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SOME SUBGROUPS OF THE AUTOMORPHISM GROUP .
OF A HNITE ABEUAN GROUP

Thesis for the Degree of Ph. D.
MiCHiGAN STATE UNIVERSITY
WILLIAM L. HIGHTOWER
1970

w m "HI”lilllllflflfllfllllllllljflljjflflfl HWUIUJIILHI

3 12

This is to certify that the

thesis entitled

Some Subgroups of the Automorphism Group
of a Finite Abelian Group

presented by

William L. Hightower

has been accepted towards fulfillment
of the requirements for

M__ degree mmfica

aka/gt

Date—Aw

0-169

 

 

ABSTRACT

SOME SUBGROUPS OF THE AUTOMORPHISM GROUP
OF A FINITE ABELIAN GROUP

By

William L. Hightower

A main objective of this thesis is to determine the
structure of certain subgroups of the automorphism group of
a finite abelian p-group. This is done by analyzing their
action on certain series of subgroups of the abelian p—group.

For s:G=GO:GI:'°':Gt=l a chain of subgroups of the

group G, define

80(8) ={aeAut(G)|61“=G for i=0,l,...,t} and

i

_ -l a
Si+1(s)-{aesi(s)lg g eG1+1 for all geGi} ,

i=0,l,...,t-l.

The chain So(s)gsl(s):f°'gst(s) is known as the
stability series of s and St(s) as the stability group of s.
P. Hall has shown St(s) to be nilpotent and it is well known
that, for s a composition series in a solvable group, 80(3)

is solvable.

William L. Hightower

The following two theorems from Chapter II on
stability series are of some interest in their own right:

If G is a finite abelian p—group and if

1 t

elements aieGi and subgroups HfEGi+l such that Gi=<a1>xHi

i+1=<aip>xH1, i=0,l,...,t-1. If, for some 1 with

szG=GO>G >'°'>G =1 is a composition series, there exist

and G

o:i<t, <ai> is not a direct factor of G or if there is an

element he¢(G) such that aiheG1+1 1+1(s).

If the finite group G is supersolvable and

then Si(s)=S

s:g=GO>Gl>"'>Gt=l is a chief series, then So(s) is

supersolvable.

The following notation is used throughout Chapter III:

d

th where H1

with

G is a finite abelian p-group and G=H x

1

n
is homocyclic of exponent p 1 and [Hi:¢(Hi)]=p 1

for i=l,...,t, taking n =0. Define f0=0 and

”1> n1+1 t+1

i t
f =2 d for i=l,...,t and d=f =2 d =[G:¢(G)]. Let
1 3:1 3 t i=1 1

>x---x<a > where ord(a )= °°° = ord(a ) for
1_1+l f1 fi_l+1 f1
i=l,...,t. Then G=<a

>X°°°x<a > and ord(a1):°°': ord(ad)> O.

l d

For qip, 0<rgd and m = qd+r, define Gm=lg+l(<al>x---x<ar>) x

z[q(<ar+l>x-°'x<ad>) and GO=G. Let sl denote the series
GOZ'GIZ... . FOI‘ (1:0, O<rit and m = qt+r’ define Mm ___
Uq+1(HIX“°xHr) x Uq(Hr+lx---xnt) and MO=G. Let s2 denote

>000 .

the chain MO_>__M1

William L. Hightower

Using this notation and that for stability series given

above, the following theorems are established in Chapter III:

(1) Sn d(s1) is a p-Sylow subgroup of Aut(G) with
1

order pn where n = (fiZEni-ni+l] - d1(d1+1)/2).

HMd‘

=1
(2) NAut(G)(Snld(Sl)) = 30(81) which is supersolvable

and has order (p—l)dpn with n as in (l),

(3) If H1 is cyclic, CAut(G)(Sn d(31)) =

l

{oGAut(G)Ia1a =aim for (m,p) = l and lgigd} which has order
(n -l)

(p—l)p l

(u) If H1 is not cyclic, CAut(G)(Snld(sl)) =

m

{aeAut(G)|a1“=a mh; aia=ai if 1<ige, with (m,p)=1 and

l

n
hte(<ad >)} which has order (p-1)p 1.
l

(5) The intersection, W, of the p—Sylow subgroups of

Aut(G) is equal to Sn t(s2) which has order pn where
l

2 2
(fi [hi-ni+l] - di ).

:5
l-“MC‘I”

=1
(6) If G is not elementary abelian, CAut(G)(w) =

and a a=a m for d <i§d where

m
h for liigdl i 1 1

“—
{aeAut(G)la1 --a1 1

(nl+d12-2)
(m,p)=1 and h enl(Hl) for liigdl} which has order (p-Dp .

i

William L. Hightower

Also in Chapter III, the Fitting subgroup of So(sl),
the Fitting subgroup of Aut(G) and the maximal normal
solvable subgroup of Aut(G) are characterized by showing
how their elements act upon the elements a1. In each case
the order of the subgroup is obtained, giving criteria for
the nilpotence or solvability of Aut(G).

It is then shown that Aut(G) is supersolvable if, and
only if, G is elementary abelian of order 4 or ord(al) >
ord(a2)>"'>’ord(at), i.e., the invariants of the abelian
p-group G are all distinct.

The final section of Chapter III contains the
following two theorems on the existence of normal Hall sub-
groups of the automorphism group of a finite abelian
p-group:

If Aut(G) is not supersolvable, then Aut(G) has no
proper non—trivial normal Hall subgroup.

If G is not elementary abelian of order H and if
Aut(G) is supersolvable but not nilpotent, with ord(Aut(G)) =
An where (l,u) = 1,11% 1 # A and p divides A, then Aut(G)
has a normal subgroup of order A and does not have a normal
subgroup of order u.

Chapter IV consists of the extension of some of the
results of Chapter III to the case where G is a finite

abelian group and not necessarily of prime power order.

SOME SUBGROUPS OF THE AUTOMORPHISM GROUP

OF A FINITE ABELIAN GROUP
By

William L: Hightower

A THESIS

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

DOCTOR OF PHILOSOPHY
Department of Mathematics

1970

 

Gr 9545/
/ -' 03-0 " 7/

ACKNOWLEDGMENTS

I would like to express my deep feeling of gratitude
toward my major professor, Dr. J. E. Adney, for his patience
and encouragement in the preparation of this thesis. His
cheerful willingness to give of his time in spite of a busy
schedule has been especially appreciated. For his guidance
and direction in the research leading to this thesis I am

greatly indebted.

11

TABLE OF CONTENTS

Page
ACKNOWLEDGMENTS. . . . . . . . . . . . . . 11
INTRODUCTION. . . . . . . . . . . . . . . 1
Chapter
I. BASIC DEFINITIONS AND PROPERTIES. . . . . . N
1.1 Abelian Groups . . . . . . . A
1.2 p- Groups , . . . . . . . . A
1.3 Abelian p- Groups. . . . . . . . . 6
l. A Stability Series. . . . . . . 15
1.5 Homocyclic Abelian p- -Groups . . . 17
II. STABILITY SERIES
2.1 Some Sufficient Conditions for . . . . 23
S i(s)=Si+l(s) . . . . . . . . . 23
2. 2 A Sufficient Condition for So (3) to be
Supersolvable. . . . . . . . . . 25

III. THE AUTOMORPHISM GROUP OF AN ABELIAN p-GROUP. . 29

3.1 Notation and Basic Properties . . . . 29
3.2 The p-Sylow Subgroups of Aut(G). . . . 3A
3.3 The Intersection of the p-Sylow

Subgroups of Aut(G). . . . . . . . an
3.“ The Fitting Subgroup . . . . . . . 51
3.5 Sol(Aut(G)) . . 63
3.6 A Necessary and Sufficient Condition for

Aut(G) to be Supersolvable . . . . . 65
3.7 Normal Hall Subgroups of Aut(G). . . . 68

IV. THE AUTOMORPHISM GROUP OF AN ABELIAN GROUP . . 7H

“.1 Necessary and Sufficient Conditions for
the Nilpotence, Supersolvability or

Solvability of Aut(G) , , , , 7H

u.2 Normal Hall Subgroups of Aut(G) , , . 75

INDEX OF NOTATIONS. , , , . . . . . . . . . 79
BIBLIOGRAPHY. . . . . . . . . . . . . . . 82

INTRODUCTION

All groups referred to herein are finite.

An abelian group G is the direct product of its Sylow
subgroups and its automorphism group is the direct product
of the automorphism groups of the Sylow subgroups of G.
Hence many results concerning automorphism groups of abelian
p—groups can be extended, in some fashion ,to automorphism
groups of abelian groups in general.

A major objective of this thesis is to determine pro-
perties of the p-Sylow subgroup of the automorphism group
of an abelian p-group.

In section 1.3 the order of the automorphism group of
an abelian p-group is quoted from P. Hall, [A], p. 65. Also
in this section are a number of conditions related to a cyclic
subgroup of an abelian p-group G being a direct factor of
G.

Section 1.“ deals with the definition and elementary
properties of the stability series of a chain of subgroups
of a group.

Section 1.5 contains some elementary properties of
homocyclic abelian p-groups and the result that if G is a
homocyclic abelian p-group, then any automorphism offiuR(G)

or of Gflyk(G) can be extended to an automorphism of G.

l

Section 2.1 gives conditions for the coincidence of
consecutive terms of the stability series of a composition
series of an abelian p-group. In the following section the
supersolvability of the group of automorphisms which fix a
chief series of a supersolvable group is established.

Throughout Chapter III G denotes an abelian p-group.
In the first section two series, 31 and 82’ of subgroups of
G are defined. 31 contains a composition series and is a
refinement of 52 which is characteristic. The stability
group of S1 is shown to be a p-Sylow subgroup of Aut(G).
Its normalizer is shown to be the group of automorphisms
which fix $1. The centralizer of this p-Sylow subgroup of
Aut(G) is also characterized as is Z(Aut(G)).

The intersection, W, of the p—Sylow subgroups of
Aut(G) is proven to be the stability group of $2. C(W) is
characterized, showing in the process that it is super-
solvable if G is not elementary abelian.

Characterizationscfl‘the Fitting subgroup of the group
of automorphisms which fix s1 and of the Fitting subgroup
of Aut(G) show these to be the product of Z(Aut(G)) with the
stability group of s1 and 52 respectively, excepting the
latter when G is elementary abelian of order A or 9.
Necessary and sufficient conditions for Aut(G) to be nilpotent
are obtained.

The maximal normal solvable subgroup of Aut(G) is

characterized and in the process is shown to be supersolvable if

p>3. Necessary and sufficient conditions for Aut(G) to be
solvable are obtained.

Section 3.6 contains the result that Aut(G) is super-
solvable if, and only if, all of the invariants of G are
distinct or G is elementary abelian of order A.

In the final section of Chapter III it is shown that
if Aut(G) is not supersolvable, then Aut(G) has no proper
non-trivial normal Hall subgroup and if Aut(G) is super-
solvable but not nilpotent and if G is not elementary abelian
of order A, then Aut(G) has a normal Hall subgroup of each
possible order divisible by p but no non-trivial ones of
orders not divisible by p.

In Chapter IV some of the results of Chapter III are
extended to Aut(G) where G is any abelian group. In partic-
ular, necessary and sufficient conditions for Aut(G) to be
solvable, supersolvable or nilpotent, respectively, are
given as well as necessary and sufficient conditions for
Aut(G) to have a normal Hall subgroup of a given order.

The reader is asked to consult the index of notation
for identification of symbolic notations of groups, sets

and relations.

CHAPTER I

BASIC DEFINITIONS AND PROPERTIES

This chapter contains definitions and properties which

are used in the chapters which follow.

1.1 Abelian groups.

If G is an abelian group, it is the direct product of
its Sylow subgroups and Aut(G) is the direct product of the
automorphism groups of the Sylow subgroups of G (of. M. Hall
[3], p. 85). Hence an analysis of Aut(G) can be undertaken

by analyzing the automorphism groups of abelian p-groups.

1.2 p-groups.

Throughout this section G is a p-group. ¢(G), the
Frattini subgroup of G, is the intersection of the maximal
subgroups of G. The first part of the following theorem is

the Burnside Basis Theorem (cf. M. Hall [3], pp. 176-178).

Theorem 1.1: The factor group G/¢(G) is elementary

 

abelian. If its order is pr then every set of elements

zl,...,zS which generates G contains a subset of r elements

xl,...,xr which generate G. In the mapping G+G/¢(G), the

elements x1,...,xr are mapped onto a basis of G/<I>(G). Con-

versely any set of r elements of G which, in G+G/¢(G) is

mapped onto a set of generators of G/¢(G), will generate G.
The order of Aut(G), where ord(G)=pn, divides pr(n-r).

r
ifll(pr-pi_l) and the order of the group of automorphisms

which fix G/¢(G) elementwise, is a divisor of pr(n-r).

Definition 1.2: Where the order of G/¢(G) is pr and

 

the set of r elements xl,...,xr generates G, x1,...,xr is

called a minimal basis of G.

 

For the following theorem G need not be a p-group.

Theorem 1.3: If G=HxK, ¢(G)=¢(H)x¢(K), (of. Scott,

 

[10], Theorem 7.3.23, p. 16“).

Lemma 1.4: If H is a subgroup of G and K is a sub-

 

group of H, with [HzK]=p, then [H¢:K¢]=p or 1. If [H¢:K¢]=p,

then [Hfl¢:Kfl¢]=l and if [H¢:K¢]=1, then [HrlozK/7¢]=p.

P_r_-_g_c_>_§: [H<I>:K<I>]=(|H|-|Kn<l>l)/(IK|-|H/l<l>|)=p|K/l<l>I/|Hn¢|.
Since KCH, Kntenno and hence lKfl<D|/IHI‘<I>I51. Thus
.[HozKGJsp and since [H¢:K¢] is a power of p, [H¢:K¢]=p or 1.
Also, [Hfldell¢]=p/[H¢:K¢]. Hence, if [H¢:K¢]=p, then

[HHQan¢]=l and if [H<D:K<l>]=l, then [HflGzK/l¢]=p.

Lemma 1.5: If H is a subgroup of G and K is a subgroup

 

of H, with [HzK]=p, and if H¢=K¢, then H=K(Hf\¢).

Proof: (HA¢)K£_H. If an£x, then Hnd>£Knd> and
hence K(l¢=H{1¢ contrary to Lemma 1.“. Therefore, HllofliK

and KC(Hn<b)K_c_H. Since p is a prime, (Hn¢)K=H.

Definition 1.6: If A is a group of automorphisms of G

 

-1
and H is a subgroup of G then [H,A]=<{h halheH and aeA}>.

1
For any positive integer n, [H,An+ ]=[[H,An],A].

Theorem 1.7: If A is a p’-group of automorphisms of

 

G, then [G,A2]=[G,A] (cf. Gorenstein, [2], Theorem 5.3.6,

p. 181).

Theorem 1.8: If A is a p’-group of automorphisms of

 

the group G, then [G,An]=[G,A] for each positive integer n.

Proof: The conclusion holds for n=1 since [G,A]=[G,A].

k+1

1r [G,Ak]=[G,A], then [G,A ]=[[G,Ak],A]=[[G,A],A]=[G,A2]=

[G,A] by Theorem 1.7.

1.3 Abelian p-groups.

Throughout this section G is an abelian p-group.

Theorem 1.9: There exist elements of G aifl, i=1,...,d,

 

such that G=<a1>x...x<ad>. The number d is invariant as are
the numbers ord(ai), i=1,...,d, except for their arrangement.

(of. M. Hall, [3], Theorems 3.3.1 and 3.3.2, pp. A0 and U1).

 

Definition 1.10: If each aifl, for Isisd, and if

G=<a1>x...x<a then a1,...,a is a direct basis of G.

d>’ d
From the definition and Theorem 1.1, if a1,...,ad is

 

a direct basis of G and if [G:¢(G)]=pr, then d=r and al,...,ad
is a minimal basis of G. In general, however, it is not the

case that every minimal basis of G is a direct basis.

Example 1.11: Let G=<a>x<b> where ord(a)=p2 and

 

ord(b)=p. Then a, ab is a minimal basis of G but not a
_direct basis since ord(a)=ord(ab)=p2 and the order of a
direct product is equal to the product of the orders of the

direct factors.

Definition 1.12: An abelian p-group is homocyclic if

 

 

it has a direct basis each of whose elements has the same

order.

 

Theorem 1.13: If G=H1x...th where H1 is homocyclic
of exponent pni and [Hi :<I>(Hi )]=pdi with ni>n1+1 for 1$i<t
and nt+1= -O and if fi = jildj for lsist, then ord(Aut(G))=mpn

t
where n=1;1(fi ZEni'ni+1]-di (di+1)/2) and m=inl(Jn11(pJ-1))
(of. P. Hall, [A], p. 65).

Definition 1.1“: For a p-group H and a non-negative

 

i i
integer 1 define 91(H)=<{h€H|hp =1}> and 1/1(H)=<{hp |h€H}>.

91 (H) and 1J’(H) are characteristic subgroups of H.

p1
If G is ani abelian p— group, then {21(G)={g€G|gp =1} and
18“" -{SP IgeG}.

