A STUDY OF SCHUNCK CLASSES AND RELATED CLASSES

OF FINITE SOLVABLE GROUPS

by

Ali M. Sager

B.Sc., Al-Fateh University, Tripoli, Libya, 1979

A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF

THE REQUIREMENTS FOR THE DEGREE OF

MASTER OF SCIENCE

in the Department

0 f

Mathematics and Statistics

@ Ali M. Sager, 1986

SIMON FRASER UNIVERSITY

April 1986

All rights reserved. This thesis may not be reproduced in whole or in part, by photocopy

or other means, without permission of the author.

APPROVAL

Name : SAGER, Ali

Master of Science Degree :

Title of Thesis: A Study of Schunck Classes and Related Classes

of Finite Solvable Groups.

Examining Committee:

Chairman: A.R. Freedman

J. L. ~Frggren Senior Supervisor

S.K. Thomason

N.R. Reilly

K. Heinrich External Examiner

Department of Mathematics and Statistics Simon Fraser University

Date Approved: April 30, 1986

(ii)

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T i t l e o f Thesi s/Project/Extended Essay

A STUDY OF SCHUNCK CLASSES AND RELATED CLASSES OF FINITE -

SOLVABLE GROUPS - "-2 .

Author:

( s i gna tu re )

A l i M. S a n e r

(name)

APRIL 30,1986

(da te)

ABSTRACT

In t h i s t h e s i s we introduce the concepts of closure

operators and X-covering subgroups and we use these concepts t o -

invest igate the proper t ies of ce r t a in c lasses of f i n i t e solvable

groups.

Let G be any f i n i t e solvable group. Then G has X-covering - subgroups i f f X is a Schunck class 'and a l l the X-covering subgroups of - -

G a r e conjugate. In the case when X is a formation we show t h a t X i s - -

a Schunck c l a s s i f f X i s saturated. -

GaschZtz asks whether every {Q, closed class i s

a Schunck c l a s s . Hawkes (1973) gives a negative answer and produces a

- - closure operator R such t h a t D 5 R 5 R . He shows t h a t X is a

0 0 0 0 - Schunck c l a s s i f f

An important t oo l i n the study of c lasses i s the study of

methods of producing new classes from old. ~ e t X be a c l a s s and Y a - -

formation. Then XY = {G: G y f XI i s a saturated formation when X - - - - - and Y are. Also we extend t h i s r e s u l t i n t o non-solvable groups. -

Composition of c lasses allows us t o study the l a t t i c e of c lasses . We

invest igate Schunck c lasses with a spec ia l property known a s the

D-property. The main r e s u l t i s t h a t the s e t of these c lasses is a

complete, complemented l a t t i c e . We a l so show these c lasses a r e

characterized by an avoidance property.

Although formations were the f i r s t c lasses t o be introduced,

other c lasses , such a s extreme and ske le t a l c lasses , have appeared, and we

investigate some properties of these classes . Here the main r e su l t

i s the following: Let - F be the local formation - F( f ) where f i s a

f u l l formation function, and l e t - X be an extreme c lass . Then

F'X = F(f*) where f* ( p ) = ( f (p) ) 'X f o r a l l prime p . - - - -

DEDICATION

To my parents .

ACKNOWLEDGEMENTS

I wish t o express my deepest g r a t i t u d e t o

Professor J . L . Berggren, my Senior Supervisor , f o r suggest ing t h e

t o p i c of t h i s t h e s i s and f o r a l l h i s he lp during i t s prepara t ion . I

wish t o thank Professor N.R. Re i l ly and Professor C. Godsil f o r t h e i r

valuable comments and suggest ions during t h e seminars. My thanks a l s o

go t o t h e Chairman and t h e r e s t of t h e f a c u l t y and s t a f f members of

the Department of Mathematics and S t a t i s t i c s , Simon Frase r Univers i ty ,

f o r t h e i r kindness. I would l i k e t o thank a l l my family members and

a l l my f r i e n d s f o r t h e i r moral support .

F i n a l l y , I would l i k e t o thank my Universi ty i n Libya f o r

the f i n a n c i a l suppor t , without which t h i s t h e s i s would never have been

accomplished. I a l s o thank Sylvia Holmes f o r h e r e x c e l l e n t typing of

t h i s t h e s i s .

