelian.

II. The Structure and Propgrties of the Group A L

Introduction

 

we will discuss propertiesfland the structures of A0

 

for abelian groups, p-groups and nilpotent groups, and derive
a few properties of A0 for solvable groups.
If G is an abelian group, then every subgroup of G

is normal in G. Consequently A0 must fix all subgroups

 

of G and therefore A0 - B0(G) = A(G). We have, therefore,

 

that all the conclusions of Theorem 2.11 hold if we replace

B1 by A1.
Theorem 2.30: If G is a direct product of groups H

and K, and E is a subgroup of A0; if EH and EK are
the restrictions of E to H and E to K respectively, then:

(1) M(05E) - M(HsEH) x M(KsEK)-
(2) F(G;E) - F(H3EH) x F(K;EK).

Proof: The proof follows the same pattern as that of
Theorem 2.1.

Before we go any further, it might be worthwhile to
illustrate the difficulties that will be encountered with

an example.

 

38
Example I: G =‘<a, b, c/a3 = b3 = c3 = l;
b-lab a c-lac = a; c-lbc a be) .

A0 - an, s, e / a3 - s3 - 92 = 13131.5.) = [a.e] = [as] = 1>
we observe that 2‘ JAOI although 2 1 IGI.

The previous’example shows that although G is a
p-group, A0 neednhot be a p-group. If G is a p—group,
and pl IAOI, we let A; denote a Sylow p-subgroup of A0.
If p I IAOI, we let A; = 1.

Theorem 2.31: If G is a p-group of order pn, then:

(1) A6 is a p-group of class_< ri-l and A55 An.

(2) The Sylow p—subgroup A3 of A0 is normal in A(G).
(3) A; - An.

(4) A p'-subgroup of A0 is abelian.

(5) A0 Splits over A3.

M: Since G is a p-group, all chief factors of
G have order p. Since IGI - pm, a chief series of G ‘
must be of length n. Let s: e = Go 2 cl 2 G22-~lzcn = 1
be such a chief series. The group A0 must fix all normal
subgroups of G. Therefore, every 9 e A0 must induce an auto—
morphism on each chief factor Gi/ (3}1+1 i = O, l, ..., n—l.
Since A(Gi/ am) is cyclic of order p-l, A") must induce

the identity on G1 / G for i = O, l, ..., nV—l. Consequently,

1+1 .
Ac', < A(s). By Theorem 2.10, A(s) is a p-group and therefore
A5 is a p—group. By Theorem [T-S] , A5 is a p-group of

class 5 ml;

 

 

 

 

39

Let A3 be a Sylow subgroup containing A6 . Since any
subgroup containing the commutator subgroup is normal, we have
that A3 is normal in A0. The Sylow subgroups of a group
are conjugate: therefore A3 is characteristic in A0. Since
A0 is normal in A(G), A: must be normal in A(G).

If s is a chief series for G, then Sn(s) - A(s).
By Theorem 2.10, Sn(s) is a p-group. Since An - s £~$(G)Sn(s),
we must have that An isla p-group.

Let s: G - G02 Gig...an - 1 be a chief series
for G. If 9 6 A3 , then G? - G1. Therefore: 9 must
induce an automorphism on Gi/ Gi+l‘ Since A: is a p-Sylow
subgroup of G, [6| - pt. 0n the other hand, A(G1 / Gi+l)
is cyclic of order p~l. By Lemma 2.7, G/IG1 //ai+l - l ,
hence a e Sn(s) . Since 3 was arbitrary, 9 e 8 g'B(G)sn(s) - An;
therefore A3€%. Since An is a p-group and A; is a
D-Sylow subgroup of A0, we must have Ag - An.

Let H be a p'-subgroup of A0, then H’s A6 .
Since le' - pt and H' is a p'-subgroup of A0, we
must have 3' - l or H is abelian.

we may now apply a theorem by Schur-Zassenhaus [T-6]
from which we obtain A0 - A; H and Ag!“ H - l: where H
18 a p'-subgroup of ‘0‘ I

If G is a p-group, A0 need not be a p-group, as

may be seen from Example 6. we have, however, the following

result:

Theorem 2.32: Let a be a p-sroupo The”?!

 

40
(1) If H is a normal nilpotent subgroup of A0, then
H is a p-group.
(2) 4(Ao) is a p-group.
(3) If A0 is nilpotent, then A0 is a p-group.

2299:: Let Q be a Sylow q-subgroup of H, where q f p.
Since H is nilpotent, Q is a characteristic subgroup of
H. Since H is normal in A0, we must have that Q is
normal in A0. The inner automorphism group I(G) is con-
tained in A0, and therefore [Q,I(G)]5 [Q,A0] 5 Q. Since
I(G) is a normal subgroup of A(G), we must have [Q,I(G)]
S I(G). Therefore [Q,I(G)] s Q n I(G) - 1, since Q is
a q-group and I(G) is a p-group. Therefore Q and I(G)
permute, and by Theorem [T-7] we have that M(G3Q) g z(G).
Since (|Q|,p) - l, we have by Theorem T-8 that‘ “'2’ 2(0).
The group M(qu) g 2(0) 5 G' and consequently Q must
induce the' identity on e/¢(o). By Theorem [T-a], this im-
plies that [Q] - pk for some positive integer k. There-
fore Q - l and all q—Sylow subgroups of H are equal to
the identity. Consequently H is a p-group.

Result (2) follows from (1), since :HAO) is a normal
nilpotent subgroup of A0. If A0 is nilpotent, then a
Sylow q-subgroup Q of A0 is normal in A0 and by the

first result Q - l and therefore A0 is a p-group.

Theorem 2.33: If G is a solvable group, then:
(1) Every normal subgroup of I(G) is normal in A0

 

41

(2) Bo normalizes every subnormal subgroup of I(G).

P_rgo_i_‘: Let a 6 A0 ; e e I(G). Let H g I(G). By the
Homomorphism Theorems, there is a one to one correspond-
ence between the normal subgroups of I(G) and the normal
subgroups of G, which contain the center of G. Let H
correspond to H g G. Let 6 e H and let g e H such
that the inner automorphism induced by g is 6. If 'x e G,
then xa-lea - (g-l(xa-l)g)a' = (ga).l x g“, but g e H I

and H 4 G; hence 3“ e H. Hence the inner automorphism
' l

 

induced by g“ e H. Hence a“ an e H and H 4 A0 and
(1) follows.

Let H be a subnormal subgroup of I(G). Let H be
the subgroup of G that corresponds to H under the homo-
morphism of G onto I(G). Since H44 I(G). we have HQQG.
If 9.6 H, let 9 be induced by g e G. If x e G, then
a-lea a-l 9a -l “-1 a a - a _
x - (x ) - (g x g) a (g ) x g . Since g e H
and a 6 Bo’ then g“ e H hence a-le a e H and (2) follows.