Theorem 1.15: If G=H1x...th is an abelian p-group,

then Qm(G)=Qm(Hl)x...me(Ht) and “(G)=%(H1)x”‘x%(flt)’

 

for each non-negative integer m.

t
mProof:m If g=1glhieflm(G) where hieHi’ lslst, then

m
l= gpm =inlhi and hence hip =1 for all i, lsist, whence

geflm(Hl)x...me(Ht).

t
If g=iglhieflm(Hl)x...me(Ht) where hieQm(Hi) for
pm pm t m
lsist, then hi =1 for i=1,...,t. Hence g =iglhip =1 and
seflmm).
9 £2 9
Therefore, In(G) m(H1)x...x m(Ht).
If ge‘Z/m(G), then there is an element x€G such that

pIn t t pm
g-x . x-iglhi for some hieHi’ lsist. Hence g=inlhi

eVmCH )x. uxV(Ht ).

If g=inlh1e1/;H(H )x. . 'me(Ht ) where h1€.Vm(Hi ) for

15iSt, then there are elements xieHi such that hi=xipm for
t

lsiSt. Hence g=(iglxi)pm€1fm(G).
Therefore, 'U’m(G)=“)fm(Hl)x.. .an(Ht).

Theorem 1.16: If G is an abelian p-group and isj
are non-negative integers, then Vim/741 (G) )=-Ui(G)/7/:j (G) .

 

Proof: Let xV(G)e"Ui(G)/'U3(G). Then there is an

element geG such that gp1=x. Hence x'I/j(G)= (g'Z/‘(G))p1 e

vim/741m»
Let x'U'(G)€‘U'1(G/'U3(G)). Then there is

1
an element geG such that (g'l/J(G))p =x‘l/j(G). Hence

pi P1 DJ
x€g 15(6) and there is an element heG such that x=g h =
(J-i) 1 *
(ghp )9 cal/3(a). Thus x7/J(G)‘e'v;<e>/VJ(G).

Therefore, Vim/16 (G) )=‘U3_(G)/‘UJ(G) .

 

Theorem 1.17: If G is of exponent pn and if r and s

 

are integers with oSrgsSn, then 1/B_r(G)§;QS(G). (of. P. Hall,

[5]. p. 79).

Theorem 1.18: '2/:(G)=¢(G), since G is an abelian p-

 

group. (of. Huppert, [7], Satz 3.1M, p. 272).

The following two theorems give the relationship between

the Frattini index of G and that of a maximal subgroup.

Theorem 1.19: If H is a maximal subgroup of G, then
[G:¢(G)]3[H:¢(H)].

 

£3933: Let [G:t(c)]=pd. Since t(Gth, H/¢(G) is ele-
mentary abelian of order pd-l. Let a;"°"ad-1 be represen-
tatives in H of a minimal basis of H/¢(G). Let adeG-H.
Since HEB¢(G), ad? H¢(G) and a1¢(G),...,ad¢(G) is a minimal

basis of G/¢(G). Hence a1,...,a is a minimal basis of G.

d
P p
<al,...,ad_1,ad>5;H since adeH. Let er. Then
(1 1 ad-l
x=al '°'ad-l m where me¢(G), since al¢(G),...,ad_l¢(G)

is a minimal basis for H/¢(G). By Theorem 1.18

= p p - P P
me‘Lf(G) (31"'“’ad>5’<a1’°"’ad-1’ad>‘ Thus XG<a1""’ad-l’ad>
and HSI<al,...,ad_l,ag>. Therefore, H=<a1,...,ad_l,ag>. By

Theorem 1.1 [H:¢(H)]spd.

Theorem 1.20: If H is a maximal subgroup of G, where

[a o(e)]=pd, and if ¢(G)=¢(H), then [H:t(H)]=pd‘1. 1r

 

¢(H)<:t(G), then [H:¢(H)]=pd.

10

Proof: If ¢(H)=¢(G), then [H:¢(H)]=[H:¢(G)]=
[G:¢(G>J/[G:H1=pd/p=pd‘1.

If ¢(H)<:¢(G), then [H:¢(H)]>[H:¢(G)]=pd-l. By Theorem
1.19, [H:<I>(H)]spd and since [H:¢(H)] is a power of p,
[H:¢(H)]=pd.

It is not generally the case that for an arbitrary
subgroup of G one can choose a direct basis for G in such a
way that the subgroup is the direct product of subgroups of
the cyclic groups generated by the elements of the direct

basis. However, as indicated by the next theorem, this can

be done when the subgroup is maximal.

Example: Let G=<a>x<b> where ord(a)=p3 and ord(b)=p.
Let H=<apb>. If H is the direct product of subgroups of the
cyclic groups generated by the elements of some direct basis,
then, since H is cyclic, H is a subgroup of the cyclic group
generated by some element, say c, of a direct basis. By
Theorem 1.9, ord(c)=p or p3. Since ord(H)=p2, ord (c)=p3.
Hence H=<cp>. But cpe¢(G) and apb is not an element of the
maximal subgroup <a> and hence apb6¢(G). Hence H is not a
subgroup of the cyclic group generated by an element of a

direct basis of G.

Theorem 1.21: If H is a maximal subgroup of G, where

 

[G:¢(G)]=pd, then there is a direct basis a1,...,ad of G

such that H=<a€>x<az>x---x<ad>.

11

Proof: Let g1,...,gd be a direct basis of G with
ord(gilsord(gi+l) for i=1,...,d-1. Let j=min{ilgi¢H}.
{ilgiEH}#¢ since H<:G. Since [G:H]=p, there is an integer

n1 _ -
n1, with lsni_p and hieH such that gJ h i-gj+i for i-1,...,d-j.

Since ord(gJ)<ord(gJ+i) and <gJ>r7<gJ+i>= 1, ord(hi )=

ord(gJ n1gJ+i)= ord(gJ+i) for lsi<d— j. Hence ord(G)=

d j d- J
iglord(gi)=(iglord(gi))(iglordwi)).

d m1 j- -1 mi d-j -ni

Let rec. Then x=iglg11%[i 1gi 1][121(gj 83+1)mj+i]
dEJ ( dEJ )
m + = n m j-l m m + = n m d-J m
[SJ j i 1 i J+i]=[12181 iJESJ j i 1 i j+i JEiglhi J+iJ

e<gl""’8j’h1"'°’hd-j>° Hence G=<gl’°°"gj’h1’°"’hd-j>

‘ J d-J
Since also ord(G)=[iglord(gi)][iglord(hi)J. 81,...,gJ,h1,...,

h is a direct basis of G.

d-J
By definition of j, gieH for i<j and hiGH for
i=1,...,d-j. Since [G:H]=p, K=<g1>x---x<gJ_l>x<g§>x<hl>
x~nx<hd_ J>_¢_.H. Since ord(K)=ord(H),K=H. The result now

follows upon relabeling.

Corollary 1.22: If H is a maximal subgroup of G,

 

where [G:¢(G)]=pd, then there is an element geG and a sub-

group K of H such that G=Kx<g> and H=Kx<gp>

Proof: By Theorem 1.21 there is a direct basis
a ,...,a of G such that H=<a p>x<a >x°.°x<a >. Let g=a
l d 1 2 d 1

and K=<a2>x---x<ad>.

12

The next series of theorems deals with conditions

relating to a subgroup being a direct factor.

Lemma 1.23: If G=Kx<g> and if g€¢(G), then g=1.

 

'(cf. M. Hall, [3], Theorem 10.“.1, p. 156).

 

Theorem 1.2“: If H is a maximal subgroup of G, then

¢(G)=¢(H) if, and only if, H is a direct factor of G.

Pgoof: Suppose H is a direct factor of G. Then, since
[G:H]=p, G=HxK where K is cyclic of order p. Hence ¢(K)=1
and by Theorem 1.3, ¢(G)=¢(H)x¢(K)=¢(H).

Suppose ¢(G)=¢(H) and [G:¢(G)]=pd. By Theorem 1.21,
there is a direct basis al,...,ad of G such that
<a p>x<a2>x---x<ad>=H. alp€¢(G)=¢(H). By Lemma 1.23, a1p=1.

Q
‘

Hence H=<a2>x---x<a > which is a direct factor of G.

d

Theorem 1.25: If H is a subgroup of G, with G=<g>xH

 

and if h€¢(G), then ord(ghlzord(h).

Proof: By Theorems 1.3 and 1.18, since he¢(G), there

is an element k€¢(H) and an integer n such that h=gpnk.

(pn+1)mkm.

Let m=ord(gh). Then l=[g(gpnk)]m=g Hence

g(pn+1)m=1 and km=l. Let ord(g)=pi. Since (pi,pn+1)=l,
pi divides m and hence gm=1. Thus, gpnm=l and km=1. Hence
(gpnk)m=hm=1. Therefore, ord(h) divides m and ord(h).<m=

ord(gh).

13

Theorem 1.26: If H is a subgroup of G, with <g>xH=G

 

and h€¢(G), then <gh> is a direct factor of G if, and only

if, ord(h)$ord(g).

Pgoof; By Theorems 1.3 and 1.18, since he¢(G), there

is an element ke¢(H) and an integer n such that h=gpnk.
Let ord(g)=p1. Since (pn+l,p1)=l, there is an integer j
such that j(pn+1)sl modpi. Let m be an integer and yeH.
Then gmy=[g(gpnk)Jka-mjy=(gh)ka-mJye<gh>H. Therefore,
G=<gh>H.

Suppose ord(h)$ord(g). Then since G is an abelian
p—group, ord(ghlsord(g) and ord(<gh>(lH)=ord(<gh>) ord (H)/
ord(G)§ord(<g>) ord(H)/ord(G)=ord(<g>(7H)=1. Hence
ord(<gh>f7H)=1 and G=<gh>xH.

Suppose <gh> is a direct factor of G. Then by

Theorem 1.25 ord(g)=ord(ghh-l)zord(h-1)=ord(h).

Theorem 1.27: If geG, then either ge¢(G) or there is

 

an element h€¢(G) such that <gh> is a direct factor of G.

Proof: Let a1,...,ad be a direct basis of G with

d n
- a i
ord(ai)5ord(ai+1) for i-1,...,d-l. Then g iglai . If

g¢¢(G) then there is an i, lsisd, such that p does not divide

n1. Let j=max{i|p does not divide n1}. Let A={ilp does not
n
= = n i
divide hi} and B {ilp divides n1}. Let a 18Aa1 and

b Then g=ab and by definition of B, b€¢(G).

= H n1
ieBai '
Claim that al,..

.,a is a direct basis

.,aJ_1,a,aJ+l,.. d

of G.

1”

By definition of j, ord(a)=ord(a ). Hence ord(G)=

3
d
iglord(ai)=ord(a)i§Jord(ai).
d mi
Let xeG. Then x=1gla1 Since p does not divide n3,
there is an integer k such that knJEmJ mod ord(aJ). Then
d m m (m -kn ) kn kn
= i= i i i i j
x iglai [iGBai JEigA-{j}ai JEigA-{j}ai 3&3 e
H kn1 k

<a1,...,aJ_1,a,aJ+l,...,ad> since ieAa1 =a . Hence
G=<a1""’aj-1’a’aj+1"‘°’ad> and since ord(G)=

ord(a)i§Jord(ai), a1,...,aj_l,a, aj+1’°°"ad is a direct
basis for G.
Thus <a> is a direct factor of G. Take h=b-le¢(G).

Then gh=a and <gh> is a direct factor of G.

Corollary 1.28: If geG and <g> is not a direct factor
of G, then g€¢(G) or there is a ke¢(G) and heG such that <h>

is a direct factor of G, ord(k)>ord(h) and g=hk.

Proof: If g¢¢(G) then by Theorem 1.27 there is an

element fe¢(G) such that <gf> is a direct factor of G. Since

1> is not a direct factor of G, ord(f-l)>ord(gf) by

l

<gff-
Theorem 1.26. Take k=f- and h=gf.
The following theorem gives an aid in the construction

of automorphisms of G.

are direct

 

Theorem 1.29: If a1,...,a and bl,...,b

d d
bases of G with ord(a1)=ord(bi) for i=1,...,d, then the map

n d n
i a_ i
) 'ifllbi

d
a defined by (iglai is an automorphism of G.

 

15

d n d m d n d m
, i a i a= i i _
35993? (iglai ) (iglai ) (iglbi )(iglbi )
d n +m d n +m d n d m
i l_ l l G_ i i a
iglbi -(1Ela1 -U&fllai )(iglai )] . a is defined

d n1
on G since al,..r.l,ad is a direct basis for G. iglai -1

if, and only if, a i=1, for i=1,...,d, if, and only if,

1
n1
bi =1, for i=1,...,d (since ord(ai)=ord(bi)) if, and only
d n1
if, iglbi =1. Hence a is well defined and one to one. a

is onto since b1,...,b is a direct basis of G.

d

1.“ Stability Series.

Definition 1.30: Let s:G=G 2G go-ozG 2G =1 be a
o 1 t-l t

chain of subgroups for an arbitrary group G. Define

 

I So(s)={eeAut(G)|Gie=G for i=0,1,...,t} and Sk(s)=

i
{083 (s)|g-lg6€G vgeo } for 1$k<t The series
k-l k’ k—l ~ °
S°(s)2SI(s)2---2St(s) is called the stability series of 3.
Each 81(3), osist, is a subgroup of Aut(G) and
31(S)£ So(s), osist (cf. Polimeni, [8], Theorem 1.1, p. “).
St(s), the stability group of s, has been shown by P. Hall
[6] to be nilpotent. It is well known that if s is a com-

position series in a solvable group, then 30(5) is solvable.

Theorem 1.31: Let s :G=G 2-'-2G =1 and s :G=
1 0 t 2

 

H02°--2Ht=l be two chains of subgroups of G such that sle=s2

for some eeAut(G). Then e"sk(s1)e=sk(s2) for k=0,1,...,t
(cf. Polimeni, [8], Theorem 1.2, p. 7).

 

16

Theorem 1.32: If s:G=Gdelz-'-3G 2Gt=l is a chain of

t-l
subgroups of an arbitrary group G, and if p is a prime divid-
ing ord (St(8))’ then p divides ord(G). (Cf. Feit-Thompson,

[1], Lemma 8.1, p. 795).

 

Theorem 1.33: If S:G=Ga>Gi>°°°>Gt-1>Gt=l is a com-

position series in the p-group G, then [81(3):S (8)] divides

1+1
p—l for osigt-l and St(s) is the p-Sylow subgroup of So(s).

Proof: Define oi:Si(s)+Aut(Gi/G ), oSiSt-l by

1+1

¢ ¢
)a i a for a€Si(s), geGi. a 1GAut(Gi/G

=8 Gi+l ) since

(gGi+1 i+l

' t¢i ¢i 01
a fixes G1+1' A computation shows that (as) =0 8 so oi

is a homomorphism. Ker(¢i)=Si+l(s) so Si(S)/Si+1(s) is

isomorphic to some subgroup of Aut(Gi/G Since 8 is a

i+l)'

composition series, ord(Gi/G =p and hence ord(Aut(G1/G

1+1) i+l))=

p-l. Therefore, ord(Si(s)/S (s)) divides p-l.

1+1

¢
Let 0680(3) and ord(a)=pn. If aesi(s), then ord(a i

)
n ¢i ¢i

divides p and ord(a ) divides p-l. Hence ord(a )=1 and

desi+1(s). Therefore, aeSt(s). Thus St(s), in view of

Theorem 1.32, is the p-Sylow subgroup of 80(3).

t-l“ t
of subgroupSOf the p-group G and if a p‘-e1ement aeSk(s)

 

Theorem 1.3“: If s:G=GozG1z---2G >G =1 is a chain

for some k, oskgt, then g‘lgaeok for all geG.

Proof: If a is a p’—e1ement of Sk(s), then <a> is a

-l

p’-subgroup of Sk(s). Hence g gae [G,<a>]=[G,<a>k] (by

Theorem 1.8) s [G, Sk(s)k]SG by definition of Sk(s).

k

17

 

Theorem 1.35: If s:G=Go>G1>°°->Gt_l>Gt=l is a com-
position series in the p-group G and if for some i, osigt,
a€Si(s), then apn is a p’-e1ement of Aut(G), where
pn=ord(St(s)).

Pooof: By Theorem 1.33, St(s) is the p-Sylow subgroup
of So(s) and hence of Si(s). Therefore there is a positive
integer m such that ord(Si(s))=pnm and (p,m)=1. Hence
(apn)m=apnm=1 and ord(apn) divides m. Therefore, apn is a

p’-e1ement.

1.5 Homocyclic abelian p-groups.

This section contains the result that if the abelian
p-group G is homocyclic, then the automorphisms of'1/1(G)
and G/ 711(G) can be extended to automorphisms of G.

Throughout this section G is a homocyclic abelian p-

group.

Theorem 1.36: If exp(G)=pn and if r and s are integers

with osrsssn, then on_s(o)§'Z/;(G).

Proof: Let a ,...,a be a direct basis of G. Then
‘“—“ 1

d n
n ‘ d mip 1
ord(ai)=p for i=1,...,d. Let xefln_S(G). Then x=1I_Ilai

(n +n-s) _
d mip i - p(n s)_

with (p,m1)=1 and 15131 -x —1. Hence
(n1+n—S)
m P
i - Th f “ divides
a1 -1 for all i, Isisd. ere ore, p
(ni+n-s)
mip and n3n1+n-s. Thus nizszr and hence

n
d mip 1 crzj'
x=1g1a1 e?U;(G). Therefore, an-s(G)" r(G)'

 

18

is a minimal basis of G,

 

Theorem 1.37: If b1’°'°’bd

it is also a direct basis of G.

Proof: As remarked following Definition 1.10 a direct
basis of G has the same number of elements as a minimal basis.