TABLE OF CONTENTS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A p p r o v a l ( i i )

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . A b s t r a c t ( i i i )

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . D e d i c a t i o n ( v )

. . . . . . . . . . . . . . . . . . . . . . . . . . ~cknowle d g e m e n t ( v i )

. . . . . . . . . . . . . . . . . . . . . . . . . T a b l e of C o n t e n t s ( v i i )

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

. . . . . . . . . . . . . . . . . . . . . . . CHAPTER 1 . PRELIMINARIES 5

. . . . . . . . . . . . . . . . . . . . . . . 1 . F r a t t i n i S u b g r o u p -6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . E x t e n s i o n s 1 4

. . . . . . . . . . . . . . . . . . . . . . . . . 3 . H a l l s u b g r o u p s 16 . . . . . . . . . . . . . . . . . . . . . . . . 4 . P r i m i t i v e G r o u p s 2 2

. . . . . . . . . . . . . . . . . . . . . . 1 . C l o s u r e O p e r a t o r s 27

. . . . . . . . . . . . . . . . . . . . . . 2 . C o v e r i n g S u b g r o u p s 37

CHAPTER 3 . SCHUNCK CLASSES AND FORMATIONS

. . . . . . . . . . . . . . . . . . . . . . . 1 . Schunck C l a s s e s 43 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . Formations 50

. . . . . . 3 . N o r m a l i z e r and C l o s u r e O p e r a t o r s on S c h u n c k C l a s s e s 69

. . . . . CHAPTER 4 . A COMPOSITION AND A LATTICE OF SCHUNM CLASSES 79

. . . . . . . . . . . . . . . . . . . . 1 . C o m p o s i t i o n of C l a s s e s 79 . . . . . . . . . . . . . . . . . . 2 . A L a t t i c e of Schunck C l a s s e s 93

CHAPTER 5 . EXTREME CLASSES OF F I N I T E SOLVABLE GROUPS . . . . . . . 1 0 2

. . . . . . . . . . . . . . . . 1 . E x t r e m e and S k e l e t a l C l a s s e s 102 . . . . . . . . . . . . . . . . . . . . . . . . 2 . M a i n T h e o r e m 114

. . . . . . . . . . . . . . . . . . . . . . . . . . . . BIBLIOGRAPHY 120

INTRODUCTION

The purpose of t h i s t h e s i s i s t o study the proper t ies of

ce r t a in c lasses of f i n i t e solvable groups using the notions of closure

operators and - X-covering subgroups. Our primary i n t e r e s t i s the

s t ruc tu re and proper t ies of t he c lasses themselves.

Chapter 1 contains the preliminary mater ia l needed i n t h i s

research. We a l s o introduce the concept of Hall subgroups along with

the theorem of P. Hall about t he existence and conjugacy of Hall subgroups

i n f i n i t e solvable groups. The proof of Ha l l ' s theorem is bas ica l ly the

one given by P. Hall , though it has been expanded somewhat and rearranged

(following Godsil (1974) t o demonstrate the s i m i l a r i t i e s between t h i s proof

and the proof of the Schunck's theorem.

In Chapter 2 we introduce t h e concepts of closure operators and

%covering subgroups. A s a general izat ion of t he notion of covering sub- -

groups we study b r i e f l y t h e notion of projectors .

In Chapter 3 we use t he previous concepts t o introduce the notion

of Schunck classes and t o inves t iga te the proper t ies of these c lasses .

Schunck (1967) proves the following character izat ion, which gives Schunck

classes t h e i r importance: Every f i n i t e solvable group G has - %covering

subgroups i f f - i s a Schunck c lass . In t h i s case - %covering subgroups

of G a re conjugate. Also i n t h i s chapter we introduce the notion of

formation and sa tura ted formation. Then we prove the following

character izat ion: L e t - X be a formation. Then the following a re

equivalent :

1. - is a s a t u r a t e d formation;

2. - X i s a Eg-closed;

3 . X i s a Schunck c las s . -

A c h a r a c t e r i z a t i o n of s a t u r a t e d formations, which allows us

t o cons t ruc t examples of such formations, i s i n terms of t h e i d e a of a

l o c a l l y def ined formation. Gaschctz (1963) descr ibes an important method

of cons t ruc t ing s a t u r a t e d formations, and he proves t h e following:

I f - F ( f 1 is defined l o c a l l y by f (p) then - F ( f ) i s a s a t u r a t e d

formation. Car te r and Hawkes (1967) a l s o show t h a t every l o c a l l y defined

formation can be defined uniquely by f u l l i n t e g r a t e d formations. We a l s o

use t h e not ion of - A-covering subgroups t o def ine t h e n o m a l i z e r of a

Schunck c l a s s : For any Schunck c l a s s - X , N (XI - = {G: G has a normal

%covering subgroup), and we show that N ( X ) is a Schunck c l a s s . Since - -

Schunck c l a s s e s a re {Q, EB, Do}- c losed, ~ a s c h i i t z asks whether every

closed c l a s s i s a Schunck c l a s s . Hawkes (1973) g ives a negat ive

- - answer and produces a c losure opera to r

Ro such t h a t D 5 R 5 Ro .

0 0

In t h e end of Chapter 3 w e w i l l show t h a t - X i s a Schunck c l a s s i f f

An important t o o l i n t h e study of c l a s s e s i s t h e s tudy of

methods of producing new c l a s s e s from old. In t h e beginning of

Chapter 4 we study s e v e r a l methods of combining c l a s s e s i n t o a new one.

Thus ~ a s c h c t z (1979) definesXY - - = IG: G C X I , and he shows Xy i s Y - - - - s a t u r a t e d formation i f X - and - Y a re . Erickson (1982) extends the

previous r e s u l t i n t o non-solvable groups. I f - X and - Y a r e c l a s s e s

of groups, the usual c l a s s product i s given by:

X * y = {G: G h - - as a normal sub.gxoup N E X with G/N C y} . The - -

following questions seem qui te na tura l f o r formations - X and - y . 1 When is X * Y a formation? - -

( 2 ) When does X * Y = Xy ? - - -- (3 ) When i s X * Y a Schunck c lass? - -

There i s a l so a construction f o r Schunck classes corresponding

t o subgroup generation. Blessnohi (1966) had already defined the

composition of two classes by

< -9 X 'Y 'L > = (G: G = < SIT > f o r every - X-covering subgroup S and every

Y-covering subgroup T of G 1 . We show < X,Y > i s a ~chunck c l a s s - - - i f X and Y are. - -

The following approach i s another way of combining two classes .

Let - X be a c l a s s and Y a Schunck c lass . Then -

X = {G: an X-covering subgroup E of G belongs t o Y). We show t h a t - Y - - - L

X is a formation when Y i s a formation. However, unlike t h e composition -Y - -

< - X,Y - >, Xy need not be a Schunck c l a s s even when X and Y are . - - -

Thus it i s remarkable t h a t a t l e a s t t he following holds. Let p E TT

and - X be a formation. Then (Sill i s a sa tura ted formation, P ' P-

when ST i s t h e c l a s s of f i n i t e 'IT-solvable groups.

Composition of c lasses allows us t o define a l a t t i c e of c lasses .

In the l a s t sect ion of Chapter 4, we consider a paper of Wood (1973)

inves t iga t ing Schunck c lasses with a spec i a l property known as the

D-property. The main r e s u l t i s t h a t the s e t of these c lasses i s a complete,

complemented l a t t i c e . We a lso show. tha t these classes are characterized

by an avoidance property.

Although formations were the f i r s t classes t o be introduced,

other classes , such as extreme and skele ta l classes, have since appeared

and been investigated. In the f i n a l Chapter we w i l l consider a paper

on t h i s subject written by Carter, Fischer and Hawkes (1968). Here the

main re su l t i s the following: Let - F be the local formation - F ( f 1 ,

where f is a f u l l formation function, and l e t - X be an extreme

class. Then - - F'X = F ( f *I where f * (p) = (f (p) - for a l l primes p .

CHAPTER 1

PRELIMINARIES

A l l groups mentioned a r e assumed t o .be f i n i t e . Most of t h e

following information may be found i n Gorenstein (1968) , Robinson (1982) ,

~ a s c h u t z (1979). A l i s t of symbols used and t h e i r meanings is provided

be low:

Notation :

H < G max

H Q G -

H Q G

: the i d e n t i t y subgroup

: t h e i d e n t i t y element of a group

: g is an element of G

: Card ina l i ty , number of elements of a set X

: isomorphic

: H i s a subgroup of G

: H 5 G and H # G

: H < G; H 5 H* < G impl ies H = H*

: H is a normal subgroup of G

: H d G - and H # G

: index of H i n G

: subgroup generated by X

: < x-ly-lxy 1 x x x, y C Y >

: c e n t r a l i z e r of H i n G

: commutator subgroup of G

$(GI

F ( G ) o r F i t (G) :