We will now generalize Theorem 2.31 for nilpotent
groups, the proof of which follows the same pattern as
that of Theorem 2.31. l _

If G 15 a w-group, and A0 possesses w-elements, we

let A; denote a maximal w-subgroup of A0. If A0 .is a

v'-subgroup, we let A; - 1.

Theorem 2.34: If G is a nilpotent w-group, then

(1) AG has unique maximal w-subgroup Ag, which

42
contains every r-element and every r-subgroup of A0.
(2) A3 is the Hall r-subgroup of A0.
(3) A0 splits over A3.

(4) A6 is a r-group and a 1'-subgroup of A0 is abelian.

We shall now proceed to derive a few properties for

A0 in case G is supersolvable.

Theorem 2.33: If G is a supersolvable w-group, then
(1) A6 is a r-group and A6 is nilpotent

 

(2) A0 is solvable
(3) A w'-subgroup of A0 is abelian.

2292;: If G is supersolvable, then every chief series
of G has chief factors of prime order. Let s be such
a chief series, i.e., s: G - Go > 91> ... >Gh - 1. Then
G: - G1 for all i, for all 9 6 A0. Moreover 9 5 A0
induces an automorphism on Gf/Gi+l’ Since lA(Gi/Gi+1)l
' P-1 and A(Gi/Gi+l) is cyclic, A6 must induce the

identity on Gi/G for i - 0,...,n—l. Hence A6 5 A(s).

1+1
By Theorem 2.10, A6 is a regroup, and by Philip Hall‘s
Theorem T-5 , A6 is nilpotent.

Since A6 is nilpotent and Ab/Aé is abelian, the
group A0 is solvable.

Let H be a 1r'-subgroup of A0. then H' 5 A5:
but A6 is a w-subgroup of A0. Consequently H' - 1 and

H is abelian.

CHAPTER III

In this chapter we will try to answer the following
question: If a group G admits an automorphism that
fixes all subnormal subgroups, what conditions, if any,
does this impose on the structure of G? We will re-
strict our attention to solvable groups.

In particular we will investigate how the groups
F(G32) and M(ij), for a subgroup E of Bo’ are imbed-
ded in G. Furthermore, we will place conditions on the
groups n(ezs) and F(G3E) and see what this must imply

 

about the structure of G. we will begin the chapter with
two results which hold for arbitrary automorphisms of the
group G.

Next we will focus our attention on automorphisms in
30 for which Mme) n M(age) ‘- l. we saw in Chapter II
that although G is a s-group, Bo need not be a r-group.
For the w'-elements of ED, as well as a more general class
of automorphisms, we obtain the condition that F(G;9)f\ M(Gge)
- 1. For a group G, it is not only possible that Bo may
contain a w'-element, but that a subgroup of 30’ or even
the whole group Bo, is a w'-group. The condition that Bo
be a v'-group if G is a v-group, places strong conditions
on the group G.

We will next turn to the inner automorphisms of G

which are elements of Bo, and in particular, we will try

‘43

 

44
to determine some properties of the group H, which has
the property that if x e K, then the inner automorphism

induced by x fixes all subnormal subgroups.

Lemma 3.1: Let E be a subgroup of the automorphism
group of a group G. Let M = M(G7E) = <xex"1 / X e G and
6 e E) and F = F(G;E) = [g e G / g9 = g].

If W is the subgroup generated by M and F, then

M(W:E) = M(MyE).

Proof: By Theorem 1.9, we have that for any subgroup E
of A(G), M is normal in G. Hence if W is generated

by M and F, then W must be the product of M and F;

 

therefore w = M . F. If w e w, then w - mf with m e M

—1 e —1 _
and f e F. Therefore if 9 E E, wew = (mf) (mf) -

6 6 -1 —1 9 —1
m

—1
f f m = m m . Hence wew

e M(M7E) for all e e E

_ 6 -1
and for all w e w. Consequently, M(W7E) — (w w /w e W
<

and e e E> : M(M;E); but M i w and therefore M(M7E) _

M(W;E); hence, M(w;E) =M(M7E).

Theorem 3.2: Let 6 be an automorphism of G and
M = M(G76), F = F(G;e). If F(M;6) = 1, then G is gener—

ated by M and F.

Proof: If F(M76) = 1, consider the map a 2
for x c M. Then Q is a map from M into M(M;6). Now

- - o _ d
o is a one to one map, for if x — y , then x x

yey-1 or (y-lx)e = y_1x. Hence, y_1X E F(M76) = 1 or

45
- 1 _ .
y x - 1 or x = y. Since M(M76) : M(G;6) and the group

G is finite, we must have M = M(M76). In particular, if

—1
x e M, then x = yey for some y e M. Let Mg

9 _

be a coset of M in G. If g — g, then let g be the

coset representative of M. If ge # g, then geg_1 e M

9 _1 / .
and g g x 1. Hence there ex15ts an x e M such that

-1 9-1 9-

xex = g g . Therefore, (x‘lg) — (x-1

g) and x—lg e F;
but x-lg 6 Mg. In this case, let x-lg be the coset
representative of Mg. Hence there exists a collection of
coset representatives for M in G that is fixed element-

wise by 9. If g e G, then g e Mx with x e F; hence

 

g = mx with m e M and x e F, or G is the product of

M and F.

The hypothesis of the previous Theorem that F(M;6) = 1
is equivalent to F n M = 1.

Let us now turn our attention to automorphisms that
fix all subnormal subgroups of G. As in Chapter I and II,
we will restrict our attention to solvable groups. The
next two theorems will show that the action of such an auto—
morphism is to a great extent characterized by its action

On the Fitting subgroup F(G) of G. Since the work in

this chapter depends upon some of the results of Chapters I ‘

and II,we will summarize these results’as in the following
two theorems:
Theorem 3.3: Let G be a solvable group and E a

subgroup of B0(G); then:

 

46

(1) If H is a subnormal nilpotent subgroup of G,
then H and all subgroups of H are fixed by E.
(2) If H is a subnormal nilpotent subgroup of a,
and if 6 is an element of E, then G restricted
to H is a dilation of H. Hence if h e H,

9 s(h39),

h - h where (s(h36), |h|) - 1.

(3) If H is an abelian normal subgroup of G
and 9 is an element of E, then 0 restricted
to H is a power automorphism of H and he a

h8(°) for all h e H where (8, exp H) - 1.

 

Theorem 3.4: Let G be a solvable group. Let F*(G)
be the Fitting subgroup of G. Let E be a subgroup of
30(0). Then:

(1) n(cns) - 4 xex‘l/x e a; e e 13> is contained in

the norm of P*(G).
(2) If 9 e E is a power automorphism on F*(G), then
14(039) g z(s*).

2:32;: (1) follows from Theorem 2.23.