Let a1,...,a be a direct basis of G with ord(a1)=pn for

d

d

i=1,2,...,d. Then ord(G)=pr1 and exp(G)=pn. Hence

ord(bi)$pn for i=1,...,d. Since G=<b1,...,b
d - nd d
1I_llord(bi)$p =ord(G). Therefore, ord(G)=iglord(b1). Since

d>, ord(G)s

 

b ,...,bd is a direct

d
ord(G)=iglord(bi) and G=<b1"°"bd>’ 1

basis of G.

Lemma 1.38: If exp(G)=pn, n>l and [G:¢(G)]=pd, then

1 and [15(G>:¢(‘1/;<G>>J=pd.

 

'Lf1(G) is homocyclic of exponent pn-

Proof: Let a1,...,a be a direct basis of G. By Theorem

d
1.18, ¢(G)=17:(G)=<alp,...,adp>. Since ord(a1)=pn,

0Pd(aip)=pn'l for lsisd. 0rd(—L€KG))=ord(¢(G))=pnd/pd=

d
d n-l .
p ( )=1fllord(aip). Since 1/;(G)=<a1p,...,adp> and
d P p p P
ord(l/l(G))=iglord(ai ) and a1 #1 for lsisd, al,...,ad is a

direct basis of 111(6). Hence 'Z/‘l(G) is homocyclic of

exponent pn-l with [1}:(G):¢(]D:(G))]=pd.

Theorem 1.39: If exp(G)==pn and [G:¢(G)]=pd, then 271(G)

 

is homocyclic of exponent pm":L with [U1(G):¢(‘Ui(G)]-=pd for

osi<n.

19

Proof: The conclusion holds for i=0 since [10(G)=G.

k with

Suppose le(G) is homocyclic of exponent pn-
[°7fl’{(G):<l>(7fk(G))]=pd for some k, oSk<n-l. Then n-k>1 and
by Lemma 1.38 “(m(G))=-Ul;+l(6) is homocyclic of exponent

pn‘k'l with [79;+1<G):¢(79;+1<G)>J=pd.

Lemma 1.NO: If exp(G)=pn, n>1 and [G:<I>(G)]=pd and

 

if xl,...,xd is a direct basis for 121(G) with a1p=xi for

lsigd, then a1,...,a is a direct basis of G.

d

Proof: By Lemma 1.38,'2/: is homocyclic of exponent

pn_l. Hence ord(x1)=pn-l for lsigd. Thus ord(ai)=pn for
d ni . d n1 p d ni
lsisd. If inlai =1, then (iglai ) =ig1xi =1. Hence
x1 1=1 for all i, lsisd. Therefore ni=pn-lmi, lsiSd. Since

d n d (n-l) d (n-2)m

= i= p m = p 1
n>1, 1 iglai iglai i iglxi . Hence

n-2

x p( )
1

Thus p divides m

2

(“‘1’ divides pn‘ m1.

m1=1 for all i, lsisd and P

i and pn divides n1 for all i, lsisd. Hence

n

a1 1=1 for all i, lsisd. Therefore, <al,...,ad>=<a1>x°--x<ad>.
d

Also ord(<a1,...,ad>)=iglord(ai)=pnd=ord(G). Because n>1,

ord(a1)=pn>1 and al,...,ad is a direct basis of.G.

Theorem 1.u1: If exp(c)=pn and [G:¢(G)]=pd and if

 

x1,...,xd is a direct basis for 'VJXG) with aipd=x1 for

lgisd, then al,...,a is a direct basis for G, osJ<n.

d

 

20

Proof: For j=0, a and a1,...,a is a direct basis

i=xi d

of G.
Suppose that if y1,...,yd is a direct basis of‘Z/E(G)

k
with b p =yi for lsisd, then bl,...,b is a direct basis of

i
G, for some k, osk<n-1.

d

Let x1,...,xd be a direct basis of ‘V’k+l(G) with
k+l
ai =xi for lsisd. By Lemma 1.HO and Theorem 1.39

k k

p ,...,adp is a direct basis of‘1flE(G) and hence by the

a
1

 

induction hypothesis al,...,a is a direct basis of G.

d

Theorem 1.N2: If exp(G)=pn and [G:¢(G)]=pd, then,

 

for lsjsn, G/Z/3’(G) is homocyclic of exponent pJ and

[G/v§(e):¢(G/v;<e)>1=pd.

Proof: Let a1,...,ad be a direct basis of G. If
m (m +n-J)
i i
a1p e 151G), then a1p =1 since, by Theorem 1.39

l/J(G) is homocyclic of exponent pn-J. Hence mizj since
ord(ai)=pn. Therefore, ord(ai‘1/3’(G))=pJ for lsigd. By
Theorem 1.39 ord(‘V"(G))=pd(n-J). Hence [G:V(G)J=pd3.

J J
Since <a1'b516).---.ad‘b3(G)>=G/”LSKG) and ord(G/”D3(Go)=pd3=
d
1.Illord(ajl_'l/3(G)) and a17/J(G)#'?/3(G) for lsisd,
al‘U‘g (G),...,ad‘U3(G) is a direct basis of G/7/:j’(G) and
hence G/‘US(G) is homocyclic of exponent pJ with
[G/ Wm.) :<I>(G/"Z/'(G) > ]=pd.

J J

21

Lemma 1.u;; If exp(G)=pn and [G:<1>(G)J=pd and if

 

all/:(G),...,ad2/:(G) is a direct basis of G/Qy:(G), then

al,...,.,a.d is a direct basis of G.

Proof: By Theorem 1.18, Z/1(G)=¢(G). Hence, by

Theorem 1.1, a1,...,a is a minimal basis of G and, by

d

Theorem 1.37, al,...,a is a direct basis of G.

d

Theorem 1.NH: If exp(G)=pn and [G:<I>(G)]=pd and if

 

a;1/3(G),...,ad7{3(G) is a direct basis of G/Y{;(G), then

a1,...,ad is a direct basis of G, lsjsn.

Proof: By Theorem 1.42 G/TVFKG) is homocyclic of

exponent pJ. By Theorem 1.39 '7/3’_1(G/7/3’(G)) is homocyclic
of exponent p with ['vjzlm/‘VBWH:<I>(-U:_l(G/7/3’(G)))]=pd.
- pJ-l pJ-l
Since also 2/3_l(G/'2/3’(G))=<al '1/3(G),...,ad ‘1/3(G)>,
J-l J-l
a? '2/3(G),...,af '2r3(G) is minimal basis and hence, by

Theorem 1.37, a direct basis of 1/3_1(G/79F(G)). By Theorem

1.16, 7/;_1(G/‘D3(G))="L3;1(G)/123(G). By Theorem 1.39:

'U3_1(G) is homocyclic and since V:('1/J’_1(G))=7/'(G),

J
3-1 J-1
? ,...,ag is a direct basis for 1/3-1

1.h3. Hence, by Theorem 1.u1, a1,...,ad is a direct basis of

a (G), by Lemma

G.

Theorem 1.45: If exp(G)=pn and [G:¢(G)]=pd and if,

 

for some J with lsJ<n, BGAut(7UF(G)), then there is an de

Aut(G) such that “IV/(G)=B'
J

22

Proof: Let x1,...,xd
yi=xiB for lsiSd. Then y1,...,yd is a direct basis for

be a direct basis of 7{I(G) and

J J
P = P =
'1/3(G). There exist a1,bi€G such that ai x1 and b1 yi

for lsisd. By Theorem 1.41, a1,...,a and b1,...,b are

d d

direct bases of G. Define a by d:ai+b1 for lsisd. By
a: 93 a: a P = P3: = 8

Theorem.1.29 ,deAut(G). xi (a1 ) (a1 ) bi yi x1 ,

for lsisd. Hence GIV(G)=B.
J

Theorem 1.H6: If exp(G)=pn and [G:<I>(G)]=pd and if,

 

for some J with lsjsn, BeAut(G/7/3’(G)), then there is an

deAut(G) such that a induces B on G/1{3(G).

Proof: Let a.l UB,...,ad"zg be a direct basis of

G/‘U3(G) and (ai‘U3(G))B=(bi‘Z/3(G)) for 1519. Then

bl‘LfJ(G),...,bd“UJ’(G) is a direct basis of G/‘l/3(G). By

Theorem 1.HN, al,...,ad and b1""’bd are direct bases of

G. Define a by d:a1+b1 for lsisd. deAut(G) by Theorem 1.29.

(a11/3(G))a=aia‘bg(G)=b17U3(G)=(aijyg)8. Hence a induces
B on G/773(G).

 

CHAPTER II
STABILITY SERIES
For basic definitions see section N of Chapter I.

2.1 Some sufficient conditions for Si(s)=Si+l(s).

This section gives some sufficient conditions for

Si(s)=S (s) when G is an abelian p-group and s is a

1+1
composition series.

Throughout this section G is an abelian p-group and

s:G=Go>G1>--->Gt_l>Gt=1 is a composition series. For

osi<t, aieG1 and H1 is a subgroup of G1+l(the existence of
which is guaranteed by Corollary 1.22) such that G1=<ai>xH1

=<a p>xH .

and G1+1 i 1

Theorem 2.1: If, for some i with oSi<t, a1€¢(G), then

31(S)=Si+1(s)'

Proof: Suppose aesi(s). Let ord(St(s))=pn (cf. Theorem

n
1.33). By Theorem 1.35, up is a p’-element of Aut(G). Hence,

n
-1 up

by Theorem 1.3M, g g 6G for all geG. If aie¢(G), then,

13
by Theorem 1.18, there exists gieG such that g1p=a1. Since

n n n

P _
)p=a l a 6G

— p -1 a
661’ (31 31 i 81

1 a
g1 g1 1+1 and hence

23

 

2“

1'1

up esi+l(s). By Theorem 1.33, 0rd(Si(s)/S (s)) divides

1+1
n -
p-l. Since (p ,p—l) l, aesi+1(s). Hence Si(s)-Si+1(s).

Theorem 2.2: If, for some i with osi<t, <a > is not

i
a direct factor of G, then Si(s)=Si+1(s).

Proof: Suppose <a1> is not a direct factor of G and

let aesi(s) with ord(St(s))=pn. Then, by Theorem 1.35,

n
up is a p’—e1ement of 81(8)“

If aie¢(G), then, by Theorem 2.1, Si(s)=Si+l(s).

Suppose a1£¢(G). Then, by Corollary 1.28, there

exist be¢(G) and gEG such that <g> is a direct factor of G,
- pn
ord(b)>ord(g) and a1=gb. By Theorem 1.3h, g 1g“ 8G1=<ai>xHi.
n

_ P
Hence g lg“ =a1mh for some hEH1 and m an integer.

n
If ’1 “p - P -
g g ¢G1+1-<a1 >xHi, then (p,m)-1 and
ord(aimh)2ord(a1m)=ord(ai). By Theorem 1.25, ord(a1)=
n

-1 up _ mh
ord(gb)zord(b)>ord(g). Hence ord(g g )-ord(a1 )>ord(g).

n
-1 up
But ord(g g )sord(g).

n

_ P
Therefore, g 1g“ €Gi+1. Since b€¢(G), by Theorem 1.18,

there exists xeG such that xp=b. Hence, by Theorem 1.3h,
n n n n
-1 up

D p
-1 a p_ -1 a -1 a =
x x 801 and (x x ) -b b eGi+l. Thus, a1 a1
n n

'1 “p b—lbap eG Therefore apnes (s) B Theorem 1 33
8 3 1+1' ’ 1+1 ' y ‘ ’

ord(Si(s)/S (s)) divides p-l. Since (pn,p—1)=1,desi+1(s).

1+1
Hence Si(s)=81+l(s).

25

Theorem 2.3: If, for some i with o5i<t, there is an

 

element he¢(G) such that a heG then Si(S)=S

1 1+1, i+1(s)'

Proof: Suppose there is an element he¢(G) such that

hes Let 0881(8) and ord(St(s))=pn. Since a has

n 1 1+1
“la apnhIIhap 6G Since he¢(G)
i i 1+1° ’

by Theorem 1.18, there exists geG such that gp=h. By Theorem

ai 1+1“
and deSi(s)¢_-_’-SO(S), a

n ,
1.35, up is a p’-e1ement of 81(3) and by Theorem 1.34 g

E
n n n 1
_ -1 up _ -1 up p
n n n n

1 up -1 up -1 up —1 up
h h 6G and h h €G1+1, a1 a1 1+1.

1+1
n
Since G =<a >xH =<a p>xH up esi+1(s). By Theorem

1 1 1 1+1 1 1’
1.33, ord(Si(s)/Si+l(s)) divides p-l. Since (pn,p-1)=1,

-1 up
g g 8G1. Since [Gi:G

Since a - 8G

1 a1
and G

desi+1(8). Hence 81(S)=Si+1(s)°
2.2 A sufficient condition for 80(3) to be supersolvable.

In this section it is shown that 80(3) is supersolvable

when s is a chief series in a supersolvable group.

Theorem 2.4: If H is a subgroup of the group G and K
is a subgroup of H with [G:H] and ord(K) both prime (not
necessarily the same) and if C={aeAut(G)Iha=theH

-1 d

and g g 6K7geG}, then C is a subgroup of Aut(G) and either

ord(C) = ord(K) or ord(C) = 1.

Proof: 160. If d,B€C, then h°‘8=h8=h,VheH and

8-18a8=(g-lga)((g-l)a(ga)8)81(, VgeG. Hence 0:860 and Since

Aut(G) is finite, C is a subgroup of Aut(G).

26

Since [G:H] is prime there is an element gIGG—H such
that Vfififl,.3heH and m an integer such that g=g1mh. Hence
if d,Bec and g1a=g18, then c=s. Thus if use and afl, then
gla=glk with kfl and kGK. Since ord(K) is prime, <k>=K.

Hence if BSC, 3 an integer m, 15mgord(K) such that g18=

m
g km=g1a . Since C is a group, amGC and, by the above,

1
dm=8. Hence C=<a> and ord(C)=ord(d)=ord(k)=ord(K).

Theorem 2.5: If G is solvable and S:G=Go>Gl>°°°>Gt-l>

 

Gt=1 is a composition series, then (80(3))’€:St(s).

Proof: Since G is solvable and Gi/G is a composition

1+1
factor, Gi/G1+1 is cyclic of prime order, s:y p1. Srppoie
- a_ 1 a .g 2
Gi/Gi+l'<giGi+li' Let a,B€So(s) and g1 -g1 a1,g1 gi la2,
B- m 8‘ ad-

m
_ a .= u = .=
g1 g1 a3,g1 g1 a“ with a1,a2,a3,a“ eGi+l' g1 g1
m1 0 m2 m1 a—l mlmz
(gi a1) =(gi a2) a1 :31 a with mes1+1 since Gi+ld G1

-1
‘1 mlmz =
and 0 380(3). Hence gi eG1+1 which gives m1m2_1 mod pi.
“-IB-lcs m2 B'laB
Similarly msmuzl mod p1. g1 -(g1 a2) -
m m -1 m ‘m m -l
[(s “a ) 2a 3. J“3=<[(s 1a ) “a “J 2a B -“)3=
1 I. 2 1 1 Io 2 _

m m B m 08 m B-1GB m m m m
(((gi 3&3) 1a1 ) “a“ ) 2a2 =g1 ‘ 2 3 “as with aseG1+1

-1 -1
= -1 a 8 GB:
since G1+11>G1. Since mlmzmamu_1 mod pi, gi g1

m m m m -1

g1 1 2 3 b a 66 G > and
s

1+1=<gi 1+1
'18'1

1+1. Since Gi/G

—1 -l

0
1+1 1+1 31

 

27

-1 -l
g-lgcl ’8 “BGGHJ, VgeGi. Since this is the case for

i=0,1,...,t-1, a-lB-laBeSt(S) and (80(3))’§;St(s).

Theorem 2.6: If G is supersolvable and s:G=G0>Gl>--->Gt_1

 

>Gt=1 is a chief series, then So(s) is supersolvable.

Proof: Since G is supersolvable and S is a chief

series, [Gi G ] is prime for i=0,1,...,t-1. Hence, by

1+1

Theorem 2.5, (30(3))’S;St(s). For kzt let Gk=l° Define

Bi 3’ for i a non-negative integer and osjst-l by B1 J:
’ 3

{cos (s)|for osksj,g lgaeck+i+1,‘vgsok and for oskst-l

—l d
g g €Gk+1,V$GGk}. Then Bo t- 1= St(s) and if i >i2 or if

11:12 and J 12>3 , B1 J 5; B1 E’Jz. For osj<t-1, Bi:J+l=

 

{aeBi Jlgl g :GGJ+1+2,‘YEQGJ+:} and for 120, Bi+l,o=
d
{a631,t-llglg GG1+2,‘Y@£Gh}. Each Bi J is a group since

e31 J and s 1s°3=s'1s“[<s’: >“ (3“ >8 1. Let _393 (3)8 and

86B1 3' Let g€Gk. Then gl lmg le=[(g—1)a (ga ) 80‘] e
’
-1_Ba
Gk+i+l if lsksj and g ga 6Gk+i if lskst-l. Hence
-1
d BGGBi’J and 131,3 vSo(s).

For o$J<t—2 and izl, the elements of B1 J induce
’

automorphisms on G Let ¢ be the map from

3+1/GJ+1+2'

to Aut(G for aEB ¢ is the auto-

Bi,J 3+1/GJ+1+2) "here’ 1,3’“
morphism induced on GJ+1/G J+i+2 by a. ¢ is a homomorphism.