CorGH

'n

An

'n

v

Q8

0, CG 1

7r (GI

F r a t t i n i subgroup of G

F i t t i n g subgroup of G

the core of H i n G = i l E H ~ : g C G I , where H 5 G

the symmetric group on n symbols

a l te rna t ing group on n symbols

a cyc l ic group of order n

four group

quaternion group

the l a rges t normal IT-subgroup of G

t he set of prime d iv i sors of I G 1.

The following is the well-known modular law which is bas ic t o

our invest igat ions . Let A , B , C 5 G , i f A 5 B , then A ( B ~ C ) = B n AC.

The f i r s t sec t ion on t h i s Chapter contains most of the bas ic

r e s u l t s on F r a t t i n i subgroups and F i t t i n g subgroups that we need.

1. F r a t t i n i Subgroup.

Definition 1.1.1. The in te rsec t ion of a l l maximal subgroups

of a group G i s ca l led the F r a t t i n i subgroup of G and is denoted

Lemma1.1.2. (i) G = < x . i = 1 , 2 ,..., n > i f f i '

G = < $ ( G I , x . i = 1,2 ,..., n > . In pa r t i cu l a r i f G = @ ( G ) H f o r i '

some subgroup H of G , then G = H .

(ii) I f H - 4 G , then H has a p a r t i a l complement i n G i f f H ) @ ( G ) .

Proof. Suppose G = < @ ( G I , x € G I 1 I i 5 n > , and suppose i

Go = < x x C G, 1 5 i I n > f G . Then there e x i s t s a maximal sub- i' i

group M # G such t h a t Go I M . Also @ ( G I I M , s o

G = < @ ( G I , Go > 5 M , a contradiction. Thus < x x E G , 1 I i 5 n > = G. i ' i

The converse is obvious.

NOW i f G = $(G)H, then G = < @ ( G I , hi, 1 I i 5 m > where

(ii) I f H - Q G with H ) @ ( G I , then there i s a maximal subgroup M of

G such t h a t H ) M. But HM I G since H - < G and a l s o M < HM s ince

H M . Thus HM = G , by t h e maximality of M , s o M is a p a r t i a l

complement of H i n G . Conversely, suppose H I @ ( G I , and H has

a p a r t i a l complement K . Then G = HK I (P(G)K = K , a contradict ion,

s o (ii) holds.

Note. @ ( G I is a cha rac t e r i s t i c subgroup of G , because any - 8 € Aut sends a m a x i m a l subgroup i n t o a maximal subgroup, s o the s e t

G

of a l l maximal subgroups is invar iant under automorphisms. Hence

Lemma 1.1.3. ( F r a t t i n i argument) . I f G is a group, H a G

and P a Sylow p-subgroup of H , then G = HNG (PI . More generally,

if Q is any subgroup of H such t h a t f o r a l l g E G , Q~ i s conjugate

t o Q i n H , then G = HNG (Q) . X

Proof. I•’ x c G, pX z H = H as H Q G . NOW IpXI = / P I and s o pX i s a l s o a Sylow p-subgroup of H . Hence by Sylow's theorem

there is an element y i n H such t h a t (pXIY = P . Thus f o r a l l

- 1 x C G , there e x i s t s y € H such t h a t xy C NG(P) i . e . , x € NG(P)y .

?his implies x C N G ( p ) ~ f o r a l l x i n G . Thus G = HNG(P).

Lemma 1.1.4. d(G) i s ni lpotent .

Proof. d(G) is cha rac t e r i s t i c , s o + ( G I is normal i n G .

Let K = d(G) and l e t K be Sylow p-subgroup of K , then P

G = KN (K 1, by a F r a t t i n i argument. G P

Then K - d G . P

So K 2 d(G). Then any Sylow p-subgroup of d(G) i s ~ o r m a l i n + ( G I . P

Therefore 4 (GI i s n i lpo ten t .