If 9 e E, restricted to F(G), is a power auto-
morphism, let x e O, f e F*(G). Then (x-lfx)9 a
(x-lfx)s(e), since 6 is a power automorphism on F*, and

x'lrx e F*(G). Therefore (x-lfx)e - x-1f8(9)x -

(x9)-lf°xo - (x0)'1f8(9)x9 and (xex'l)‘lffl(9)x9x‘l a f8(9).
Since xox'l centralizes f8(o), it must centralize every
power of 13(6), but < f5> - (f > ; hence xex'l centralizes

f. Since r was arbitrary, xex_1 e 09(F*) a z(F*). But,

 

47
x was arbitrary; consequently, M(G76) = (xex"1 / x E g)

: Z(F*).

The last two theorems show that a careful analysis of
the action of the automorphism on the Fitting subgroup of
G, should reveal information of the action of the auto-
morphism on the whole group G. In most instances, we
will place a condition on an automorphism e e B0(G), and
see what this must imply about the structure of the Fitting
subgroup. From the structure of the Fitting subgroup in

turn, we try to obtain some information about the structure

 

of the group G.

 

4
1
V

Lemma 3.5: Let E be a subgroup of B0 and let
M = M(G;E). Then:
(1) M(M;E) is an abelian subnormal subgroup of G.

(2) M(M7E) i.Z(F*), where F* is the Fitting sub-

 

group of G.

Proof: In Chapter II, Theorem 2.23, we have shown
that M(M;E) is contained in the center of the Fitting
subgroup of G. Since this is an abelian group, this must
imply that M(M7E) is abelian. In Chapter I, Theorem 1.9,
we have shown that M(G7E) is a normal subgroup of G.
Since M(M;E) is a normal subgroup of M(GyE), we must

have that M(M;E) is subnormal in G.

Theorem 3.6: Let E be a subgroup of B0(G), and

M = M(G;E), and F(G;E) = Fr then M = M(M7E) iff F(M;E) = 1.

In this case, M is an abelian group of odd order.

 

48

gagggz Let M - M(Mgfi) and let us assume that
Hum) ,1 1. We will show that this leads to a contra-
diction.

If M - n(ngs), then by the previous Lemma, M is
an abelian normal subgroup of G. By Theorem 3.3, we have
that for G a E and any 3 e M, g9 a gs(e), where
(s(9), exp M) - l.

I: 170433) f 1, then so M - F(M3E) g 1. Let x e F n M,
such that le - p. Then if 9 e E, x9 - x3(e) since

1 e M. One the other hand, xe - x since x e F. Con-

 

sequently x8(e) - x and s(e) a 1 mod p for all 9 e E.
Since 6 is a power automorphism on M, we must have for)
all g e M, g9 - g3(e), where 3(9) - 1 mod p for all
6 e E.

Let H be a maximal subgroup of H of index p.
Then H is normal in M, hence subnormal in G, or H9 = H
for all 9 e E. Consequently every 9 e E must induce an
automorphism on M/H. Let g e M and consider the coset
Hg. For 9 e 3, (fig)9 - 335(9) but, 3(9) =- 1 mod p or
3(9) - 1 + kp; therefore (Hg)9 - Hes“) = Hg:Mp = H(gp)ks-
Since the index of H in M is equal to p, (gp)k e H;
consequently H39 - H(gp)kg - H8. Therefore H89 = H8
and gag"l e H for all g e M and for all 9 e E. Hence,
M(mE) - <::°:c'l / x e M; e e E) _<_ H (FM. Since 14mm): 14,
we have a contradiction. Consequently if M(MzE) a M, then

runs) - 1.

49

Conversely, let F(M;E) = 1. By Theorem 2.23, we have
that M is nilpotent and of class 2. Therefore, M is
a direct product of its Sylow subgroups. Let M = P1 x P2
X ... X Pn where Pi is a pi—Sylow subgroup of M. If
M is not abelian, then there exists a pi—Sylow subgroup
Pi which is nonabelian. Moreover, Pi is a subnormal
subgroup of G. Let 9 e E; then P? = Pi and 6 is a
dilation of Pi' Since the dilations of a nonabelian
p—group form a p—group, we must have by Lemma 2.7, that

if x 6 P1 of order pi, then x9 = x. Since 9 was

arbitrary, we must have x6 = x for all 6 e E. There-
fore, x 6 F(M;E), a contradiction. We may assume then,
without loss of generality, that all pi—Sylow subgroups of
M are abelian, and consequently M is abelian.

Since M is abelian, we have for g e M, and 9 e E
that g9 = gs(9) where (s(6), exp G) = 1; therefore
gag—1 = gs(9)_1. Let the greatest common divisor of s(9)-1
and W be (3(9); i.e., em) = (s(e)—1, M). We will
show that the greatest common divisor of the d(9) for
9 e E is equal to 1. If the greatest common divisor of
the d(9) for 8 e E is not equal to one, then there exists
a prime p, such that p divides d(6) for all 9 e E.
Hence d(9) = p t(923 then s(e)—1 = p u(9) and
8(9) — 1 + p u(9). Since p divides the order of M, p

must divide the exp M. Therefore exp M(GyE) = pw, where

If w = 1, then M has exponent
9 Xs<9> = X1+p u(9)=

_"

p and if x e M of order p, then X ’

w is a positive integer.

50
_ u 6 _
- x(xp) ( ) - x for all 6 e E. Consequently x9 = x and
x 6 M n F = F(M;E). This is a contradiction since F(M;E) = 1a

Therefore w # 1; hence Mw = (gw / g e M(G;E)> # 1. If

gw e M(G;E) and gw # 1, then for e e E, (gw)e = (gw)1+pu(9)
w 6 9
: gwg Pu( ) : gw(ng)u( ) : gw. Therefore, gw e F n M =

F(M;E) but F(M;E) = 1, a contradiction. Therefore the ,
greatest common divisor of the d(9), for 9 t E, is equal
to one.

Since the greatest common divisor of the d(9) for

6 E E is equal to one, we must have, for every pi dividing

[M], a 6. e E such that pi I si — 1. Let M = P1 X P2 X

l
... x Pn be a direct product of its pi-Sylow Subgroups, Pi’
n. . . s.—1
_ l _ l -- z 1
where [Pi] - Pi . If gi e Pi’ then gi gi gi and
. 6i _1 si-l .
]gi gi l = igi i = ’gil Since pi 1 si—l. Consequently
9. 9.

l —1 _ l -1 =
<gi gi ) = (gi>. Therefore M(Pi79i) — (gi gi / gi c Pi>
Pi. Since Pi = M(Pi76i) i M(Pi7E) j'Pi’ we must have

M(PiyE) = pi. Therefore M(M7E) = m(91;E) x M(P2;E).x

... x M(Pn;E) = P1 x P2 x ... x Pn = M(G7E). Hence the

first part of the Theorem follows.