From the definition of B the elements of B: J fix
9

1.3’
GJ+2/GJ+1+2 elementwise and induce the identity automorphism

on (G Since [G

3+1/ GJ+i+2‘

) are both prime, ord(Bi J
3

3+1/GJ+1+2)/(GJ+1+1/Gj+i+2)°

] and ord(G )=

GJ+2/Gj+i+2 J+i+1/Gj+i+2

 

 

 

28

[GJ+1+1:GJ+1+2] or ord(Bi J¢)=l, by Theorem 2.4. By the
3

above, ker¢=Bi’J+1. Hence either [B1,J:Bi,J+1] is prime or

[B1,J:Bi,J+1J=1°

For izl, B B By an argument similar to that

i,t—2=
above, for izO [B

i,t—1'

B ] is equal to either [Gi+1:Gi+2]

i,t-1‘ i+l,0

or 1.
Thus for 1:1 and 153<t-1, [B1,J:B1,J+1] is prime or

and, for 120, Bi+1,0] is prime or

[Bi,t-l’
1. Hence from the Bi J a series
’

Bi,J=Bi.J+l

B Also B

Bi,t-1= i+l,0‘ t-1,o=
of subgroups from St(s) to 1 can be extracted, each member of
which is normal in 80(8) with factors of prime order. Since

also So(s)/St(s) is abelian, 50(8) is supersolvable.

Corollary 2.7: If the solvable group G has a character-

 

istic composition series, then Aut(G) is supersolvable.

 

CHAPTER III
THE AUTOMORPHISM GROUP OF AN
ABELIAN P-GROUP
The notation developed in the first section of this

chapter will be used throughout the chapter.

3.1 Notation and Basic Properties.

(3.1.1) G is an abelian p—group. G=H1x~°-th where
n d
Hi is homocyclic of eXponent p 1 and [H1:0(H1)] u p 1 for
i=1,...,t. n1 > n1+1 for i=1,...,t with nt+l = 0.
1
(3.1.2) f = O and f = 2 d for i=1,...,t.
o i J
3‘1
‘3 d
d=ft = 2 d1 ; [G:¢(G)]=P .
i=1

(3.1.3) af1_l+1,...,af1 is a direct basis of H1 for
i=1,...,t. Hence al,...,ad is a direct basis of G and ord

(a1) ?_ ord (a1+1) for i=1,...,d-1.

(3.1.4) K a <al>x°"x<a

1 >x°°°x<a > for

>X<a1+1 d

1-1
i=1,ooo’d.
(3.1.5) For q 3.0, o<r§d and m = qd+r, GIn =
Uq+l(<a1>x---x<ar>) XUQ (<ar+l>x--°x<ad>) and Go = G.

29

 

30

From this definition and Theorem 1.15 it follows that
Gm =U§<Gr) and for i_>_o, [aim Jip and GH d = 1.

1+1 1

(3.1.6) 51 denotes the chain G :,G Z.°.°°

o 1

(3.1.7) For q 1.0, o<rit and m = qt+r,

M =7};+1(Hlx- - ~xHr) x Uq(H

m x-o-th) and Mo = G.

r+l

From this definition and Theorem 1.15 it follows that

Mm =U;(Mr) and Mnlt = 1.

(3.1.8) 5 denotes the chain Mo Z.M 3""

2 1

From the above definitions it follows directly that 81

is a refinement of 32.

Theorem 3.1: $2 is a characteristic series.

Proof: It suffices to show that, for o :_q and

“pit, (Hle'XHr) XUq(Hr+lx---th) a Uq-i-lm).

q+l

[nnfirqmml’émn where 9km) = 1 if k g 0.

Let x evé+l(HlX°-°xHr) and y evqm °°xH

r+lx° t)'

Then there are elements g and h with g e Hlx---xH and
q+1 r

such that x = gp and y = hpq. Since

t
n n _q=
exp (Hr+lx---th) = pnr+l’ hp r+1 = 1. Hence yp r+1

h e Hr+IX°'°xH

n
h 1 an y e nnr+Iq( ), if nr+l Z_q and y = 1 if

nr+l < q. Also xe1g+1(G) and y€76(G). Thus xy€7§+l(G)‘

 

31

[9nr+Iq(G) n U;1(G)]. Therefore, UA+1(Hlx- . oxHr) x

U;(Hr+lx--°th) gyaflmmnrqqmmqlmn.

Let x evéflm) and y eUé(G)n nnrfiqm). Then

there are elements h h with h e H x-ooxHr

1 ’ E51 1+
q+1 q l
t p 82p

q nr+lnq nr+l nr+l

q
and y = hlp th and yp = h p h p

2: 81, 82
such that x = g1

13
and h2, g2 e Hr+lx---xH

= 1 or

= 1. By Theorems 1.15 and 1.36, and since nr > nr+l Z O,

 

y
9n (Hlx-HxHr) SDi(Hlx---xHr). Therefore, Since

r+1

nr+1

hlp a 1, there is an element h3 e Hlx-“xHr such that

q+1 q
= p = p p p 0..
h1 h3 . Hence xy (glh3) (g2 h2) €27q+1(H1x xHr)

xv;(Hr+lx-°-th). Thus Ué+l(G)[n (G)an’(G)] g

nr-l-l-q
Hulmlx-o-xHr) x Ué(Hr+lx-~th) and the conclusion
follows.

The next two theorems provide an aid in the construc-

tion of automorphisms.

Theorem 3.2: If when lgjgt and f <i<f

 

3’1 "J’
g1 e GinU;lJ(G) and o<mi<p, then the map a, defined by
m
dzai+a1 1gi for i=1,...,d, is an automorphism of G.

Proof: By Theorem 1.29, the map 8, defined by

Bzai+a1mi, o<mi<p, i=1,...,d, is an automorphism of G.

32

Since Gd =‘L§(G) = ¢(G), G = <a1,...,ad_1,adgd>.
ord(adgd) i ord(ad) since gd 6 ant(G). Hence al""’ad-1’

adgd is a direct basis of G and ord(adgd) = ord(ad).

Suppose al,...,ai,a1+lg1+1,...,adgd is a direct

basis of G and ord(ai+Jgi+d) = ord(a1+J) for lgjfid-i.

Then g = (i a r3)(: a rd g rJ). Since g e G =
1 i=1 3 J=i+l J J 1 1

1+1 d

 

ZE(<al> x...x<a1>) x (<a1+1 gi+l>x°°°x<adgd>), p divides

r1+1 i-1 rJ d rJ rJ
r1. Hence g1 a (31:1(8.J ))(1Jr=1 aJ gJ ) e

<al"°"ai-1’ a1g1,..., adgd> and p does not divide r1+1.
Thus g1 and hence a1 e <a1,...,a1_l,aig1,..., adgd>.

Therefore G = <al,..., a1_l, a1g1,..., adgd>. Since

ord(aigi) g_ord(ai) and ord(ai+Jg1+J) = ord(a1+J) for

lgjgd-i, a ,...,a1_1, a1g1,..., adgd is a direct basis

1
of G and ord(aigi) = ord(ai).

It follows that a1g1,..., adgd is a direct basis of G

and ord(a ) = ord(ai) for lgigd. By Theorem 1.29, the

igi
map 7, defined by y:a1+a1g1, is an automorphism of G.

The conclusion follows since the G1 are the same
m m
whether obtained from al,...,ad or al 1,...,ad d for

O<Hli<p o

33

 

Theorem 3.3: If when lgjit and fJ_1<i§fJ,

g1 6 G10 an (G) and o<m1<p, then the map a, defined by
J
m

1

aza +a1 g1 for i=1,...,d, is an element of 80(31) and if

1

all miEl mod p, then cesn1d(s1).

$2.93.: By Theorem 3.2, one Aut(G). Since Gd_1 =

<ad, Gd> = <ad,¢(G)>, afixes Gd-l' Suppose a fixes G1

for some i with lfigd. Since G = <ai, G >, a

i-l i
as well. Hence a fixes Gi for i=0,1,...,d.

 

fixes Gi-l

Since if i = qd+r where o<r_<_d, G1 =U;(Gr) and a fixes Gr’
therefore a fixes the characteristic subgroup G1 -‘U;(cr)
of Gr for i = 0,1,... . Hence 0.330(81).

Suppose each miil mod p. Since G‘—1 = <a1, G1> for

-1 a

lgigd and G “= 61 and a1p e ¢(c) = G 591, g g as for

i d 1
each geG1_l when lgigd. Since if i=qd+r where o<rgd,

_ -1 a
G1 0:103!) and Gd -Ul(GO), g g es1 for each ge (3.1”l

and for i=1,2,... . Hence one Snld(sl).

Lemma 3.4: For lgjgt, Aut (MJ_1/M ) is a

 

J
homomorphic image of Aut(G).

Proof: By Theorem 1.46, any automorphism of HJ/Zfi(HJ)
can be induced by an automorphism of H3. Hence any auto-

morphism of MJ_1/MJ can be induced by an automorphism of G.

34

3.2 The p-Sylow Subgroups of Aut(G).

This section contains characterizations of a p-Sylow
subgroup of Aut(G), its normalizer and centralizer in
Aut(G) and, as a by-product of these, a characterization of
Z(Aut(G)) and the fact that the normalizer in Aut(G) of a

p-Sylow subgroup of Aut(G) is super301Vab1e.

Theorem 3.5: If a e Aut(G) and a E 30(81), then there

 

is an element 8 e S d(s ) such that sea-1 9 S (s ).
n1 1 o 1

 

Proof: Suppose a e Aut(G) and d 9 So(s1). Let

m a qd+r, where o<rgd, be the smallest integer such that

O. . a
Gm # Gm. Since Gm 'LZ(Gr), Gr # Gr and r m by the

P P
. < > ... < > < > --- < >,
choice of m Gr = a1 x x ar x ar+1 x x ad

For 1_<_i§_r, a1p eUl(G) gar“. Hence there is an integer u,
r<ugd, such that au e or“.

a a
c: a
Gr -Gr-l Gr-l

a . .000. p-l a =
GruarGrU Uar Gr' If arGr n Gr fl,

= P
<ar,Gr> and ar e Gr' Hence

a
c:
Gr —-Gr-1

then ariorrrcr“ = s for all i with o<i<p which would imply

Gr.cl E Gr which contradicts the assumption that Gr“ ’5 Gr'
0 m a
Hence arGr/IGr # fl. Suppose ar heGr where p does not

divide w and heGRWKr (cf.3.1.4ixlsection 3.1).

B _ B _
Define the map 8 by a1 a1 if i i r and ar - arau.

Since r<u, aueGriinnJ(G) where fJ_l<rng. Hence, by

Theorem 3.3, B€Sn1d(sl).

35

w d m -1 w B = w d
Since ar hGGr and (ar h) (ar h) au dGr ,

dBd-l dd-l
# 31

a8 a
sl # sl . Hence 31 Thus

31.
asc‘l e s (s )
o 1 '

Corollary 3.6: NAut(G)(Snld(sl)) = 30(81).

Corollary 3.7: Sn d(sl) is a p-Sylow subgroup of
l
Aut(G).

Proof: By Theorem 1.33, Sn d(S1) is the p-Sylow
l
subgroup of 30(81). Hence if Sn1d(sl) were not a p-Sylow
subgroup of Aut(G), there would be an element of Aut(G),
outside of 80(31), which would normalize Sn d(sl). This

1
would contradict Theorem 3.5.

Corollary 3.8: The normalizer in Aut(G) of a p-Sylow

 

subgroup of Aut(G) is supersolvable.

Proof: By Corollaries 3.6 and 3.7, the normalizer
of a p-Sylow subgroup of Aut(G) is conjugate to 30(81).

By Theorem 2.6, So(sl) is supersolvable.

Definition 3.9: G is multicyclic if each H1 is

 

cyclic, for lgigt.

Theorem 3.10: Aut(G) has a normal p-Sylow subgroup
if, and only if, G is multicyclic.

 

36

Proof: If G is multicyclic, then 31 = 32. By Theorem
3.1, $2 is characteristic. Hence 30(81) is Aut(G). Hence,

by Corollary 3.6, N 1)) a Aut(G). Thus Aut(G)

Aut(G)(Sn1d(s

has a normal p—Sylow subgroup Sn d(81) (cf. Corollary 3.7).
1

If HJ is not cyclic, then afJ—l+l # afJ and ord(a

B
1 if 1 # f

)
fJ_1+l

). Define the map 8 by a +1 and

J
B 8
and a = a _
fJ-l+1 fJ and afj — afJ-1+l. By Theorem

1.29, BeAut(G). Since afJ-1+l E GfJ-1+l and afJeG

d(31)) # Aut(G). Therefore,

= ord(a = a

f i J—l

1 # fJ

f 1’

J-l+

B¢So(sl) Hence NAut(G)(Sn1

Aut(G) does not have a normal p-Sylow subgroup.

Lemma 3.11: Let a and B be the maps defined by

 

m r
. 1 . 1

d.a1+a1 g1 and B.ai+ai hi where o<m1,r1<p and

g1, hieGil) QnJ(G) when fJ_l<ing, for i=1,...,d. Then

a =8 if, and only if, m1 = r1 and g1 = hi for i = 1,...,d.

Proof: By Theorem 3.2, d,B€Aut(G). If m1 = r1
and g1 = h1 for i = 1,...,d, then a1“ = a18 for i = 1,...,d.
Hence a = B.
m r
- a _ B i = i for
If a - B, then ai - a1 and a1 gi a1 h1

i=1,...,d.

 

37

m-r

Hence a1 1 1GG1 for i = 1,...,d. Thus p divides mil-r1
for i a 1,...,d. Since o<m1, r1<p, -p<m1-ri<p and

m r
therefore m1 r1 for i 1,...,d. Since ai g1 a1 h1

and m1 = r1, g1 B h1 for i = 1,...,d.

 

d
Lemma 3.12: n ord(Qn (G)) = pn where 3(1) is
1=l 3(1)
. t
_ 2
related to i by fJ(i)-1<1irJ(i) and n - islfk (nk-nk+l).
f1 f1 f12nl
Proof: n ord(Q (H )) = [ord(H )] = p .
————— n l 1
i=1 1
Suppose, for some k with lgk<t, that
fk r k-l 2
"-1 ord(Qn (H1X° ka)) = p where r = (2 fJ (nJ-nJ+1))
1-- J J=1
2 fk+l
+ fk nk. Then n ord(an (Hlxo-oka+1)) =
i=1 J
fk fk+l
n [ord(a (H x--°xH )) ord(H )1). (n ord(a
i=1 “J l k k+l i=fk+1 nk+1
oo- = n = —
(Hlx XHk+1))) p where n r+fk(fk+l fk)nk+l +
(fk+1'fk)fk+1 nk+l ”‘(§=1 f3 (“J‘nJ+1)) + fk (nk‘nk+l) +
2 k 2 2
fk+1 nk+l = (§=lfJ (nJ-nJ+l)) + fk+l nk+l°

 

38

f
d t
Hence fl ord(fl (G)) = n ord(Q (G)) = pn where
i=1 “3 i=1 "3
t'1 2 2 t 2
n = (gglf J (nJ 'nJ+l)) + ft nt = §=lfJ (nJ-nJ+l) since nt+l
= 0.

Theorem 3.13: Let P = {alaia = aigi, where

gieG/lfl (G) whenf -1<i<f

1 n3 1&3, for i=1,...,d }. Then

P= nld(s ) and ord(Sn d(Sl )) = pn where n =
n1 nl

C—I-Mcf

2
=1[fJ (nJ-nJ+l) _ dJ(dJ+l)/2].

Proof: By Theorem 3.3, PgS (s1). If aes

nld n1d(sl)’

-1a a
1 8‘1

1a10‘ 6G1 n {In (G) when

n3
nld

then a a - a1(a1) and a

1

<isf for i=1,...,d. Hence deP and S (sl)gP.

fJ-l 3’
Therefore Sn d(sl ) = P.
n1

d
By Lemma 3.11, ord(P) = n ord(Ginfln (G)) where J
i=1 J

is related to i by f <igf Using Lemma 3.12 and the

J-1 J'

definition of G , ord(P) = pn where

1

d (d

t
— 2 - -
n - 32:1[fJ (nJ nJ+1) J +1)/2].

J

m
Theorem 3.14: Let A = {alaia=a1 1g1, where

 

giGBGfflfln (G) when fJ_l<ing, and o<mi<p, for i=1,...,d}.

39

d
Then 30(81) = A and ord(So(sl)) = (p-l) pn where

[r32(n ) - d (d +1)/2].
=1

-n

:3
ll
LuMc-r

J 3+1

3 J

Proof: By Theorem 3.3, AgSo(sl). Let aeSO(sl).

a . 3
Then a1 6G1_1 for i=1,...rad. Since G111 <a1,G1> and
a i

P =
a1 €61 for liiid, a1 a1 g1 where gie 61“ QnJ(G) when

a —m

f3-1<1Sfj’ and o<migp, for i=1,...,d, since ord(a1 a1 1)
n

:_ord (a1) = p J. If, for some i, lgifid, minp, then

“6G and hence G

a1 1 contrary to the fact that G C

“1G 1

1-1 1

Thus o<m <p for i-l,...,d and aeA and So(sl) £_A.