Lemma 1.1.5. Let G be an elementary abelian p-group. Then

+ ( G ) = 1.

proof. G = C x C2 x . . . x Cn , where each 1 'i

is cyc l ic

of order p , s ince G is an elementary abel ian p-group. Let

Mi = R c . ?hen IG: M ~ I = Icil = p , so M~ is maximal f o r a l l i . j P i j

Hence Q ( G ) 5 M f o r a l l i , s o Q ( G ) 5 Mi = < 1 > . i

Lemma 1.1.6. I f N 4 G , then + ( N ) I + ( G I .

Proof. I f the lemma i s f a l s e , then there i s a maximal subgroup

M of G such t h a t + ( N ) $ M . Since d(N) i s cha rac t e r i s t i c i n N

and N 5 G , 4(N) -Q G . But G = ~ ( N ) M by maximality of M . Now

N = ~ f l G = N fl $ ( N ) M = + ( N ) ( N n M) by the Modular Law, s o N = N fl M

by Lemma ( 1 . 1 . 2 ) CiI,and N 5 M I a contradiction. Thus 4(N) I + ( G I .

Lemma 1.1.7. I f N/+ ( G ) is a n i lpo ten t normal subgroup of

G / ~ J ( G I , then N is n i l po t en t . In p a r t i c u l a r i f G/& ( G I i s n i lpo ten t ,

then G i s ni lpotent .

Proof. Let P be Sylow p-subgroup of N . Then + ( G I P/4 ( G )

i s a Sylow p-subgroup of t h e n i lpo ten t group N/4(G), from which it

follows t h a t 4(G)P 5 G . P i s a Sylow p-subgroup of +(G)P, s o by a

F r a t t i n i argument G = ( G ) P N ~ ( P ) = $ ( G ) N ~ ( P ) = N G ( P ) by (1.1.2) (i) . Hence N (PI = G , s o P 3 G and P 3 N . Therefore every Sylow subgroup

G

of N is normal i n N , s o N is n i lpo ten t .

Lemma 1.1.8. I f G i s p-group, then G/d (GI i s an elementary

abelian group.

Proof, For any maximal subgroup M , M 5 G , because G i s

p-group . Also J G / M I = p , s o G/M i s abelian. ' Hence G' 5 M f o r

a l l maximal subgroups M of G , s o GI 5 fl M . Thus G' 5 4(G), and

therefore G/&(G) is abel ian. To show it is an elementary group, we

need t o show the pth powers of elements of G/d(G) are t r i v i a l . It

su f f i ce s t o show f o r a l l x € G we have 2 € $(GI , and f o r t h i s it

Suf f i c e s t o show xP C M f o r a l l maximal subgroups M of G . But

G/M has order p , s o we a r e done.

Lemma 1.1.9. The f i n i t e group G i s n i lpo ten t i f f 4 ( G I 2 GI .

Proof. By t h e same argument as i n (1.1.8) , we can show t h a t i f

G i s n i l p o t e n t then + ( G I 2 G I .

Conversely, suppose G I 5 + ( G ) , and we have t o shcw G i s

n i l p o t e n t . L e t G be a Sylow p-subgroup of G . I f N (G 1 # G , then P G P

N (G 1 5 M , where M is a maximal subgroup of G . Also $(GI 5 M G P

and by t h e hypothesis G ' 5 + ( G I s o G ' 5 M . Therefore M 9 G .

But any subgroup conta in ing N (G 1 i s se l f -normal iz ing , s o M cannot G P

be normal i n G , a con t rad ic t ion . Then N (G ) = G and G 5 G . G P P

Therefore any Sylow p-subgroup of G is normal, s o G is n i l p o t e n t .

Proof. Let M1,M2,. . . ,Mn be t h e maximal subgroups

of G . Then M ~ / Q ( G ) , M ~ / ~ ( G ) , . . . , M / o ( G ) a r e a maximal subgroups n

The product AB of two normal n i l p o t e n t subgroups A and B

of a group G i s again a n i l p o t e n t subgroup. This means t h a t every group

G has a maximal n i l p o t e n t normal subgroup, which is c a l l e d t h e F i t t i n g

subgroup of G and is denoted by F i t ( G I .

Lemma 1.1.11. I f G i s a f i n i t e group then t h e fol lowing ho ld

(ii) it (G) /+ (G) i s abel ian .

Proof. (i) - f o l l a v s from Lema (1.1.4) and maximality of F i t (G) .

(ii) Since F i t (G) i s n i l p o t e n t ( F i t (GI) ' 5 4 ( ~ i t ( G I ) and s i n c e

it (G) G,4(Fi t ( G ) ) 2 4 ( G ) . Hence ( F i t (G) ) ' 5 + ( G I . Thus F i t (GI / + ( G I

i s abel ian .