, let

 

If we assume that M(M7E) = M and 2 divides [M
2.

X a M(G7E) such that |x| - Then, since every 9 e E

9 _
must fix the subgroup generated by x, we must have x — x

for all 6 e E. This leads to x e F(M;E), a contradiction.
One would be inclined to prove Theorem 3.2 for an arbi-

trary subgroup E of B0(G). That this cannot be done can

be seen from the following example:

 

51
Example 8: Let G - 83, the symmetric group on three
letters. Then G is generated by a and b, subject to

2-l and b‘1

defining relations a3 - b ab - a2. The only
composition series of G is the chain G ai<a> > 1. Since
‘:a> is the only subgroup of G that is normal in G, we
must have that < a> is characteristic. Therefore B0 =
Ito-A(G). Let E-A(G)-<a,B/a3-Baal; B‘laB=
a2) 3 then F(G;E) - l, M(GxE) =<’a> . Hence F(G;E) n

M(GgE) - l, but it - F(G3E) . M(GgE) = A3 f, c.

 

Theorem 3.2: Let G be a nilpotent group and E be a
subgroup of 130(0). If F(G3E) - 1, then a -= n(cm), and
G is abelian of odd order.

2222;: In Theorem 3.6, we showed that the condition
F(M;E) - 1 implies M(MjE) - M, where M =- M(GgE). To
prove this result, consider the following: the only property
or M - M(ms) that was used, was that n(cm) was nil-

potent. Therefore, if we assume that G is nilpotent, then

 

,F(GIE) - 1 will imply that G - M(GgE), by the same argu-

 

 

ment as was used in Theorem 3.6.

Definition 3.8: A group G is said to be a semi-
direct product of its subgroups H and K iff H is normal

in G, G - HK and H n K = 1.

Lemma . 1 Let E be a subgroup of BO(G). Let
M - M(GgE) and F - F(G;E). If G is generated by M

and F, then:

 

52
(1) F H M = 1 and G is a semi—direct product of
M and F.
(2) M = M(M;E) = M(F*;E), where F* is the Fitting

subgroup of G.

Proof: If G is generated by F and M, by Lemma 3.1,
we must have that M(M7E) = M and, by Theorem 3.6, this
implies that F(M;B) = 1; but, F(M;E) = F n M = 1. From
the normality of M in G, we obtain that G is a semi»
direct product of M and F.

The Fitting subgroup F* is contained in G. Hence

 

M(F*;E) i-M, but M < F*; therefore, M(M7E) 1 M(F*;E).

Since M = M(M;E), we must have M(F*;E) j-M _ m(M;E) :

M(F*;E) or M(F*;E) = M.

Theorem 3.10: If 9 e B0(G) and F = F(G;G) and
M - M(Gye), then the following are equivalent:
(1) G is generated by F and M
(2) G is a semi—direct product of M and F
(3) M = M(M79) = M(F*;9) and is an abelian group of

odd order

(4) F n M = 1.

Proof: The Theorem follows from the previous Theorems

and Lemmas: (1) => (2) by Lemma 3.9, (2) => (3) by Lemma

3.9 and Theorem 3.6, (3) => (4) by Theorem 3.6, (4) => (1)

by Theorem 3.2.
For a subgroup E of B0(G), we have the following

result.

 

53

Theorem 3.11: Let E be a subgroup of B0(G) and

 

let F = F(G;E), M = M(G;E) and let F*(G) denote the
Fitting subgroup of G. If F n M = 1, then

(1) M 1 Z(F*(G)).
(2) M is a Hall subgroup of F*
(3) F* is generated by M and F(F*;E).
<4)

Every 6 6 E is a power automorphism on F*.

Proof: By Lemma 3.9 we have that M = M(F*;E). We
will show that M(F*;E) is a Hall subgroup of F* and
that M(F*;E) i Z(F*).

Since M(F*;E) fl F(F*;E) : M n F I 1, we have that

M(F*;E) fl F(F*;E) = 1. Now M(F*;E) :.N(F*), the norm of
F*. Hence if x e M(F*;E), then the inner automorphism
induced by x is a dilation on F*. Since the dilations

of a group commute, we have for x e M(F*;E), y e F(F*7E)

and 9 E E, that (ye)X = (yx)9. Since y e F(F*;E), we
have that y6 = y; therefore (ye)X = yx° On the other
hand:
<y9>x = (yx>9

= (x_ YX)

= (X9)-1y9X9

= (Xe)_1yxe
Therefore x_1yx = (xe)_1yxe; hence xexmlyl(xexul)_1 = y.
Consequently x9x_1 e CF*(<Y>)' Since x, y and 9 are

arbitrary, M(F*;E) = (x9x_1 / x e F*, 6 e E) i.CF*(F(F*;E)

)
We are now in a position to show that (|M|, |F(F*7E)I) = 1.

 

54
If p is a prime divisor of IM(F*;E)[ and [F(F*7E)[,
let X E M(F*;E), y e F(F*;E), such that [X] = [Y]
If e e B, then

to)6 = <xy)S<XY’9)

= XS(XY:9)YS(XY79)

I since xy 6 F*

/ since x centralizes y.

On the other hand, (xy)9 3 Keys
= Xys(y;9)I since x e F(F*7E)o
Therefore 5(XY76) YS<Xy;e) Z XYs(y;e) or
X—1Xs(xy;6) = yS(y79)y—S(XY79)°

Since F(F*7E) fl M(F*;E) = 1, we must have X-1Xs(xy;9) = 1

and ys(xy;9) y—s(y;9) = 1. The former gives 5(XY76) = x
1

or S<XY79) 5 mod p; the latter gives ys(xy;e) :

s(Y79) or 8(XY79) E 5(y;9) mod p. Consider the congruence

Y
system:
s(xy;9) E 1 mod p

s(xy;9) E s(y;6) mod p.

Consequently; s(y;6) E 1 mod p or
ye = y “(y 6)

y1+kp

= (y p)ky

y' since yp = 1.

Hence y e F(F*;E) H M = 1, a contradiction. Hence

(|M|, |F(F*7E)|) = 1.
Let P be a p—Sylow subgroup of F*. If x e P such
that x9 = x for all 9 e E, then x e F(P;E). Consequently

M(P7E) : M(G7E) must be equal to the identity, since

 

(IF(F*:E)I, lM(F*;

then the p—Sylow subgroup of F*

 

55

EH)

1. Hence, if p divides [F(F*;E)[,

is contained in F(G;E).

Hence F(F*:E) is a Hall subgroup of F* and is character—

istic in F*. Hence F(F*;E) has a complement K in F*.

We claim that M(G;E) is contained in K, for if g E F*,

g = fk with f E F(F*;E), k e K; then for 9 e E, geg-1 =
e _1 _ 6 _1 < .