91-1“ 1

Therefore So(sl) = A.

d
By Lemma 3.11, ord(A) = (p-l)d 1r ord(GinQn (G))
181 J
where J is related to i by fJ_l<i§fJ. As in the proof of

Theorem 3.13, ord(A) = (p-l)d pn where n =

2

cared

=1

Corollary 3.15: The number of p-Sylow subgroups of

t d3‘1 1

Aut(G) is n [n (2 pk)].
J-l i=1 k=o
Proof: By Corollaries 3.6 and 3.7, the normalizer

in Aut(G) of a p-Sylow subgroup of Aut(G) is conjugate to

no

30(51). By Theorems 3.1“ and 1.13 the index of So(sl) in

t d3'1 1 k
Aut(G) is n [n (2 p )1.
i=1 i=1 k=o

Theorem 3.16: Let C1 = {alaiaaa m for (m,p)=1 and

 

1

liigd} and let 02 a {alalaaa mh; aiagaim 1f 1>1, with (m,p)

1
>)}. IfH1 is cyclic, then c
(ml-1)

1 and hefll(<ad Aut(G)(Sn1d(sl))

1
Cl and ord(CAut(G)(Sn

. If H is

1

not cyclic, then CAut(G)(Sn d(31)) = 02 and

1 n

1
ord(CAut(G)(Snld(sl)) a (p-l)p .
Proof: Let aeCAut(G)(snld(Sl))' By Corollary 3.6,
- a—
NAut(G)(Sn d(sl))-So(s1), so deSo(sl). Hence G1 -G1. Since

1

G = <a1,G1> and G“ Q’G there is an integer m with (p,m)=l

1’
and an element heKl (of. 3.1.“ in section 3.1) such that

a _ m

Let k be an integer with l<kgd. Define the map 8

a1 h.

by a18 = alak and a18 = a1 for l<i§d. By Theorem 3.3,
m m = m B _ 08 = 30 -
Besnld(s1). Hence al ak h (a1 h) - a1 a1 -
a _ m a a = m
(alak) — al h ak . Therefore ak ak . Thus, for each

0 m
i with 1<igc, a1 a1 .

1:1

If there is an integer r and an integer k fldl

with 1<kid and an element hlexln Kk such that akr :3 1 and
h - akrhl, define the map 7 by a1Y - 31 if i i k and

n -n
ak akad where fJ_1<kng. If nl-nd, then k<d

1 l

nl-nJ
and adleGlpnnJW). If n1>nJ, then adlp 6 Wimmnnjce) .

J(G) Eij/lnnJ(G). Hence, by Theorem 3.3, yesn

cw-(a1mh)7-a

Gdn an <1(81).

1

nl-nJ
81 lmakrad rp hi and a ya_a G-a m rh
rpnl-n n
adl " 1 since 9 3 does not divide r. Hence alaYyalY“

contrary to the assumption that “ecAut(G)(Sn d(81)).
1

Therefore h e<a ‘>.
‘11
If H1 is cyclic, then d1 - 1 and heKad SIIK1 - l.

C ..
Thus (:601 and Aut(G)(Snld(81)) S-Cl'

1

If H1 is not cyclic, then d1>1. Define the map 6 by

6.a

a1

5. p P .
1 if 1 ,4 <11 and adl halal1 . Since 3.1 swim)
I'1
(s ) by Theorem 3.3. If h - a , then
‘1 1 d1

06 rlp do a m r1
a and a -a -a a . Since
1 . 1 1 1 d
r p 1 n -1 1
a6 Ga 1 1 -
a1 -a1 ,a1 -1 which implies that p divides r1. Hence

G EEG , 668
d d1 n1

r

6
_ m 1
-(almh) a1 ad a1

’1
en _1(<ad

‘11 n1 >)gal(<ad1>). by Theorem 1.17. Thus 0:602

1
d(81)) g 02.

and CAut(G)(snl

 

M2

Cl 5 Z(Aut(G)) SC (31)). Hence if H

Aut(G)(Sn d 1

l

is cyclic, CAut(G)(Snld(Sl)) = Cl'

Suppose H1 is not cyclic and (1602 with a1a=a lmh,

a
a1 =aim for l<i§d where (m,p)=1 and henl(<ad1>)' By

Theorem 3.3, aeso(sl). If BGSnld(sl), then B fixes

nl-l)d+d -1 ‘ Unl<<ai>r ' '“adl-f) x

G
( 1

>) elementwise since

‘0; -1(<ad >xszax<ad>) =vl; -l(<a l

l 1 l

G(nl-l)d+dl "Ur-11(<al>x ° ° °x<ad 1)) x 7151-81.11.31“ ' 'X<ad>)

d

_ 3,
- 1. Hence h h, since $21(<ad >) 9% l(<ad >) by

l ' l- 1 .
- a8: Ba 8 m 3: PP+1
Theorem 1.36. Thus a1 (almh) (al ) h. Let a1 a1 h1

where h GK Then a 8)m hrp+l =

803a m(rp+1)h(rp+l)
l 1 1 1

m
h1 ‘(al
(a 8)m h since hen (<a >). Therefore a “g=a Ba.
1 1 d1 1 l
I'i

8.1)
For i>l, let a1 -a1 h1 where hieKl' This can

= P
always be done since a1€G1_1_CG1 <a1 >xKl and

8 Ba pr1 a mpr1 pr1 m =

G -
1-1 = Gi-l' ai =(a1 hi) -al h h1

pr
(a18)m h 1 = (a18)m since henl(<ad >). Since
1

aia8=(a1m)8'(a18 )1“, 3108:8183 .

43

Hence a 68,3180 for lgigd and a8=8a. Therefore

aeC(Aut(G)(Sn

1 l

02 = CAut(G)(Snld(sl))'

Suppose d:ai+a1m and Bzai+a1n for (n,p)=l=(m,p)
n
and i=1,...,d. If mEn mod p 1, then d=B. If d=8, then

nl n1-1

n and hence mEn mod p . Thus ord(Cl) = (p-1)p .

l
n
From this and Lemma 3.11, ord (02) a (p-l)p 1.

m
3
a1 a

Corollary_3.l7: If H1 is cyclic, Z(Sn1d(sl)) =

a m = a
{ala1 a1 where m-1 mod p, i 1,...,d} and ord(Z(Sn d(sl)))

l
nl-l
= p . If H1 is not cyclic, then Z(Sn1d(sl)) =
{alala-almh, aim-aim, mil mod p, henl(<ad:), i=2,3,...,d}
n1
and ord(Z(Sn d(sl))) = p .

1

 

Corollary 3.18: Taking C1 as in Theorem 3.16,

n -l
Z(Aut(G)) = c1 and ord(Z(Aut(G))) = (p-1)p 1 .

Proof: Cl §Z(Aut(G)) ECAut(G)(Snid(sl))' From

is cyclic C = Z(Aut(G)) g_c

Theorem 3.16, if H l

1 2'

nu

Suppose H1 is not cyclic and let an(Aut(G)) 5 02,

as in Theorem 3.16. Then alaaalmh, aiar-aim for l<i:d, (m,p)
= 1 and hen (<a >). Define the map 8 by a B=a ,
1 d1 1 d1
a B=a and a B==a if i # l and i # d B Theorem 1 29
d1 1 i i 1' y "

BQAut(G). Since, by Theorem 1.36, Ql(<adl>) QUnl_l(<adl>),

(n -l)
d rp 1 . Then ala8=(almad rp
l 1

(n -1)
let h = a 1 )3 =

m
l

andaea=a “ea m. Since d€Z(Aut(G)),
1 d1 (1

(nl-l)

alaB=alBaand a1rp = 1. Hence p divides r and h = 1.

Therefore if (36 Z(Aut(G)), a EC and h=l. Thus 016C and

2 l

Z(Aut(G)) 5 cl.

Therefore, in either case, Z(Aut(G)) = C

nl-l

1. ord(Cl) =

(p-1)p as in the proof of Theorem 3.16.

3.3 The Intersection of the p-Sylow Subgroups of Aut(G).
Throughout the rest of this chapter W is the inter-

section of the p-Sylow subgroups of Aut(G). w and CAut(G)(w)

both characteristic subgroups of Aut(G), are characterized

in this section.

 

 

45

Theorem 3.19: Let T = {alaia=a where gieMJflQn (G)

igi
J
when fJ— 1<1<fj’1 =1 ,...,d}. Then Snlt(s2) = T and

_ n _ 2 _ 2
ord(Sn t(s2)) - p where n - (fi [n1 n1+1]—di ).

1

HMd‘

=1

Proof: Let GET. Since gieMJnDn (G) g Gifmn (G
“3
when fJ -1<1<fj’ aesn1d(sl) by Theorem 3.3. By theorem 3.1,

each MJ is characteristic. For lgjgt, M = <af +1,...,af ,

3'1 J-1 J

-l

M >. Hence for each geMJ_1.8

J
where o<r§_t, MJ = UC;(Mr). Also Mt - U1(Mo)' Hence

g°‘eMJ for l_<_J_<_t. If J=qt+r

-1 a

g g GMJ for each gGMJ_1 and for J=l,2,... . Thus
aesn t(82) and T SSn t(82)'
l 1
Let aesn1t(s2). If f3- l<1<f3’ then a1€MJ_l. There-
-1 d a_ -1 a
fore a1 a1 GvaflnJ(G) when fJ_l<i§fJ, and a1 -a1(a1 ai )
for i—1.,,,.d. Hence aeT and S (s ) CIT. Thus
nlt 2 —
S (s ) = T.
nlt 2
t (11
By Lemma 3.11, ord(T) = fl [ord(Mfflfl (G)] =
n
i=1 1
t d d t
u [ord(fl (G))/p 1] 1 = p“ where n = z (fi2[“i‘“i+1]‘d12)
i=1 ni i=1

(cf. Lemma 3.12).

M6

Theorem 3.20: W = Sn1t(32).

 

Proof: Let Q be a p-Sylow subgroup of Aut(G). By

Corollary 3.7, there is an element BGAut(G) such that

B _ B a B
(Snld(sl)) - Q. By Theorem 1.31, (Snld(sl)) Sn d(sl ).

1
Since 31 is a refinement of s2 and s2 is characteristic, by
Theorem 3.1, s B is a refinement of s . Hence S (s ) CI
1 . 2 nlt 2 —

B _
Snld(sl ) - Q. Therefore Sn1t(s2) <5 w.

Let dew. Then by Corollary 3.7, deSn d(sl) and hence
I

a for lfifid. Suppose for some k,

i igi
likfid, gkgMJ when f

a = a where gieG

1

<k§_f M 3 G and

J' J-1

Hence for some u, with k<u§f

D D
k-l-Gk—M

J-l .1

since g 8G G 3M

k k’ k 3‘ J’
- r -
3k - au hk where (r,p)-l and hkexu. Define the map 8 by
a B = a if i # k and i # u and a B = a and a B = a
i i k u u k'

Since f <k<ufifJ, ord(ak) = ord(au). Thus, by Theorem

j-l

> < >
x au )

B = .0.
1.29, BGAut(G). Gk ‘Ui(<a1>x x<ak_1

r
x<a >x-°°x<a >. Since au ¢<aup> and

>°°'°< ><
x x au-l x au+l d

B

k

-1 a B a
hkeKu’_gquk . Hence ak ak 9Gk whereas akeG

k-l

G 3. Therefore a¢Sn 318) contradicting the fact that

k-l d(

l

‘ B _ 8
w 5 (Snld(sl)) - Sn1d(sl ), by Theorem 1.31.

“7

Hence, for i = 1,..., d, gi 6 MJ when

 

fJ_1<i_<_fJ and aesnlt(sz), by Theorem 3.19, since also
gieQnJ(G) when fJ_1<i_<_fJ. Therefore Snlt(82) 2W and it
follows that w a Snlt(82)'
a mi

Theorem 3.21: Let v = {alai =a1 h1 for lgigcl

a m1 "1'1
and a1 -a1 for dl<igd where (m1,p) = l and mizm1 mod p
for zgigcl and hienl(H1)nKi for l§i§d1.}. Then CAut(G)(w)=v

‘ (n1+d12-2)

and ord(CAut(G)(W))=(p—l)p , if G is not elementary
abelian.

Proof- Let aec (W) The a “-a mih wh r h ex

-————-' Aut(G) ' n i i i e e i i'

Let k be an integer with lghgd Define the map 8

l.

(n -1)
8. pt 8.
by ak akad and a a

i for i # k and liigd. If

1

then a 8M

d>dl’ d 1'

If d=d1, then, since G is not elementary
(nt-l)

abelian, nt-lil and a p GM

d Thus, by Theorems 3.19

m m p(nt-l) mk
k k _ B _
k hkad - (a hk) -

1.

and 3.20, sew. Hence a k

(nt-l) (n -l)

t
a °B=akBa )0 _ mk )a. Therefore

d

3 P
k (aka

( -1)
mlp nt (nt-l) (mt-1)

_ P a P
ad — (ad ) . Since ad # 1,

48

(n —l) t
t a mkp
(adp ) # 1 and hence ad # 1. Therefore,
(p,mk) = 1. Thus (p,mi) = l for i=1,...,dl and, in

particular (p,ml) = 1.
Let k be an integer with lgkgdl. Define the map y

Y a D Y a
by 3.1 alak and a1 a1 for Ziigd. By Theorems 3.19

ml mkp p p a
and 3.20, yew. Hence a1 hlak hk = (alak ) =
m m m p
yd: aY = 1 Y = l 1 p =
al al (a1 hl) a1 ak h1° Thus bk 1 and
mkp mlp nl-l
ak = ak . Therefore, for liigdl, miEl mod p

and hienlmmxi.

Suppose for some k with l§k$dl, hkeHl' Then there is

an integer u with dl<ugd and fJ_l<ugfJ and J>l and there is

an integer r with (P,r)=l and an element gknglKu such

(“3'1)

- I") :3 =
that hk - au gk (hkealmmxk Ql(<au>xKu)ka

6F
Ql(<au>)x91(KunKk) g nJ_l(<au>)x 91(KunKk), by Theorems

1.15 and 1.36). Define the map 6 by an6 = auakp

and a16 = a1 for i i u and liigd. Since J>1, n1>nJ and

(nl-nJ)

a p GM

G:
k t—M

1' Hence, by Theorems 3.19 and 3.20,

“9

(n -1) (n -1)
69 mk rp J rp l
w. Therefore, ak gk au ak =
(n -1) ' (n -1)
m J J
k rp 6 3 a6 _ 6d _ a a mk rp
(ak au 3k) ak ‘ ak ' ak ak 3k au
(nl-l)
Hence akrp = l contradicting the fact that (r,p) = 1.

Thus, for 1_<_if_dl, hiefll(Hl)nK1.

If d <d, let k be an integer with d

1
e a e _
alak and 8.1 - a

1<kgd. Define

the map 8 by al for 2iigd. Since

1

d <k, a GM and, by Theorems 3.19 and 3.20, sew.

l k 1
m m m
1 1 = 1 B 3 a8 =
Thus, a1 ak hl (al hl) al
m m m m
Ba = a _ l k 1 _ k
al (alak) - al hl ak hk' Hence ak - ak and
a m1
hk = 1. Therefore, for dl<igd, a1 = a1 .
Thus, CAut(G)(w)'§ V.
Let aeV. Since G is not elementary abelian,
dGSo(sl) by Theorem 3.3. Let BGW. By Theorem 3.20,
B -
Besnlt(82) and, by Theorem 3.19, a1 - aig1 where giGMJ
when fj-l<1-<-f,j° M(nl-l)t =U(n1-l)(G) = 521011) by
Theorems 1.15, 1.17 and 1.36. Since also M(n -l)t+1 = l,
1

B fixes 91(H1) elementwise. Hence, for liigdl,

50

m

 

m m m
OLB_I 1 6= 1 1 Ba: Q: i o
a1 (a1 hi) a1 81 hi and a1 (3131) ai higi °
d; pri (1 r1 a
For geMl, g-(igla1 )(igdl+lai ). Hence g -
d; r m p d m r n -l
i i 1 i = 1
(iglai )(1I=Idl+lai ). For liiidl, mi_m1mod p .
' n1 mip mlp a m;
Hence pmispm1 mod p and a1 =a1 . Thus g =g . There—
a mi , ml-m
fore, for liiid, g1 =g1 . For liiidl, gi =1 since
_ (1 I'll-1 M _ I'll—l H 1111- m1 _ a d
mizm1 mo p and exp( 1)-p . ence g1 —g1 .81 an
dB_ Ba
a1 -a1 for liifid .
m1 m1 m1
a8= B= Ba= 0.
For d1<i:d, a1 (a1 ) a1 g1 and a1 (aigi)

m1 m1 1 en Th r 1<i<d “B= 8“ Th
ai gi s nce g1 1' us, or ‘_‘_ , a1 a1 . ere-

: C
fore, 08 Ba and aecAut(G)(w)' Hence V..C )(W) and it

Aut(G
follows that VaCAut(G)(W)'

r
i

m
1 1
i hi’ a1
“1"1 n -1
(m1.p)=l=(rl,p), mismlmodp and risrlmod p l and

Let d,B€V and a a=a 8=

a g1 for lfiifidl, with

1

In; 8 Pl

G_ =
hi’gieQ.(H.)n K1 and a1 -a1 ,a1 a1 for d1<i:d.
mi ri
If a=B, then a1 h1=a1 gi for lfiifidl. Hence h1=gi
n
z 1
and mi-r1mod p for liifidl. n
Suppose h1=gi and misrirmxip 1 for lfiiid. Then
m1 r1 m r
a =a for 1<i<d and a l=a 1 for d <i<d. Hence
1 1 '-‘- l 1 1 1 —
a a=a Bfor 1<i<d. Therefore a=8.
1 1 ‘—'—
(nl-l) (dl-l) (dl-l)dl
Thus ord(CAut(G)(W))=ord(V)=(p-l)p p P “
2
(nl+dl -2)

(p-1)p

51

Corollary 3.22: If G is not elementary abelian, then

 

= a: "11 a: 111‘
Z(W) {ala1 ai h1 for liiidl, a1 a1 for d1<igd, where
n -1
mil modp, miEm1 modp and hiefll(Hl)/)K1 for l_<_i_<_dl}
2
(nl+dl -2)

and ord (Z(W))=p

Corollary 3.23: If G is not elementary abelian, then

 

CAut(G)(w) is supersolvable. F

9
1... A. 1

Proof: By Theorems 3.1“ and 3.21, and since G is not 1

kl

elementary abelian, )(W)£;So(sl) which, by Theorem 2.5,

CAut(G
is supersolvable.