Lemma 1-1 -12 . For any group G,Fi t (G/rb(G))= Fit(G)/Q(G).

Proof. Let N/+ (G) = F i t (G/+ (G) ) s o N/+(G) 2 G/+(G) and

s o N 5 G I b u t N/+ ( G ) i s n i l p o t e n t , s o by (1.1.71 , N is n i l p o t e n t .

Then N 5 F i t (GI . ~ l s o F i t ( G ) /4 ( G ) i s a n i l p o t e n t normal

subgroup of G / 4 ( G ) , s o F i t ( G ) /+ ( G ) 5 it (G/+ ( G I ) = N / + ( G ) , s o

it ( G ) 5 N . Therefore F i t (G/+(G) ) = it ( G ) /+ (G) .

Lemma 1.1.13. I f G i s solvable and + ( G I = < 1 > then

F i t (G) is t h e d i r e c t sum of minimal normal subgroups of G .

Proof. L e t L be maximal among a l l subgroups of F i t ( G ) which

can be expressed a s t h e d i r e c t sum of minimal normal subgroups of G . Note t h a t L Q G and L 5 it ( G I . W e w i l l f i r s t show t h a t t h e r e is a

subgroup S of G wi th LS = G and L n S = < 1 > . Let S be a

minimal element of t h e s e t {T: TL = G I T 5 GI. Now S n L 4 S as L P G

and S fl L 5 L s i n c e L i s a b e l i a n , because L i s a d i r e c t sum of

abe l i an normal subgroups of G .

T ~ U S s n L ~ S L = G . ~f s f l L # < 1 > , then s fl L p + ( G I ,

s i n c e +(G) = < 1 > . So t h e r e i s a maximal subgroup M of G such

t h a t G = M(S n L ) . W e now see that S = S n G = S n M ( S n L) =

(S n M) ( S n L) and, s i n c e M does no t conta in S fI L, S n M < S . However, G = SL = (S fl M ) (S I-I L)L = (S n MIL. This c o n t r a d i c t s the

choice of s . Thus s fl L = < 1 > .

W e next cons ider t h e subgroup S n F i t (GI . F i r s t ,

S n F i t ( G ) 5 S a s F i t (G) 5 G. Also by (1.1.11) and (1 .1 .12) , F ~ ~ ( G / $ ( G ) ) i s

abe l i an , b u t + ( G I = < 1 > , s o ~ i t ( G ) i s abe l i an . Then

S n F i t ( G ) 9 F i t ( G ) , then S n F i t (G) C' F i t ( G ) *S = G . I f

S n F i t ( G ) f < 1 > , then t h e r e i s an abe l i an minimal normal subgroup H

of G w i t h H C s n F i t ( G ) . AS s fl L = < 1 > w e conclude t h a t

L < L H , which c o n t r a d i c t s t h e choice of L . Then S 0 F i t ( G ) = < 1 >.

It now fol lows t h a t L = F i t ( G ) , and t h e r e s u l t fol lows.

Lemma 1-1-14. For a so lvab le group G , F ~ ~ ( G ) / $ ( G ) i s t h e

d i r e c t sum of minimal normal subgroups of G / + ( G ) .

Proof. The r e s u l t fol lows from (1.1.12), (1.1.13) and (1.1.10).

The fol lowing r e s u l t i s a very u s e f u l and important cha rac te r i za t ion

of F i t ( G I .

Lemma 1-1 -15 . F i t ( G ) = fl {cG (H/K) : H/K is a ch ie f f a c t o r of GI , '

where CG (H/K) i s t h e c e n t r a l i z e r of H/K i n G and def ined as

-1 -1 CG(H/K) = {x: x k xh C K f o r a l l h C H ) .

Proof. Let D denote t h e right-hand s i d e , C G ( H / ~ ) -4 G , s o

D S G . I f < l > = H o < H 1 < H~ < . . . < H = G i s a ch ie f s e r i e s of G I r

then < 1 > = D n H 5 D n H1 C D n H2 -C ... C D n H = D i s a normal ( * I 0 r

s e r i e s f o r D , because D fI H 5 D f o r a l l i = 1 , 2 , . . . 1 r . Now i

contained i n each C ( H /H 1. Then [D n Hi1D] 5 D fI Hi-l. Thus (*I G i i-1

is a c e n t r a l s e r i e s for D . Hence D i s n i l p o t e n t . Now D is a

n i l p o t e n t and normal subgroup of G , s o D 5 F i t (G) .