(kf) (kf) - k k Hence M(F*;E) = M(K;E) _ K. USing

F(K;E) = 1 by Theorem 3.7, and the fact that K is a nil-

potent group, we h

ave M(K;E) M(G7E) K and F

F(F*;E) >< M(G7E). (2) and (3) follow.
Since M(F*;E) is abelian, we must have that every
6 e E must induce a power automorphism on M(F*;E). If
x E M(F*;E) and y e F(F*;E), then x9 = xs(ez where
8(6) E 1 mod exp M and ya = y. If g e F*, then g = xy,
with x e M, y e F(F*;E). Therefore:
<36 = (XY)e
. xs<e>y
___ Xtyt
= (Xy)t and e is a power automorphism
on F*. Here, t is the solution to the congruence system
x = s(e) mod exp M
x — 1 mod exp F(F*7E).
Since (exp M, exp F(F*;E)) = 1 and exp M exp F(F*7E) =

exp F*, the above

congruence system has a unique solution,

modulo the exponent of F*.

In Chapter I;

automorphisms 9 e

we saw that a solvable v—group may admit

Bo/ whose order is a w'-element. For

 

 

56
these automorphisms,as well as a more general class of auto-
morphisms/we are in a position to apply some of the pre—
vious results. We will start with a fundamental theorem

which will be needed in what follows.

Theorem 3.12: (H. J. Jaschke) Let B be a group of

 

automorphisms of a group G, U a solvable subgroup of G,
and let M be a collection of cosets of U in G. If

(IBl, IUI) = 1 and B leaves U invariant and permutes

the cosets of M, then there existsacxfllaflion of coset repre—
sentativesfor the cosets of U in G which remains fixed
elementwise by B; and the B—fixed subgroup F(G;B) is a

supplement for U in G.

Proof: For the cosets of M let K — (r1, r2, ..., rn)
be a corresponding collection of coset representatives.
3 E K will denote the coset representative of the coset
Ug e M, which contains the element g of the complex UK

Of G. Since B permutes the cosets of M, we have for

d e B, g 6 UK:

(Ug)0 = ugo‘ = U90 with 90‘ 6 K-

d _ d
Hence there exists ug,d e U such that g - uglag
For Q,B e B , r e K we have:
a d
: r
r ur,d
r _ ur.dBr (Ur/a ) r'a 7
r ,5

By the equality of coset representativesflwe have:

 

57
u 1 = 5
rte: ur,o u—— .
r°,5
-1 _1
Hence (1) u5 = u US
r105 INC —0 °
r ,5

Consider (1) for all S e B and fixed a e B and fixed
r e K. As 6 runs through B/ we obtain [Bl equations

of type 1. If we multiply these [Bl equations and let

u = I u: . we Obtain
g l
66B g’t
-1 -1 -1
r r, I I __ °
sea 05 sea r QB BeB r 0 rd B

The elements ug y with g 6 UK, y e B, permute modulo

_1
U'. Hence,letting u* = H u8 ,we obtain
9 fits 9'5

—1
Q -1
(2) u* = F u uB E u‘61 mu
r.d r,d

__ E ulB; u:_ mod U'°
fieB r0,6 r [B r

G

Since (]B|, LUI) = 1, we have that .B] has a unique in—
—1
—1 modulo the order of u [and since uglfil

verse |B| r,a

E u mod U' we have:
g /
_ _ ..1 —1 1
1r)a = u a r = u u u r0 = u r mod U'.
r r —Q. 1:10! rIQ —a_
r r

(3) (u

for a e B , r e k

U‘ is characteristic in U; hence U' remains invariant

under B. Hence B leaves U' invariant and permutes

according to (3)/the cosets:

— -1 .
U' ur1 r1 , U' ur r2 ... of U' in G.
1 2

 

58
Apply induction on the solvable length of the subgroup U,
assuming the theorem holds for all solvable subgroups whose
derived length is smaller than that of U. Then there
exists for the cosets U'u'lr1 , U'u;l , ... of U' in G,
r1 2
a system of coset representatives K* -(rf , r* , ...),

1r C Ur

which is mapped onto itself by B. Since U'u;
for every r c K, then K“ is also a collection of coset
representatives K which is mapped onto itself by B. If
B leaves every coset of U in G invariant, then K
remains fixed elementwise and hence K g F(G:B). Hence,
for every 3 e G, we have g e Uk or g - uk with u e U,
k e K 5 r(c;s); hence a - Uo F(c;s) and the theorem

follows.

Theorem 3.13: Let E be a subgroup of 30’ such that
('3'; IM(G:E)I) - 1: then:

(1) G is the semi-direct product of M(GJE) and

F(c;s). .

(2) “(033) is a normal complement for F(G3E).

(3) n(cxs) - u(n(c;s);s) - M(F*3E).

(4) M(GJE) is an abelian group of odd order. V

(5) Every 9 e E induces a power automorphism on F*.

2222;: By the previous Theorem, we have that G is
generated by r(a;s) and n(axs). By Lemma 3.9 F(G:E) n
"(9:3) - 1 and M(G32) - s(sc;n);s) - n(r*;s). By
Theorem 3.6, we obtain (h) and by Theorem 3.11, we obtain (5).

 

59

At this point it might be worthwhile to give an example.

Example 9: Let G1 = <a,b,c / a3 = b3 = c3 = 1} [a,b]

= [a,c] = 1; [bIC] a), [Gfi = 33, G1 is nilpotent and
G1 = o(G1) = z(cl) = (a). If 6 e B0(G1)/ then 9 must
fix all subgroups of G1; moreoven since G1 is a non—abelian
3—group, we must have that B0(G1) is a 3—group. Hence 6
has order a power of 3. Since |<a>| = [<b>| = I<c>[ = 3,

9 6 9

then a = a ; b = b 7 c - c. We must have that 9 in—

duces the identity on the subgroups (a) , (b) and (c).

Hence 9 is the identity on G1 and B0(G1) = 1. Let
G = <G1 , d | d2 = l: [a,d] = [b,d] = 1 ; [d,c] = c);
then BO(G) = 1. For if 6 e B0(G), then 9 restricted

to G1 is a dilation of G1: hence 9 must induce the
identity on G1. Since lG/Gll = 2, 9 must induce the
identity on G/Gl. Hence 9 must belong to the stability
group of the chain G > G1 > 1. Consequently X9X_1 e z(Gl)

= (a) for all x E G. If d6 = d, then 9 is the identity

 

on c. If ded‘1 g 1, then ded—1 = a or dad"1 = a2;
then d6 2 ad or d9 = a2d; then |d9| = |a[ |d| or {d9[
= |a2j [d . At any rate, |d9| = 6; but [del = [d] = 3,

a contradiction;hence d9 = d and a is the identity on

G.