Theorem 3.2“: If G is elementary abelian, then CAut(G)(w)=

 

Aut(G) and Z(W)=W=l.

Proof: By Theorems 3.19 and 3.20, and since M1=l,

W=l.
3.“ The Fitting Subgroup.

This section starts with a characterization of the Fitting
subgroup of the normalizer of a p-Sylow subgroup of Aut(G) and
ends with a characterization of the Fitting subgroup of
Aut(G).

Definition 3.25: The Fitting subgroup of a group H,

 

denoted Fit(H), is the maximal normal nilpotent subgroup of
H. (of. Gorenstein, [2], p. 218).

52

Theorem 3.26: Let F={a|a1a=a mg1 for liiid where

1

o<m<p and gieGil)Qn (G) when fJ _1<i<fJ}. Then Fit(So(s1))=F
t

and ord [Fit(So(sl))]=(p-l)pn where n=1;1(f12[n1-n1+1]

—d1(d1+l)/2).

Proof: By Theorem 1.33, Sn d(s ) is the (normal) p-
———_' 1 l

C
Sylow subgroup of So(s1). Hence Sn1d(s1)"Fit(So(81))'
Let c8Fit(So(sl)). Then aeso(sl). By Theroem 3.1a,
a a=a m1g for 1<i<d and o<m <p and g 6G /?9 (G) when
i i 1’ —~— ’ i i i n

J
f3-1<1if3-

If aflF, then for some k, 1<kid and fJ_1<K§fJ,
m17‘mk. Define the map Bby a18=a1(g1 l)‘ll for lgiid. By

m
Theorem 3.3, 388 (s ). Hence y=8a€F1t(So(s )). a Y==a 1

l)
for lgiid. Define the map 6 by alaaa1 akp nJ and
15=a1 if 2iiid. Since k>1, ak€G1”Qn (G) and, by Theorem
1
-l

3 3, 5€Sn1d(s ). 6 fixes G1 elementwise. Hence a16’ =

p(nJ-1) and a GYG-l -;=(a akP05-1))Y5-lY-l=

1
(n -1) (n -1)
m m p 3 -l -l m [m k-m JD J ‘1
(a1 lak k )5 Y =<a! lakk )Y =
uREmk-m1]p(nd'l) -1

Y =
alak where ak a

a

alak-

“k
k
—1 -1 m -1 -1 m -1

n
u.51.mod1)J. a16Y6 ‘6"=(a1 1)6 Y ‘=(ai i)Y ”=ai for

'“k k
2§i§d. Since p does not divide u

and (p,uk)=l and

k k
p does not divide uREmK-ml]. Hence there is an integer m

“ly-l)m=6#l con-

and o<ml,m <p and mlflmk,

such that mu [m -m 351 mod p, Thus (676
k k 1

tradicting the fact that Fit(So(sl)) is nilpotent.

53

Therefore aeF and Fit(So(s1))§jh
By Theorem 3.3, FEESo (sl ). 16F and if a,B€F with
as m B
,ai a1 g1 and a1 sai rhi for 1<i<d where o<m,r<p and g1, hi
as: mr m
6G1/Ifln (G) when fJ _l<i:fJ, then a1 a1 h1

Since o<m,r<p there are integers A ,u such that A:o,o<u<p

318 for lfiiid.

= a8, ua AD 8
and mr Ap+p. Hence a1 a1 a1 him g1 . Since 8680(31) and

p . Ap m 8
a1 861 and on (G) is characteristic, a1 h1 g1 eGI/IQnJ(G)

when fJ_l<i:fJ. Therefore aBeF and F is a group. From
n1d(81)
Let aeF. For 1515a and g€G1_1, gaagmh with heG

Theorem 2.6, F’SS

1. Since

if i=qd+r with o<r§d and q3o, G1= L€(Gr) and Gd='LT(G), for

g€G1_l, gQBgmh where heG for i=1,2,....

1
Let Bk {aeSo(sl)Ig g eGi+k for each geG1 and 1 0,1,...}.

3 a G
Then B1 Sn1d(s1) and Bnld 1. Let 86Bk and a1

o<u<p and hieeillnn (G) when rJ_1<15rJ and muEl mod p. Let

a: In 8:
geG1 and g g x1 where x16G1+l. Let g gx2 where x 26G1+k

Ba a‘1= u2
xix: where x 8Gi+k+l and x2 x2 x where x 8G1+k+1.

“-1-1—1 -1

. u
a1 h1 where

and x1

a3a*;s'1

Then g m

)80- 18-

=(sm x l"”“==(s x x x )a

-l -l
um-l m a . B . umrl m a 8
g(x2 x“ 113 ) . Since (112 X“ x31)1ec:+k+1’

—1 aBa ;B-1

3 3 eGi+k+1

a (n) C
Bk-I-l and since [F,F]§Snld(sl) B1, [F,F ]_Bn. Therefore

F is nilpotent.

-1 -1 .
and hence aBa B €Bk+1. Hence [F,Bk]€3

Since (S (s ))’§S (s )QFit(S (s ))SFES (s ),
o 1 “Id 1 o 1 o 1
Fqso(s1)'

Therefore F=Fit(So(sl)).

5h

d
From Lemmas 3.11 and 3.12, ord(F)=(p-l)inlord((61)”Qn (G))

3
(where J is related to i by fJ_1<i_<_fJ)=(p-l)pn where

t
8 2 — -
n 1§l(f1 [n1 n1+1] d1(d1+1)/2).
The next series of theorems is aimed at finding the Fitting
subgroup of GL(n,p), i.e. of the automorphism group of an

elementary abelian p-group of order pn (cf. Rotman, [9], p. 155).

Theorem 3.27: The second center of GL(2,P) is equal to

Z(GL(2.p)).

 

Proof: {Z(GL(2,p))={rE|o<r<p} where E is the 2x2 identity
matrix (cf. Rotman, [9], proof of Theorem 8.18, p. 159).
Suppose the matrix (2 3) is an element of the second center

of GL(2,p). Since (i g)€GL(2,p), there is an integer r,

a b
o<r<p such that (i g)(: 3)-r(: 3)(i 2). Hence (a+c b+d).

r(::3 §)° Since barb, either r=l or b=0. If b=0, then since
a=r(a+b) and (g g)éGL(2,p), r=l. If r=l ,then since a=r(a+b),
b=0. Hence r=1 and b=0. Since a+c=r(c+d), a=d. Since

(3 i)GGL(2,p), there is an integer u, o<u<p such that

1 1 a b a b 1 1 a+c b+d g a a+b
(o 1) (c d)=u(c d)(0 1)‘ Hence ( c d ) (c c+d)° Since

_ _ a+c a = a a a
b-O and a-d, ( c a) u(c a+c). Since a ua and aio because

(2 g)¢GL(2,p), u=l. Since a+c=ua, c=0. Hence (2 3)€Z(GL(2,P))-

Lemma 3.28: If B is an nxn matrix over any field with

in the iEE-row and 323 column and if for

 

n33 and entry biJ

some u and v with 1:u,v§n, biJ=0.when iiu and Jfiv, then

det(B)=0.

55

Proof: Expanding by the VSB column each cofactor is

zero since the associated matrix contains a row of zeroes.

Theorem 3.29: The second center of GL(n,p) is equal

 

to Z(GL(n,p)).

Proof: Let E be the nxn identity matrix and let E

13
be the matrix with entry 1 in the iEE-row, J£E~column and

zeroes elsewhere.

If n=l, Z(GL(n,p))=GL(n,p) and the result follows and
if n=2, the statement is Just Theorem 3.27.

Suppose n>2. Z(GL(n,p))={rEIo<r<p} (cf. Rotman, [9],

proof of Theorem 8.18, p. 159). Let C be an element of the
second center of GL(n,p) with entry C13 in the iEE-row, JEE-
column. Let u and v be any two integ with lfiu,v§n and
ufv. Since n>2 let k be an integer with lfikfin and kfiu and

kfv. Since det(E+Eku)#0, E+E ueGL(n,p) and hence there is

R

an integer r, o<r<p, such that (E+Eku)C=rC(E+E Hence

ku)‘

ciJ=rCiJ if ifk and qu and cku+cuu=r(cku+ckk) and ckJ

rc if qu. By Lemma 3.28, and since det(C)#O and C13:

13 if ifk and qu, r=l. Hence cuu=ckk and qu=0 if qu.

In particular cuv=0' Since u,v,k were arbitrary integers

+0 =
u

J

k3
rc

between 1 and n inclusive, C€Z(GL(n,p)) and the conclusion

follows.

Lemma 3.30: If B is a solvable normal subgroup of

 

GL(n,p) such that Z(GL(n,p))S§B and if n#2 or p>3, then

B=Z(GL(n.p)).

56

23993: If n=l, GL(n,p)=z(GL(n,p))sBsGL(n,p).

If n>2 or p>3, then PSL(n,p) is simple and not of prime
order (of. Rotman, [9], pp. 161-169). Z(SL(n,p))£EZ(GL(n,p))
(cf. Rotman, [9], Theorem 8.18, p. 159). Hence BIQSL(n,p)=
Z(SL(n.p)). (GL(n,p), BJQBNSL(n,p)=Z(SL(n,p))§Z(GL(n.P)).
Hence B is contained in the second center of GL(n,p) which,

by Theorem 3.29, is equal to Z(GL(n,p)). Hence B=Z(GL(n,p)).

 

Theorem 3.31: If n#2 or p>3, then Fit(GL(n,p))=

Z(GL(n,p)).

Proof: The result follows from Lemma 3.30, since
Fit(GL(n,p)) is a normal solvable subgroup of GL(n,p) and
Z(GL(n,p) )s Fit(GL(n.p) ) .

 

, _1o 11 01
Theorem 3.32. Let Fz-{(0 1), (1 0), (l 1)). Then

Fit(GL(2,2))=F2.

Proof: GL(2,2) is particularly well known since it is
isomorphic to the symmetric group on 3 letters. It is of
order 6, is not nilpotent and has normal 3-Sylow subgroup

F2-

_10 11 22 12 21
Theorem 3.33: Let Fa—{(O 1), (1 2), (2 l)’ (2 2), (l l)’
0 1 0 2 2 0 _
(2 0), (1 0), (O 2)}. Then Fit(GL(2,3))-F3.
Proof: ord(GL(2,3))=u8. F: is the subgroup generated

by all elements of order 9. Hence F3 is normal. Since ord(F3)=

8, F3 is nilpotent. Thus F:§Fit(GL(2,3)). F3 is contained

57

in each 2-Sylow subgroup of GL(2,3). Therefore each 2-Sylow
subgroup of GL(2.3) has at most 8 elements of order 8. If
ord(Fit(GL(2,3)) were divisible by 16, then GL(2,3) would have

(1 l) (1 2)

1 1
at most 8 elements of order 8. Since (2 l), l 0 , 2 0 ,

(i i), (i 3)’ (g g). (i g),(§ g) and (g i) are all of order
8, ord(Fit(GL(2,3))) is not divisible by 16.

Ii'ord(Fit(GL(2,3))) were divisible by 3, then GL(2,3)
would have exactly 2 elements of order 3. Since (i g), (% g)
and (3 i) are all of order 3, ord(Fit(GL(2,3))) is not
divisible by 3.

Therefore ord(Fit(GL(2,3)))=8 and Fit(GL(2,3))=F3.

The following theorems deal with the homomorphic image

of Fit(Aut(G)) in Aut(M ).

J- -1/ MJ
Theorem 3.3“: If aeFit(Aut(G)), then a induces an
m

 

element of Fit(Aut(M )) on M for each J, ljjit.

J -1/ MJ J- -l J

Proof: By Lemma 3.“, Aut(MJ_l/MJ) is a homomorphic
image of Aut(G). Hence the image of Fit(Aut(G)) is normal

in Aut(MJ_l/MJ) as well as being nilpotent.

 

Lemma 3.35: If, for some J with lilit: HJ=<ak>x<ak+l>
and if deFit(Aut(G)) with aia =higi’ h 1€HJ,gi€Kk/1Kk+1 for
lii<k and k+l<i§d and a Q: A‘ a"ll and aa =a AZ a”2

k ak ak+lgk k+l k ak+lgk+l

A u
where (Au 1) eGL(2,p) and gk,gk+1€MJ, then there is an
211 2

element yeFit(Aut(G)) such that a Yex

Y: 1 Y 2 z
k+l<iid and ak ak 1a k+l

for l§i<k and

and ak+l=ak ak+l‘

58

Proof: Let i be an integer with liiid, fv_l<i_<_fv

and vfij. If lii<k, then nv>n and h en (G)f|Mv. If

J i nv
k+l<i:d, then nJ>nv and since hienn (G), hiefln -l(HJ)
v J

)ng. Hence, in either case, h can (G)/l Mv‘ Define

C.
..'U'1’(HJ -1 1 nv

the map 3 by a18=a1(h1 1)“ ' if l§i<k or k+l<i§d and

-1 -l
-1 d B _ -l a
) and ak+1'ak+l(gk+1) ‘

B —l

_ -l a
k -ak(gk )

a Since (h1

can (G)/]Mv when fv_l<i:fv for lii<k or k+l<iid and since

v .
-l -l
-l d -1
(8k ) . (3k+l)

Theorems 3.19 and 3.20. Hence y=8aeFit(Aut(G)) is the

can (Gmmyse we Fit(Aut(G)). by

required automorphism.

 

Theorem 3.36: If G is an abelian 2-group, not elementary
abelian of order N, and if deFit(Aut(G)), then a induces an

element of Z(Aut(MJ_l/MJ)) on MJ-l/MJ for each J, lijit.

Proof: Let deFit(Aut(G)). Then, by Theorem 3.3“,
a induces an element of Fit(Aut(MJ_l/MJ) on MJ-l/MJ for
lfjfit. Let J be an integer with lfjit.

If dJ#2, then, by Theorem 3.31, Fit(Aut(MJ_1/MJ))=
Z(Aut(M l/M )) since M /M is elementary abelian of order
d J- J J-1 J
2

Suppose d3=2 and HJ=<ak>x<ak+l>° Suppose (ak M3)“ =

A. A. a". «‘1. *1. "
akak+1MJ’(ak+l“J) =ak ak+lMJ’(ak NJ) ‘ak ak+lMJ and
-1

a1
ak ak+l MJ =(aak+1MJ)a where o<Ai,u1<2 for 151:“. Then
A u A 3311 A 1111 A u

), (A3 ’)eGL(2, 2) and (A “u )"1 = (A~u~)° By Lemma 3.35

(A 22u ~"~

59

there is an element YeFit(Aut(G)) such that aiyeKkliKk+l

A u
Ya a1 Y = 2 2
for l<i<k and for k+1ii<d and ak=ak1 ak+l and ak+l ak ak+l‘
111
By Theorems 3.3“ and 3.32, (1211261?2 .
A 11
Suppose J<t or nj>l. If J<t, then nJ>1. If (11u1)=
2 2
(g i),define the map 6 by a16=ai for l5i<k and for k+l<iid
5_ 2(""1) 5 _ 2(“3'1)a
and ak= aka k+1 and ak+l-ak ak+l‘ By Theorems

3.19 and 3. 20, 6GWSFit(Aut(G)). (y6y l6 --ll)[y(y6y 6"11)y
(767 16 wl) ]= 6 contradicting the fact that Fit(Aut(G)) is

A A...” .01
nilpotent. If ( 1 1) = (l 0), then (A 3) =‘(l l) which

22 ”his
leads to a contradiction as above. Hence, from Theorem 3.32,

A ul
(A2211 ) = (0 0)e Z(GL(2,2)).

Suppose J=t and n =1. Since G is not elementary abelian

t
of order a, l<d-1=k. From the part of the proof already given,

Y induces an element of Z(Aut(Mo /M )) on Mo/M1° Hence

n -l n -1 Al
2 1 )y 2 1 If (1:1) = (01), define the map 6 by

l l. 12 "2 (n 1.1-)1
by a 6=a if 1<i<k and a 6 =a 2 a and a5 =a By
1 1 — k 1 k k+l k+l

Theorems 3.19 and 3.20, 68W$Fit(Aut(G)). (767 “'15 1)

-16-l

[Y(Y6Y )Y 1(767-16-1)-1]=6 ’contradicting the nilpotence

A u A
of Fit(Aut(G)). If (Alul) = (1 0), then (A :3:) = (01) which

2 2
leads to a contradiction as above. Hence, from Theorem 3.32,

An
(,:u:) = (3 2)e 2(GL(2,2)).

60

Thus a induces an element of Z(Aut(MJ_l/MJ)) on

MJ-l/MJ for each J, lijft.

Theorem 3.37: If G is an abelian 3-group, not elementary

 

abelian of ord 9, and if GGFit(Aut(G)), then a induces an

element of Z(Aut(M /M )) on M /M for each J, lfijft.

J-1 J J-1 J

Proof: Let aeFit(Aut(G)). Then, by Theorem 3.34,

a induces an element of Fit(Aut(MJ_ 1 /MJ)) on MJ— 1/MJ’ for
lfijit. Let J be an integer with lilit°

If dJ#2, then, by Theorem 3.31, Fit(Aut(MJ_1/MJ))=
Z(Aut(MJ —l/MJ)) since MJ—l/MJ is elementary abelian of order
2

Suppose d3=2 and HJ=<ak>x<ak+l>. Suppose (ak NJ)“ =

A: “1 ax: n2 1 A3 u
3k ak+1MJ3 (ak+1M J)% 3k ak+lMJ’(ak M3)“ ‘3 3k ak+1MJ and
-1 A u

(1 , a.
(ak+1MJ) Wk ak+1MJ where o<Ai,ui<3 for iiiiu. Then

A u u _
(A “u ), (A311 H“ 3)eGL(2, 3) and (A “3) l = ( A::3) By Lemma 3.35
A“2 2

there is an element YeFit(Aut(G))such that a: 8Kk flKk +1 for

A u A u

Y= 1 1 Y = 2 2
lfi1<k and for k+l<iid and ak ak ak+l and ak+1 ak ak+1°
A u
A “u

By

Theorems 3.3M and 3.33, ( 1)GF3 .
A 1111

Suppose J<t or nJ>l. If J<t, then nJ>l. If (A :uz )

l 1 1 2 0 l G
( ), (2 2) or (2 0), define the map 6 by a1= a1 for 1§i<k

1 2
n -l n -1
and for k+1<ifid and a6=a 1+2'3 3 1+3 J in

k k for case 1, ak

 

61

n -1 nJ-l nJ-l
3 J ' d 6 =a 2.3 a1+3 in
case 2, and akak+l in case 3 an ak+l k k+1
nJ-l nJ-l n —1
case 1, ai 3 aiii 3 in case 2 and a3 ak+1 in case 3.

In each case, by Theorems 3.19 and 3.20, 68W sFit(Aut(G)). Also

in each case yGY-ly-l=6 contradicting the nilpotence of Fit(Aut(G)).

A u

If (A:u:) = <2 2), <2 1

A u
o 2 3 a = 1 1
2 1 . 1 1) 0r (1 o)’ the“ (AAuA (1 2),

(g 3) or (g 3) which leads to a contradiction as above.

3 u 1 o 2 0
Hence, from Theorem 3.33, (Alfil) €{(0 1), (O 2)}= Z(GL(2,3)).
2 2

Suppose J=t and nt=l. Since G is not elementary abelian
of order 9, 1<d-1=k. From the part of the proof already

given ,7 induces an element of Z(Aut(Mo/Ml)) on Mo/Ml. Hence
n -l n -1 - n -1
1 l A u
3 Y: 3 2'3 1 1 = l l 1 2
) al or al . If (A ) (1 2), (2 2) or

u
0 1 r a 6- r 2 : - 3(nl-l)
(2 0) de ine the map by ai-ai or 1:1 k and ak-akal
(nl-l)
6 3

and ak+1=ak+1al . In each case, by Theorems 3.19 and

3.20, 5eWSF1t(Aut(G)). Also in each case, (ysy'ly'lrl.

(al

[Y(YGY-15-1)Y_l(YGY-15-1)-l]=6 contradicting the nilpotence

A

u
of Fit(Aut(G)). If (A3u3) = (2 2). (2 1
2 2

2 1 l l

0 2

l 0), then

) or (

( 3 3) = (i g), (g g) or (g 3) which leads to a contradiction

3 u 1 o 2 o
as above. Hence, from Theorem 3.33, (Alp!) 6“0 1), ( )}=
2 2

0 2
Thus a induces an element of Z(Aut(MJ_1/MJ)) on

MJ—l/Mj for each J, lijit.

62

 

Theorem 3.38: Let U={ala1a=a mgi where o<m<p and

1

giGMJfIQ (G) when f for lfijd}. If G is an abelian

“J 3'1 J

p-group, not elementary abelian of order H or 9, then

<15;

Fit(Aut(G))=U and ord(Fit(Aut(G)))=(p-l)pn where n=

t
2 2
1§1(fi [ni‘n1+1]'d1 )°

Proof: When f <i<f 2M . Hence, by Theorem 3.26,

3-1 —'J’ 31 J
U§Fit(So(sl)). Thus U is nilpotent.

0: m a:
If an and a1 a1 g1, then, for any geMJ_l, g géh

where heM lfgjp. Hence if an and BeAut(G), then a B -“B=

J’ i

B" 8.
[(a1 )mhi] -a

m 8
hi where h GM when fJ_l<i:fJa since “3-1

1 i J
is characteristic. Since hiBeMJ when fJ_1<i_<_fJ and since

-1
ord(h18)= ord(ai-ma B as): ord(ai), B-laBeU and hence

1
U4 Aut(G) . Therefore USFit(Aut(G)) .

Let a8Fit(Aut(G)). By Theorems 3.31, 3.3“, 3.36 and

3.37, a induces an element of Z(Aut(M1_l/M )) on M

J J-l/MJ

for each J, lfjft. Therefore aia=aimjgi where o<mJ<p and

gieM flan (G) when f <iifJ for liiid. Hence, by Theorem

J J-1
J

3.3, a€So(sl). Thus a€S°(s‘)/lFit(Aut(G))éFit(So(sl)).

Therefore, by Theorem 3.26, ml=mJ for J=1,2,...,t and QGU.

Thus Fit(Aut(G)) 90 and it follows that Fit(Aut(G))=U.

t
From Lemmas 3.11 and 3.12, ord(U)=(p-1)J1_Il[ord(MJ/){2n
J

n t
(G))] J==(p-1)pn where n=i§1(f§[ni-ni+l]—diz).

63

Corollary 3.39: For p>2, Aut(G) is nilpotent if, and

 

only if, G is cyclic and, for p=2, Aut(G) is nilpotent if,

and only if, G is multicyclic.
3.5 Sol(Aut(G)).

Definition 3.h0: Sol(H) denotes the maximal normal

 

solvable subgroup of the group H.

Theorem 3.41: If aeSol(Aut(G)), then a induces an
/M

 

element of Sol(Aut(M /MJ)) on M for each J, lijit.

J-1 J-1 J

Proof:' By Lemma 3.4, Aut(M /MJ) is a homomorphic

J-l
image of Aut(G). Hence the image of Sol(Aut(G)) is normal

in Aut(MJ_l/MJ) as well as being solvable.

Theorem 3.h2: If n#2 or p>3, then Sol(GL(n,p))=

 

Z(GL(n,p)).

Proof: The result follows from Lemma 3.30 since
Sol(GL(n,p)) is a normal solvable subgroup of GL(n,P) and
Z(GL(n,p))£ Sol(GL(n,p)) .

m
Jgi where o<m <p and

1 J

glean (G)/IMJ when rJ_1<15; for 1:15q}. If p>3, then

Sol(Aut(G))=R1 and ord(Sol(Aut(G))=(p-l)tpn where n=

 

Theorem 3.33: Let Rl={a|aia=a

t
2 2
1§1(fi ["1‘“1+1]'d1 )'

 

64

Proof: By Theorem 3.3, RIEESo(sl). Therefore, by
Theorem 2.6, R1 is in fact supersolvable. If aeRl, a induces
an element of Z Aut M
BeAut(G), B-laB also induces an element of Z(Aut(MJ_l/MJ))

/MJ)) on MJ_1/M for lijit. For

on 2134/1»:J for 1535;. Hence B-laBGRl and R14 Aut(G).
Therefore HIS Sol(Aut(G) ) .

Let aeSol(Aut(G)). By Theorems 3.41 and 3.42, a induces
)) on M

an element of Z(Aut(M for lgjit. Hence

M M
3-1/ J J-l/ J
aeRl and Sol(Aut(G))s R1. If follows that Sol(Aut(G))=Rl.

t
From Lemmas 3.11 and 3.12, ord(R1)=(p-1)tig1[ord(nn (G)
t 1

d1_ t n _ 2 2
n My] -(p-l) p where n-1§l(f1 [n1‘“1+1]'d1 ).

Corollary 3.44: For p>3, Aut(G) is solvable if, and

 

only if, G is multicyclic.

1

 

m
Theorem 3.45: Let R2=falaia=a Jg1 where o<mb<p and

"1 “1

1 31+131 and

#2 and a a=a

and d i

glean (GNIM‘j when r

J J
A

o 1+1 “1+1 "1 31
a =a a g where (

31’51+1e“nJ(G)"MJ when HJ=<ai>x<ai+l>}°

Sol(Aut(G))=R2 and ord(Sol(Aut(G)))=(p-l)t[(p+1)p(p-1)Jr'

.1-1‘35-f J

)eGL(2ap) and
If p33, then

t
pn where r=ord({JIdJ=2}) and n=i;1(f12[n1-ni+l]-d12).

Proof: By Theorem 1.29, R25 Aut(G). R2(p)EW by
Theorems 3.19 and 3.20 and the fact that [GL(2,p)](p)=l

for p33. Since W is nilpotent, R2 is solvable. Let QGR2 and

65

BEAut(G). When a induces an element of Z(Aut(MJ_l/MJ)) on
MJ-l/MJ’ so does B-laB. Hence B—laBGR2 and R2<1Aut(G). There-
fore RZE Sol(Aut (G) ) .

Let u€Sol(Aut(G)). Then, by Theorems 3.41 and 3.42,
a induces an element of Z(Aut(MJ_1/MJ)) on MJ-l/MJ whenever

d #2. Hence GER2 and Sol(Aut(G))ELRz. It follows that

J
Sol(Aut(G))=R2.
For p53, ord(GL(2,p))=(p+l)p(p-l)2. Hence ord(R2)=

t n

(p-l)t[(z>+1)p(p-l)Illr'dgltonr‘dmnJ (GM MJ)]J=(p-1)t[(p+1)p(p-1)1r-
t

pn where r=ord({JIdj=2}) and n=i§1(f12[n1-n1+1]-d:).

Definition 3.46: m(G)=max{dJ|l§j§t}.

Corollary 3.47: For p33, Aut(G) is solvable if, and

only if,¢n(G)§2.

3.6 A Necessary and Sufficient Condition
for Aut(G) to be Supersolvable.

This section contains the result that Aut(G) is super-
solvable if, and only if, G is multicyclic or elementary

abelian of order 4.

Theorem 3.48: If p>2 and if Aut(G) is supersolvable,

 

then G is multicyclic.

Proof: Suppose p>3. If Aut(G) is supersolvable, then
Aut(G) is solvable and the conclusion follows from Corollary

3.44.

66

Let p=3. Suppose for some J, 1:3:t, HJ is not cyclic.
Then there is an integer k such that fj_l<k<k+1§fj. Let
H=<ak>x<ak+l> and K=kalKk+l. Then G=HxK. By Theorem 1.39,
275 _1(H) is an elementary abelian p-group of order p2. Hence
Aut(zfn _1(H))eGL(2,3). Since ord(GL(2,3))=2“'3, 1r GL(2,3)
were supersolvable, it would contain a normal 3-Sylow subgroup
(cf. M. Hall, [3], Corollary 10.5.2, p. 159) and hence exactly
two elements of order 3. But (a i), (3 i) and (i g) are
all elements of GL(2,3) each of whose order is 3. Hence
Aut(an _1(H)) is not supersolvable. By Theorem 1.45,
Aut(ZTh _1(H)) is 180morphic to a factor group of Aut(H).
Hence Aut(H) is not supersolvable. Since Aut(H)§Aut(H)x

Aut(K)S—Aut(G), Aut(G) is not supersolvable.

Lemma 3.49: If G=<a>x<b> where ord(a)=ord(b)=4, then

Aut(G) is not supersolvable.

£3293: ord(Aut(G))=25°3. Since <ab,ab2>=G, by
Theorems 1.37 and 1.29, the map a defined by aa=ab and
ba=ab2 is an automorphism of G. Similarly for B and 7 defined
by a8=ab3, bB=a3b2, eY=a3h3 and hY=a, 3 and yeAut(G). If
Aut(G) were supersolvable, then Aut(G) would have a normal 3—
Sylow subgroup and hence exactly 2 elements of order 3. But
a,B and y are all elements of Aut(G) each having order 3.

Hence Aut(G) is not supersolvable.

Lemma 3.50: If G=<a>x<b>x<c> where ord(b)=ord(c)=2

 

and ord(a)=2n with n>1, then Aut(G) is not supersolvable.

67

Proof: ord(Aut(G))=2n+u'3. abc, bc, b is a direct basis

of G with ord(a)=ord(abc), ord(b)=ordCbC) and ord(c)=ord(b).
Hence, by Theorem 1.29, the map a defined by aa=abc, ba=bc

a 02 02 02
and c =b is an automorphism of G. a =ab, b =c and c =bc.
a3=1. Similarly the map a defined by aB=a, b8=bc, cB=b is
an automorphism of G with B3=l. If Aut(G) were supersolvable,
then Aut(G) would have a normal 3-Sylow subgroup of order 3
and hence exactly 2 elements of order 3. But a,a2 and B

are all elements of Aut(G) having order 3. Hence Aut(G) is

not supersolvable.

Theorem 3.51: If p=2 and if Aut(G) is supersolvable,

then G is multicyclic or G is elementary abelian of order 4.

Proof: Suppose G is not elementary abelian of order 4

and for some 3, lfgft, H is not cyclic. Let k+1=f

J J’
H=<ak>x<ak+1> and K=kalKk+1. Then G=HxK.
If n332, then, by Theorem 1.39 and Lemma 3.49,

Aut(ZIh _2(H)) is not supersolvable. By Theorem 1.45,
Aut(Zrn _2(H)) is isomorphic to a factor group of Aut(H).
Hence Aut(H) is not supersolvable. Since Aut(H)§Aut(H)x
Aut(K)SAut(G), Aut(G) is not supersolvable.
If n =1 and d
J J

abelian of order 4. Let D1=<a

=2, then J>l since G is not elementary

>x<ak>x<ak+l> and D =

fJ—l 2

Kk/lKk+1/\Kfj_l. Then G=DlxD2. By Lemma 3.50, since nJ_1>1,

Aut(Dl) is not supersolvable. Since Aut(Dl)EEAut(Dl)x

Aut(D )QAut(G), Aut(G) is not supersolvable.
2

68

If d >2, then, by Corollary 3.47, Aut(G) is not solvable

J

and hence not supersolvable.

Theorem 3.52: Aut(G) is supersolvable if, and only if,

 

G is multicyclic or elementary abelian of order 4.

Proof: If Aut(G) is supersolvable, then ,by Theorems

3.48 and 3.51, G is multicyclic or elementary abelian of

order 4.

I If G is elementary abelian of order 4, then Aut(G) is
isomorphic to the symmetric group on three letters which is
supersolvable.

If G is multicyclic, then, by Theorem 3.10, Aut(G) has
a normal p-Sylow subgroup and hence, by Corollary 3.8, Aut(G)

is supersolvable.
3.7 Normal Hall Subgroups of Aut(G).

This section contains the results that if Aut(G) is not
supersolvable, Aut(G) has no proper non-trivial normal Hall
subgroups and if Aut(G) is supersolvable but not nilpotent,
then Aut(G) has a normal Hall subgroup for each possible order

which is divisible by p but no others.

Theorem_3.53: If Aut(G) is not supersolvable, then

 

Aut(G) has no proper non-trivial normal solvable Hall sub-

groups.

69

Proof: Suppose B is a normal solvable Hall subgroup

of Aut(G). Let k be an integer with likit such that dedJ

for each J, lfith. By Theorem 3.52, Hk is not cyclic. Hence

t+l

(p—1)2 divides ord(Aut(Hk)) and (p-l) divides ord(Aut(G)).

Suppose p>3. By Theorem 3.43, ord(B) divides (p-l)to
t
p where n 1§l(fi [ni n1+l] di ). The exponent of the highest
2
power of p which divides ord(Aut(G)) is 1§l(f1 [ni-ni+l]-

‘ 2
di(d1+l)/2). Since d >1, dk(dk+l)/2<d . Since

k
t

2 2 2
di(di+l)/2:d1 ‘for lfiiit and dk(dk+l)/2<dk ,n<1;l(fi [ni—n1+l]-

k

d1(d1+l)/2). Therefore p does not divide ord(B). Hence if q
is a prime dividing ord(B), q divides (p—l). If qr divides

r+l t rt+l

p—l and q does not, then qr divides ord(B) and q

r(t+1) divides ord(Aut(G)). Hence there is

does not. But q
no prime dividing ord(B) and B=1.

Suppose p=3. Then, by Theorem 3.45, when a prime q
divides ord(B), q=2 or q=3. Since 4 divides ord(Aut(Hk)),
4 divides ord(Aut(G)). By Theorem 3.38, 4 does not divide
the order of Fit(Aut(G)) unless G is elementary abelian of
order 9 in which case Theorem 3.33 gives B=1 or B=Aut(G).
Hence Aut(G) does not have a normal 2-Sylow subgroup. By
Theorem 3.10, Aut(G) does not have a normal 3-Sylow subgroup.
Therefore, either B=1 or both 2 and 3 divide ord(B).