We are now in a position to give an example of a super—

solvable group, admitting a subgroup E of B0, such that

(|E] , [G|) = 1.

 

60

Example 10: Let 5': G x C11, where G is the group
of the previous example and C11 is the cyclic group of
order 11. Let C11 = <e/e11 = 1). Since ([G[, [C11[) = 1,
the map, I

9: e ——> e2 x——> x for all x e G- iS an auto-
morphism of G. Since 6 is a dilation on C11 and a
dilation on G and ([G[,[C11[) = 1, 9 is a dilation of

, IEI) = 1. M(679)

 

G. Now [9| = 10; consequently ([6
= C11, F(G76) = G , G = F(G;6) x M(576) and all other

properties of the previous theorems can be shown to hold.

The previous examples show that a solvable group may
admit automorphisms 9 6 Bo, such that ([6[, [G[) = 1. The
next example shows that the whole group Bo can have order
relatively prime to the order of the group G, even though

the group is not nilpotent.

= b7 = c7 = 1;

Example 11: Let G1 = <a,b,c / a7

[a,b] = [a,c] = 1; [b,c] = a). The same argument as in
the previous example,shows that B0(G1) = 1. Let G =
(G1,d / d3 = 1, [d,c] = 1, [d,a] = 1, [d,b] = a2); again

B0(G1) = 1. Let G : GXCIl’ Where C11
of order 11 C11 = (x / x11 = 1). Then B0(G) =

<9 / X9 = x2; 96 = g for all g e G). Then [BO(G)[ = 10,

is the cyclic group

B0(G) is cyclic and ([B0(G)[, [G[) = 1.
For the case that ([B0(G)[,[G[) = 1, we will obtain

some special results about the structure of the group G.

We first have the following definition:

 

61
Definition 3.14: If 9 is an automorphism of G such

that M(G79) : z(G), then 6 is said to be a central auto—

morphism.
Theorem 3.15: If E is a subgroup of B0 such that
E is normal in A(G) and if ([E[, [M(G;E)[) - 1, then:

(1) E consists of central automorphisms.
(2) G = M(G;E) x F(G;E) where M(G;E) : 2(a) and
F(G;E) : G'.

(3) F(G;E) is a normal Hall subgroup of G.

grggfiz If ([E[, [M(G7E)[) = 1, then if e e E,
9 induces the identity automorphism on G/M(G;E). By
Theorem 3.12, G is generated by M(G7E) and F(G;E).
By Lemma 3.9, this implies that M(G;E) = M(M(G;E);E), and
by Theorem 3.6, we must have that M(G7E) is an abelian

group. Since M(G7E) is an abelian normal subgroup of G,

an automorphism 6 e E must induce a power automorphism

on M(G7E). Since the power automorphisms of a group are
contained in the center of the automorphims group, the
group [E, I(G)] must belong to the stability group of
the chain G > M(G7E) > 1. Therefore, by Theorem 2.10,
[E, I(G)] is a v—group if M(G7E) is a w—group. Since

E is a normal subgroup of A(G), we must have that

[E, I(G)] < E. Since [E, I(G)] is a y—group and E 18

a V'—group, we must have [E, I(G)] = 1 and therefore E

and I(G) permute. By Theorem T-7, this implies that

' tomor hisms.
M(GyE) i.Z(G) and that E conSlsts of central au p

 

62

If M(G;E) :.Z(G), then F(G;E) is a normal subgroup
of G. By Theorem 3.13, we have that G is the semi—direct
product of F(G;E) and M(G;E). Since F(G;E) is normal
in G, we must have that G is the direct product of
M(GyE) and F(G;E).

If F(G;E) = 1, then F(G;E) is a Hall subgroup of
G. If F(G;E) # 1, let F* = G0 < G1 < -.. < Gn = G be a [
composition series joining the Fitting subgroup and G.
If we consider E restricted to G1, then E consists of
central automorphisms on G1; moreover, G1 2 F*; therefore

G1 : F(F*;E). By Theorem 3.11, we have that F(F*;E) is a

 

Hall subgroup of F*. Since M(F*;E) = M(G7E), we must have

 

that M(GI;E) = M(G;E). If p divides ([F(G1;E)[, [M(G17E)[),

then p does not divide [Gi[ since F(F*;E) : G1 and

([F(F*;E)[ , [M(G;E)|) = 1. Let y E F(G1;E) and x E M(GlyE)
such that [x[ = [y[ = p. We may choose x and y as coset
representatives of Gl/Gi. If 6 E E, then 6 induces a
power automorphism on Gl/Gi. In other words:

(G12)e = Gizs(9) where (5(9), exp G/Gl) = 1. Hence
(Gixle = Gixe = Gixs<e> and (Giy)e = Giy = Gly, but
Y € F(G1;E); hence ya = YS(6> = y. Consequently a(e) E 1
mod p; hence (Gix)e = Gixe = Gix1+kp = Gix or x6x_1 6
G; H M(G17E) = 1. Therefore X6 = x, hence x e M(G17E) fl
F(G17E) = 1; consequently x = 1, and ([M(G17E)[,[F(G17E)[) = 1.
We may assume ([F(Gi;E)[, [M(Gi7E)[) = 1 for i =
1.2, ":, n—l. Then G' : F(G;E) and since G' :
F(Gn_1;E), we must have ([G'[,[M(Gn_17E)[) = 17

 

63
but M(Gn-17E) = M(Gn7E). Consequently ([G'[,[M(G7E)[) = 1-
If y e F(G;E) and x e M(G7E) such that [x[ =I[Y[ = P:

then the same argument as for G1 shows that x e F(G;E) fl
M(G;E) = 1. Hence [F(G;E)[ , [M(G;E)[) = 1, and the

Theorem is proven.

Lemma 3.16: Let H and K be subgroups of a group
G. Let G = H x K with ([H[ , [x[) = 1. Then:
(1) Any dilation on K or H may be extended to a
dilation for G.
(2) Any power automorphism on K or H may be ex—

tended to a power automorphism for G.

Proof: Let 6 be a dilation of K; let 6' be the
extension of 6 to G such that 6 restricted to H is
the identity. Then 6 is an automorphism of G. Let
g E G; then 9 = hk, with h e H, k e K.

E H e

g = (hk) = h k = hks(k’6)

 

where (s(k;6), k[) = 1.

Now if t is an integen

t

g = (hk)t = htkt, since H and K permute.

The congruence system

t E s(k;6) mod [k[
t E 1 mod [h[ has a unique solution,modulo
the order of [hk[= [h[ [k[. Hence 6 maps every g e G
S(g;e) ; hence 6 is a dilation of G.

onto a power g

If 6 is a power automorphism of K, then k = k,

where (5(6), exp K) = 1 for all k e K.

 

64
The congruence system
‘x 5 5(6) mod exp K
x E 1 mod exp H

has a unique solution modulo, the exp G, since (exp K,

exp H) = 1: exp K -exp H = exp G. Hence g9 = (hk)e
_ S 6) _ x _ x — .
— hk ( - hxk - (hk) and 6 is a power automorphism

on G.

In case ([80 ,[G[) = 1, the group G must have a

 

special structure as will be shown from the following theorem.

 

Theorem 3.17: If ([B0[,[G[) = 1, then

(1) G = M(G;B0) >< F(G;B0), where M(G7B0) 1 Z(G);F(G;B0)
: c-. I I

(2) F(G;Bo) is a normal Hall subgroup of G.

(3) M(G7Bo) is abelian.

(4) If Pi is a pi-Sylow subgroup of M(G7B0),
then Pi is an elementary abelian pi-group.
Moreover, (p71, [G[).

(5) B0 consists of power automorphisms.

(6) B0 is abelian.

 

(7) Bo(F(GfBo)) = 1°

Proof: Theorem 3.15 implies (1), (2) and (3).

The group M(G7B0) is a Hall subgroup of 1G. Hence,
by the previous Lemma, any dilation or power automorphism
Of M(G7Bo) may be extended to a dilation or power auto-

morphism of G. If 6 e BO(G), then 6 must fix F(G;B0)

 

65

elementwise. Moreover 6 must induce a power automorphism
on M(G;BO), since M(G7B0) is abelian: therefore all of
its dilations are power automorphisms. By Lemma 3.16, 6
is a power automorphism on G. Consequently B0 consists
of power automorphisms. Since the power automorphisms of
a group are contained in the center of the automorphism
group, we have that Bo is abelian.

Since M(G;Bo) is abelian, we have that M(GyBo) =

P1 x P2 x ...x Pn/ where Pi is an abelian pi—Sylow sub—

group of M(G;Bo). Consequently, B0(M(G;B0)) = B0(P1) x

B0(P2) x ... x Bo(Pn). Since every dilation of B0(Pi)

can be extended to a dilation for M(GyBo) and consequently
to a dilation for G, we can consider B°(Pi)° By a
previous Theorem, we have that B0(Pi) is isomorphic to

the prime residue classes module the exp Pi' Therefore if
n. n.—1
1
exp P = pif, then [Bo(Pi)[ = pi (pi-l). Consequently

if n. > 1, P. has a dilation of order pi° Therefore G
l i

has a dilation of order pi, but pi divides [G[. This

is contrary to the fact that ([B0(G)[,[G[) = 1. There-

fore n. = 1 and P. is an elementary abelian pi—group.
i ’ 1

By the same reasoning/we obtain that (p51, [G[) = 1;

therefore (4) follows.

Let F = F(G;B0). Let B0(F) denote the group of all

automorphisms of F that fix all subnormal subgroups of

G. Let 60 e B0(E) extend 6 to an automorphism 6 of

G by letting 5 be the identity of M(G;B0). Let H be

a — n
a subnormal subgroup of G, then H - H for all a e B(G)

 

 

66
Since ([BO(G)[,[M(H;B0)[) = 1, we must have that H =

F(H;BO(G)) x M(H;BO(G)). Now F(H;BO(G)) = H n F(G;Bo)

is a subnormal subgroup of F = F(G;Bo). Hence H§I=

F(H:Bo(G))g >< M(H7BO(G))6 = Forename >< wannabe. The

and since H and F(G;Bo) are subnormal in G, F(H;B0(G))

automorphism 6 fixes all subnormal subgroups of H and

is the identity on M(G;Bo). Therefore:

1:9 - F(H:Bo(G))e >< M(H7Bo<G>)§

= Feeds» >< Means);
= H.

Therefore 6 fixes all subnormal subgroups of G; con—
sequently 6'6 BO(G). If 6 e BO(G), then 6 must induce
the identity of F(G;B0). Hence 6 is the identity and
Bunsen) = 1.

Let us now turn our attention to the inner automor—

phisms of G that fix every subnormal subgroup of G.

Every inner automorphism Og of G is induced by an ele—

ment g E G. We would like to investigate the subgroup

N of G such that g e NI if Gg’ the inner automorphism

induced by g, fixes all subnormal subgroups of G.

Lemma 3.18: H = n NG(H)
HceG

Proof: If X e N3 then HX = H for all IiddG; there—

fore X e N (H). Since this holds for all IiddG, we must

G
have x e n NGKH) and N': n NG(H). Conversely/if
HddG Hang

X e n N (H)/ then HX = H for all Ii<4G. Consequently

H4 dG

 

 

67

x e N and therefore N': n NG(H).
IidQG

Theorem 3.19: If G is solvable,then N - n N (H)

 

11446 G
has the following properties:
(1) N. is a characteristic subgroup of G.
(2) If H is subnormal in N, H is normal in N.

(3) N is supersolvable.
(4) N' is abelian and N. : Z(F*), where E* is the

Fitting subgroup of G.

Proof: If 0 €A(G), and H44G, then HQ44G and

 

(H0). Hence E“: n N (HQ) = n N (H)
HaddG HddG

and N is characteristic.

If H is a subnormal subgroup of N, then, since‘ N
is normal in G, we must have Ii4dG. Consequently N-i
NG(H). Since Hi N, this implies that H is normal in N.

Since N i G and G is solvable, we must have that
N is solvable. Hence N has a composition series with
composition factors of prime order andlsince every composi—
tion subgroup of N is a normal subgroup of N, this im—
plies that N has a chief series with chief factors of
prime order. Hence N is a supersolvable group.

If X e N, and H is a subgroup of the Fitting sub"

IiddG . there—

X .
group F*(G) of G, then H = H Since ,

fore X e N must induce a dilation on F*. Since the

 

 

dilations of F* form an abelian group, then [X,y]

X—1y_lxy must centralize FT Hence [X,y]e CG(F*) = Z(F*).

 

68

Since X and y were arbitrary, this means that N” i Z(F*).

Since the subgroup N possesses such nice properties,
one might conjecture that N is nilpotent. This is not
the case,as may be seen from Example 1. In this example
G = 83, the symmetric group on three letters. Since A3,
the alternating group on three letters is the only sub-
normal subgroup of G, we must have N = G. Consequently
N is not nilpotent.

If every inner automorphism fixes every subnormal sub-

group of G, then this imposes strong conditions on the

structure of the group G, as may be seen from the follow-

ing Theorem.

Theorem 3.20: Let G be a solvable group. If 2 I G
and N = G, then:

(1) Every subnormal subgroup of G is normal in G.

(2) G is supersolvable and G' is abelian.

(3) All Sylow subgroups of G are abelian; i.e.,

G is an A—group.

(4) G=G'K, G' nK=1 and K=NG(K) =CG(K).
(5) F* (G) = G'Z(G).
Proof: (1) and (2) follow from the previous Theorem. .