Suppose p=2. Then, by Theorem 3.45, when a prime q
divides ord(B), then q=2 or q=3. By Theorem 3.38 and the
fact that Aut(G) is supersolvable if ord(G)=4, Aut(G) does

not have a normal 3-Sylow subgroup and, by Theorem 3.10,

70

Aut(G) does not have a normal 2-Sylow subgroup. Hence either
B=1 or both 2 and 3 divide ord(B).

Suppose p53. From the above paragraphs, either B=1 or
both 2 and 3 divide ord(B).

If d =2, then 2 and 3 are the only primes dividing

k
ord(Aut(G)). Hence if both 2 and 3 divide ord(B), B=Aut(G).

If d >2, then dk(dk+1)/2<d

k k
Since (2(2+l)/2)+1=2z and di(di+l)/2:d12 for liiit and dk72,
t t
2
r+1_§_1di(d1+1)/2<1;1d1 . By Theorem 3.45, the exponent of

the highest power of p which divides ord(B) must be‘: n+r

2. Let r=ord({JIdJ=2}).

t
- 2 2
where n-1§l[f1 (ni-n1+l)-d1 ]. The exponent of the highest

t
2
power of p which divides ord(Aut(G)) is 1§1[:1 (ni-ni+l)-

2

di(di+lg/2]' From the above inequality, r+1§1(f1 [ni-n1+l]—
2\ 2

(11 ,< i§l(fi [hi-n1+1]-d1(d1+1)/2). Thus, since B is a Hall

subgroup, p does not divide ord(B). Hence B=1.

Theorem 3.54: If G is not elementary abelian of order

 

4, then Aut(G) has no non-trivial normal subgroup whose order

is relatively prime to (p-1)p.

PM: Let B4Aut(G) with (ord(B), (p-l).p)=l. Then
B has odd order and by Feit-Thompson, [1], is solvable.
Hence, by Theorems 3.43 and 3.45, B=1 for p33 and for p=2
ord(B) is some power of 3, in which case Theorem 3.38 gives

B=1.

71

Theorem 3.55: If Aut(G) is not supersolvable, then

 

Aut(G) has no proper normal Hall subgroup whose order is

divisible by p.

Proof: Suppose B is a normal Hall subgroup of Aut(G)
with p dividing ord(B). Let k be an integer with lihit such

that d_>_dJ for each J, ifiJit. By Theorem 3.52, d >1. Hence

k
BfAAut(Hk) is a normal Hall subgroup of Aut(Hk) with p

dividing ord(BflAut(Hk)). By the maximality of d if q is

k,
a prime which divides ord(Aut(G)) and does not divide ord(B),
then q divides ord(Aut(Hk)) and does not divide ord(B/lAut(Hk)).
By Lemma 3.4, Aut(H /¢(H )) is a homomorphic image of Aut(H ).
The order of the kernel of the homomorphism is p k k .

Since the highest power of p dividing ordEAut(Hk)] is

2
nkdk -dk(dk+l)/2

p and d >1, the image D of B/lAut(Hk) is a

k
normal Hall subgroup of Aut(Hk/¢(Hk)) with p dividing ord(D).
Also if q is a prime which divides ord(Aut(Hk)) but not
ord(B/iAut(Hk)), q divides ord[Aut(Hk/¢(Hk))] but not ord(D).
The elements of order a power of p generate SL(dk,p)
which is of index p-l in GL(dk,p) (cf. Rotman, [9],Lemma 8.14

and Theorem 8.17, pp. 156 and 158). Since d >1, p-l divides

k

ord(SL(dk,p)). Hence the only normal Hall subgroup of

GL(dkop) whose order is divisible by p is GL(dkop).
Therefore D=Aut(Hk/¢(Hk)), B/lAut(Hk)=Aut(Hk) and

B=Aut(G).

Theorem 3.56: If Aut(G) is not supersolvable, then

 

Aut(G) has no proper non-trivial normal Hall subgroup.

72

PrOOf: Let B be a normal Hall subgroup of Aut(G). Let

k be an integer with lixit such that d‘zd

k for each J, lfiJit.

J

Then, by Theorem 3.52, d >1.

k
By Theorem 3.53, if B is of odd order, B=Aut(G) or B=1.

If d =2 and p53, then the only primes dividing ord(Aut(G))

k

are 2 and 3. Hence Aut(G) is solvable (cf. M. Hall, [3],

Theorem 9.3.2, p. 143) and, by Theorem 3.53, B=1 or B=Aut(G).
Suppose dké2 or p>3 and suppose 2 divides ord(B). By

the maximality of dk’ a prime divides ord(Aut(G)) if, and only

if, it divides ord(Aut(Hk)). By Lemma 3.4, Aut(Hk/0(Hk)) is

a homomorphic image of Aut(Hk). The kernel of the homomorphism

2 2

-l)dk nkdk -dk(dk+l)/2

and p is the highest power

(n

has order p k

of p which divides ord(Aut(Hk)) and dk>l. Hence a prime
divides ord(Aut(Hk)) if, and only if, it divides
ord(Aut(Hk/¢(Hk))). Also a prime divides ord(B) if, and
only if, it divides the order of the image of B/IAut(Hk) in
Aut(Hk/¢(Hk)).

Suppose D is a normal Hall subgroup of GL(dk,p) with
ord(D) divisible by 2. D/ISL(dk,p) is a normal Hall subgroup
of SL(dk,p) whose order is divisible by 2. Since
ord[Z(SL(dk,p))] divides p-l and ord(SL(dk,p)) is divisible
by (p-1)p(p+1), D/l SL(dk,p)¢ Z(SL(dk,p)). Therefore, since
PSL(dk,p) is simple, [DnSL(dk.p)JEZ(SL(dk.p)))]=SL(dk,p)).
Hence [SL(dk.p):D/\ SL(dk.p)]=[(D/\ SL(dk.p))(Z(SL(dk.p))):

DA SL(dk,p)] which divides ord[Z(SL(dk,p))] which divides p-l.

73

Therefore since p divides ord(SL(dk,p)), p divides
ord(D/!SL(dk,p)) and hence ord(D).

Thus p divides the order of the image of BA.Aut(Hk) in
Aut(Hk/¢(Hk)) and hence p divides ord(B). By Theorem 3.55,

B=Aut(G).

Theorem 3.51: If G is not elementary abelian of order

 

4 and if Aut(G) is supersolvable but not nilpotent, with
ord(Aut(G)=Au where (A,u)=1, uflfiA and p divides A, then
Aut(G) has a normal subgroup of order A and does not have

a normal subgroup of order u.

£3333; By Corollary 3.7, Sn d(31) is a p-Sylow sub-
group of Aut(G). By Theorems 3.521and 3.10 and Corollary 3.6
and Theorem 2.5, Aut(G)/Sn d(81) is abelian. Hence Aut(G)
has a normal subgroup of order A.

Every non-trivial subgroup of Aut(G), being supersolvable,
has a normal Sylow subgroup. A normal subgroup of order u
would have a normal Sylow subgroup which would be a normal
Sylow subgroup of Aut(G). Since Aut(G) is not nilpotent,
G is not cyclic and (p-l)z divides ord(Aut(G)). Hence Theorem
3.38 implies that the only normal Sylow subgroup of Aut(G) is
the p-Sylow subgroup. Since p does not divide u, Aut(G) can

not have a normal subgroup of order u.

CHAPTER IV
THE AUTOMORPHISM GROUP OF AN ABELIAN GROUP

Throughout this chapter G is an abelian group and, for
p a prime, Gp is the p-Sylow subgroup of G. So G is the
direct product of the Gp's for p dividing ord(G).

4.1 Necessary and Sufficient Conditions for the Nilpotence,
Supersolvability or Solvability of Aut(G).

The theorems of this section result from the fact that
the properties of nilpotence, supersolvability and solvability
are preserved by subgroups and direct products of groups ‘
possessing them as well as corresponding results from Chapter

III which will be listed following the theorem.

Theorem 4.1: Aut(G) is nilpotent if, and only if, Gp

is cyclic for p>2 and G2 is multicyclic. (Corollary 3.39).

Theorem 4.2: Aut(G) is supersolvable if, and only if,
for each p, Gp is multicyclic or elementary abelian of order

4 (Theorem 3.52).

Theorem 4.3: Aut(G) is solvable if, and only if, G

 

P
is multicyclic for p>3 and.w(Gp)§2(cf. Definition 3.46) for

p53 (Corollaries 3.44 and 3.47).

74

75

4.2 Normal Hall Subgroups of.Aut(G).

This section ends with a theorem giving necessary and
sufficient conditions for Aut(G) to have a normal Hall

subgroup of a given order.

Theorem 4.4: For q a prime, Aut(G) has a normal q-
Sylow subgroup if, and only if, Gq is multicyclic and when
q divides ord(Aut(Gp)), with piq, either Gp is cyclic or

q=3, p=2 and Gp is elementary abelian of order 4.

23233: Aut(G) has a normal q-Sylow subgroup if, and
only if, Aut(Gp) has a normal q-Sylow subgroup for each prime
p. By Theorem 3.10, Gq is multicyclic if, and only if,
Aut(Gq) has a normal q-Sylow subgroup. By Theorems 3.56 and
3.57, if pfiq and q divides ord(Aut(Gp)), Aut(Gp) has a normal
q-Sylow subgroup if, and only if, Aut(Gp) is nilpotent or Gp
is elementary abelian of order 4, which gives p=2, q=3. For
p>2, by Corollary 3.39, Aut(Gp) is nilpotent if, and only if,
Gp is cyclic. Also by Corollary 3.39, Aut(Gz) is nilpotent
if, and only if, G2 is multicyclic, in which case there is no

q¥2 which divides ord(Aut(G2)).

Corollary 4.5: Aut(G) has a normal 2-Sylow subgroup if,

 

and only if, Aut(G) is nilpotent, in which case Aut(G) is

abelian if, and only if, G2 is cyclic.

76

Theorem 4.6: For q a prime, Aut(G) has a normal q—
complement if, and only if, Gq is cyclic (or q=2 and Gq is
multicyclic or elementary abelian of order 4) and whenever

q divides ord(Aut(Gp)) with qu, Gp is multicyclic.

£3233; Aut(G) has a normal q-complement if, and only
if, Aut(Gp) has a normal q—complement for each prime p. By
Theorems 3.56 and 3.57, Aut(Gq) has a normal q-complement if,
and only if, Aut(Gq) is nilpotent or Gq is elementary abelian
of order 4. By Corollary 3.39, Aut(Gq) is nilpotent if, and
only if, Gq is cyclic or q=2 and Gq is multicyclic. For
q¥p and q dividing ord(Aut(Gp)), Theorems 3.56 and 3.57 and
the fact that GL(2,2) does not have a normal 2-Sylow subgroup
imply that Aut(Gp) has a normal q-complement if, and only if,

Aut(Gp) is supersolvable and G is not elementary abelian of

P
order 4. By Theorem 3.52, this is the case if, and only if,

Gp is multicyclic.

Corollary 4.7: Aut(G) has a normal 2-comp1ement if,

and only if, Aut(G) is supersolvable.

Corollary4.8: If q>3 is a prime and if Gq is cyclic

 

and Aut(G) is supersolvable, then Aut(G) has a normal q-

complement.

Proof: The result follows from Theorems 4.6 and 3.52

and the fact that q does not divide the order of GL(2,2).

77

Corollary 4.9: If Aut(G) has a normal 3—comp1ement, then

Aut(G) is supersolvable.

Proof: Suppose Aut(G) has a normal 3-comp1ement. Then,
by Theorem 4.6, Aut(Ga) is abelian. By Theorem 3.52, if, for
Df3, there is an Aut(Gp) which is not supersolvable, then Gp
is not multicyclic and hence (p-1)p(p+1) divides ord(Aut(Gp)).
Therefore 3 divides ord(Aut(Gp)) and, by Theorem 4.6, Gp is
multicyclic which is a contradiction. Hence Aut(Gp) is super-
solvable for all p which divide ord(G) and thus Aut(G) is

supersolvable.

Theorem 4.10: If ord(Aut(G))=Au where (A,u)=l and if

 

ord(Aut(Gp))=Apup where Ap divides A and up divides u for each
prime p, then Aut(G) has a normal subgroup of order A if, and
only if, for each prime p dividing ord(G); (1) GD is multi—
cyclic when p divides AD and up#l and (2) when p divides

u and Ap#l, Gp is cyclic or p=2, Ap=3 and Gp is elementary

P
abelian of order 4.

2329:: Aut(G) has a normal subgroup of order A if,
and only if, Aut(Gp) has a normal subgroup of order Ap for
each p dividing ord(G).

Suppose Aut(Gp) has a normal subgroup of order Ap for
each p dividing ord(G). Let p be a prime which divides
ord(G). If p divides Ap and npfl, then by Theorem 3.56,

Aut(Gp) is supersolvable. By Theorem 3.52 and since GL(2,2)

78

has no normal 2—Sylow subgroup, Gp is multicyclic. If p
divides up and Apfll, then, by Theorems 3.56 and 3.57, Aut(Gp)
is nilpotent or Gp is elementary abelian of order 4. If

Aut(Gp) is nilpotent, then, by Corollary 3.39, G is cyclic

P
or p=2 and Gp is multicyclic, the latter contradicting Ap#1.
Suppose (1) and (2) hold for each prime p dividing
ord(G). Lep p be a prime dividing ord(G).
If p does not divide Apup, then Gp is cyclic of order p
and hence Aut(Gp) has a normal subgroup of order Ap.
If up=1, then Aut(Gp) is a normal subgroup of Aut(Gp)
of order Ap. .
If p divides AD and up#1, then, by (l), Gp is multi-
cyclic. By Theorem 3.52, Aut(Gp) is supersolvable. Since
Gp is not elementary abelian of order 4, Aut(Gp) has a normal

subgroup of order Ap, by Theorem 3.57.

If Apsl, then 1 is a normal subgroup of Aut(Gp) of order

If p divides up and Ap#l, then, by (2), Gp is cyclic or
p=2, Ap=3 and GD is elementary abelian of order 4. In either

case, Aut(Gp) has a normal subgroup of order Ap.

II.

Relations:

 

Operations:

Is
Is
Is
Is
Is
Is

Is

a

a

INDEX OF NOTATION

subset of

proper subset of

subgroup of or is less than or equal to
proper subgroup of or is less than

normal subgroup of

an element of

congruent to

Image of the set G under the mapping a

Image of the element g under the mapping a

Factor group

Direct product of groups

Index of H in G

Subgroup generated by

Set whose members are

Set of all x such that P is true

Number of elements in G

ord(G)

Order of the element g

Greatest common divisor of the integers m and n

Exponent of the group G

Restriction of the mapping a to the set H

79

III.

Groups and Sets:

 

Aut(G)
2(a)
CG(H)
NG(H)
¢(G)
Fit(G)
Sol(G)

G

nk(e)
71km)
[H.AJ

[H,An+1]

8

30(8)
81(8)

ker a

G
A-B
GL(n,p)

80

Automorphism group of G

Center of'G 1

Centralizer of H in G

Normalizer of H in G

Frattini subgroup of G

Fitting subgroup of G

Maximal normal solvable subgroup of G
Commutator subgroup of G

Subgroup, of the p—group G, generated
by the elements whose pk power is l
Subgroup, of the p—group G, generated
by the pk powers of elements of G
Subgroup generated by all h'lha where
hangs and aflAffiut(G)

[[H,A“J,AJ

Chain of subgroups of the group G;
>--->G

s:G=Go>G =1

1
{deAut(G)|61“=c

t
'for i=1,2,...,t}

}

1

{cos -1g°ec ror all geG

1.1'8 1 1—1
Kernel of the homomorphism a
Empty set

Set of all x in A and not in B
General linear group of nonsingular

nxn matrices over the field of order p

81

SL(n,p) Special linear group of nxn unimodular
- matrices over the field of order p

PSL(n,p) SL(n,p)/Z(SL(n,p))

G p-Sylow subgroup of the abelian

P
group G

 

10.

BIBLIOGRAPHY

Feit, W., and Thompson, J. G. Solvability of Groups of
Odd Order, Pacific Journal of Mathematics, Vol. 13
(1963). 775-1029.

Gorenstein, D. Finite Groups. New York: Harper and
Row, 1968.

Hall, M. The Theory of Groups. New York: Macmillan,
1959. ,

Hall, P. Lecture Notes, unpublished.

. A Contribution to the Theory of Groups of
Prime Power Order. Proceedings of tthondon Mathe-
matical Society, Vol. 36 (1933), 29-95.

. Some Sufficient Conditions for a Group to be
Nilpotent. Illinois Journal of Mathematics, Vol. 2
(1958), 787-801.

Hugpert, B. Endliche Gruppen I. Berlin: Springer,
19 7.

Polemini, A. A Study of Automorphisms and Chains of a
Finite Group. Unpublished Ph.D. thesis, Michigan State
University, 1965.

Rotman, J. J. The Theory_of Groups: An Introduction.
Boston: Allyn and Bacon, Inc., 1965.

Scott, W. R. Group Theor . Englewood Cliffs, N.J.:
Prentice-Hall, Inc., 19 .

 

82

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