If G is supersolvable, then G has a Sylow Tower

for the natural ordering of the primes. In other words, G
has a normal chain:
< i :«His 5 ....S =G,where s
l—Spi SP1 SP2 .91 p2 pn pi

69
is a pi-Sylow subgroup of G and pl 1 p2 : ... :.pn.
If H is a subgroup of Sp , then H is subnormal in
1

S Since Sp1 is normal in G, H must be subnormal

P1.
in G. By (1), H is normal in G. HenceIall subgroups

of s are normal in S , and S is a Hamiltonian
Pi P1 P1

group. Since p is odd, we must have by Theorem T—9 that

Sp1 is abelian. The same argument as above shows that

i i-l i—l

.v Sp / w S is an abelian subgroup of G/ W S .

3:1 j 3:1 pl i=1 pi
i i-1

Since v s / v S is isomorphic to S , we have
i=1 Pj i=1 Pi Pi

that S is abelian for i = 1,2, '°', n. Therefore G .
i

is an A-group.

By a Theorem of Taunt [6],we have that G' can be
complemented. Therefore G = G'K with G' n K = 1.
Let x e NG(K) 0 G' , then for all y e K , [X,y] 6 G' n K
= 1. This implies that X permutes with all elements of
K. Hence X permutes with G' and with K; consequently
X e Z(G) 0 G'. By another Theorem of Taunt [6], we have

that for an A-group, G' 0 Z(G) = 1; consequently, NG(K) = K.

Since I(2§<3/G' , we have that K is abelian; therefore
NG(K) - CG(K) - K.

Since G' i F* and F* is normal in G, we must have
F* = G'(F* 0 K). Let X 6 F* n K. Since K is abelian,

X permutes with K. Moreover X permutes with all elements

Of Ffi since F* is abelian. Therefore X permutes with

G' and with K. Consequently X e Z(G). Therefore

70
F* n K :Z(G); Hence G' (F* n K) :G'Z(G). Since G' and
Z(G) are both abelian normal subgroups of G, we must have
G‘Z(G) :.F*. Therefore, F* = G'(F 0 K) i G'Z(G) :.F*.

Consequently, G'Z(G) - F*.

 

INDEX OF NOTATION

I. Relations:

I“

Is a subset of

Is a proper subset of

IA :+n

Is a subgroup of
Is a proper subgroup of

Is a normal subgroup of

A A+I\

<1 Is a subnormal subgroup of

I?

Is isomorphic to
Is an element of

Is congruent to

II. Operations:
9

G The image of G under the mapping 9
8x x-ISX
f/S Automorphism of S induced-by f
‘G/H Factor group
Ex,y] The commutator of x and y
an The nth derived group of G
x Direct product of groups
G:H Index of H in G
[E,K] Subgroup generated by all [h,k], h e H, k e K
< > Subgroup generated by
i i Set whose members are
[lej Set of all x such that P is true
[GI Number of elements in G
lgl Order of the element g

71

III.

72

Groups and Sets, and Miscellaneous:

M(G7E)

F(G;E)

Automorphism group of G
Inner automorphism group of G
Dilation group of G

Chain of subgroups of the group G;
s: G = G0 > G1 > G2 > °~- > Gn = 1
. 9 _ _ _
{6 e A(G)[ Gi — Gi for i — 1,2, ---, n}
6
[6 e Si_1[ (Gix) = Gix)
Class of all composition series of G

Set of all automorphisms of G gixing all

subnormal subgroups

O S (s)
s€C(G) 1

Class of all chief series

Set of all automorphisms fixing all normal

subgroups
0 81(5)
seD(G)

<geg_1[ g e G; 6 e E; where E : A(G))

(g e G[ g9 = g for all 6 e E; where
E : A(G)]

Norm of G

[g e G[ ag 6 B0; ag is the inner auto—
morphism induced by g]

Center of G

Centralizer of H in G
Normalizer of H in~ G

Frattini subgroup of G
Fitting subgroup of G

Symmetric group of degree n.

BIBLIOGRAPHY

Hall, P., Some Sufficient Conditions for a Group to be
Nilpotent, Illinois Journal of Mathematics, vol. 2
(1958), pp. 787—801.

Huppert, B., Zur Sylowstructur Auflosbarer Gruppen,
Archives der Mathematic, vol. 12 (1961), pp. 161-169.

Jaschke, K. H., Dilatationen Endlicher Gruppen.
(Doctoral Thesis: Kiel 1963).

 

Polemini, A., A Study of Automorphisms and Chains of
a Finite Group. (Doctoral Thesis: Michigan State
University, 1965).

Scott, W. R., Group Theory (New Jersey: Prentice-Hall
Inc., 1965).

Taunt, D. R., On A—Groups, Proceedings of the Cambridge
Philosophical Society, vol. 45 (1949, pp. 24—42).

73

APPENDIX

THEOREMS

Theorem T-l:
If G is a finite group, the following are equivalent:
(1) G is nilpotent.
(ii) If M is a maximal subgroup of G, then M 4 G.
(iii) H < a, then H g NG(H).

(iv) G is a direct product of its Sylow subgroups.

Theorem T-2:
If G is a finite p-group and c is an automorphism
of c, inducing the identity automorphism on c/¢(c), then

1

[cl - p for some positive integer i.

Theorem T-3:

If A and B are normal nilpotent subgroups of a
group G, then AB is also a normal nilpotent subgroup of

G.

Theorem T-h:

If G is a solvable group having a maximumrfiipotent character—

istic subgroup H, then H 90G(H).

Theorem T-fi: (P. Hall, 1)
If G is a group and s: G a db > Gl> ... >Gh n 1

is a chain terminating in the identity and A(s) is the
stability group of s, then:
(i) A(s) is nilpotent of classgl/e n(n-l).
74

75
(ii) If s is a normal chain, then A(s) is nilpotent

of class : n-l.

Theorem T—6:
If H is a normal Hall subgroup of a group G, then

H has a complement.

Theorem T—7:
If 6 is an automorphism of G, then the following
are equivalent:
(i) 6 is central

(ii) 6 permutes with all inner automorphisms.

Theorem T—8:
If G is a p-group and 6 is an automorphism of G

such that ([6 ,p) = 1, and 6 fixes all normal subgroups

 

of G, then the upper and lower central series coincide.

In other words, if 1=zo:z1-Z(G):~-:zn=c is
. n
the upper central series and Z0 = G 1 Z' 1 °-- 1 Z = 1.
. n—i
is the lower central series of G, then Zi = Z for

all i.

Theorem T-9: (5, pp. 253—254)
A group G is Hamiltonian iff G = A x B X C, where

A is a quaternion group, B is an elementary abelian 2—

group, and D is a periodic abelian group with all ele—

ments of odd order.

   

M'Tfiifilllfll)[[[[T[[)i[[[[[[[ll'“