T o s h o w F i t ( G ) 5 D, l e t H/K he any chief factor of G , H/K i s

F i t ( G ) *K , a m i n i m a l normal subgroup of G/K, and since K

- G/K. T h e n

F i t (G) 'K 5 G/K, so w e get e i ther H/K n F i t (G) 'K - < > or

H / K n K - K

F i t (G) 'K H/K 5 K . If w e have the first case, then H n F i t (G) 5 K, so

it (G) 1 5 K and so F i t (GI 5 CG (H/K) . I n the second case, since

F i t (GI 'K Fit O K is n i l p o t e n t H/K n z ( ) f < 1 > and H/K i s a

K F i t (G) *K

m i n i m a l normal subgroup of G/K , so H/K i Z ( . T h e n

F i t (G) 5 CG (H/K) , so F i t ( G ) 5 D. H e n c e it (G) = D .

T h e o r e m 1.1.16. If G i s a s o l v a b L e group, then

CG ( F i t ( G I ) 5 F i t (G) .

Proof. Put F = it ( G ) , C = C G ( F i t ( G ) ) , suppose t he r e s u l t

is false, so F < CF . Then C F / F i s non- t r iv ia l n o r m a l subgroup of

G/F , so C F / F contains a m i n i m a l normal subgroup Q/F of G/F. T h e n

Q/F i s abelian, since G / F i s solvalbe, so w e have Q' 5 F < Q 5 C F . L

S i n c e F 5 Q F ( Q C ) = Q C F = Q . On t h e other hand F 2 G i m p l i e s

C 5 G , and, since Q -Q G , C n Q 5 G . Now [Q fl C , Q C , Q n C ] 5

[Q,Q,c] = [Q' ,c] 5 [F,c] = < 1 > , SO Q n C is n o r m a l n i lpo ten t sub-

group of G . T h e n Q n C 5 F , so (Q n C ) F = F a contradict ion,

because (A n C ) F = Q . H e n c e CG ( F i t (G) ) 5 it ( G I .

C F

2. Extensions .

Definition 1 . 2 .l. Let A and B be groups and suppose t h a t

0 i s a homomorphism from B t o ~ u t (A). . The s e t of a l l ordered

p a i r s ( ( a , b ) : a f A, b f B) can be made i n t o a group, i f we define the

8 product by ( a r b ) (al,bl) = (a'b (al) ,bbl) . This group i s ca l led the

semi-direct product of A and B , r e l a t i v e t o homomorphism 8 .

Note. I f be = 1 f o r a l l b f B , then we obtain jus t t he - usual d i r e c t product.

I f G is a group having a normal subgroup K , then we can

fac tor G i n t o the two groups K and G/K. The study of extensions

involves t he inverse question: Given K and G/K, t o what ex ten t may

one recapture G ?

Definit ion 1.2 - 2 . I f K and Q are groups, ari extension of K

by Q i s a group G such tha t :

(i) G contains K as a normal subgroup;

(ii) G/K Q .

Examples 1.2.3.

(i) S3 i s an extension of C3 by C2

(ii) For any two groups K and Q , K x Q i s an extension of K

by Q (and a l so of Q by K).

(iii) The semi-direct product of A and B is extension of

A b y B .

Wreath Product 1.2.4. The ordinary wreath product W of two

groups A and B may be defined i n the following way:

L e t % denote an isomorphic copy of A indexed by b C B ;

thus , % C % corresponds t o a C A . Let K = 11 % . Then each b€B

n

e l e m n t b* € B induces an automorphism b* of K by defining

n

Furthermore the mapping b* + b* i s an isomorphism between B

and a group B ̂ of automorphisms of K , s o we can form the semi-direct

n b product of K by B . And we s h a l l iden t i fy B and B ; thus kl

coincides with the conjugate b-'klb of kl by b .

Therefore W = KB i s c a l l e d Wreath product of A by B , and

i s denoted by A\B = {(k.,b) : k € K , b C B ) , with mult ipl icat ion defined

IQrnrna 1.2 -6. I f A 5 A , then A1\B I A\B . 1

Proof. See Neumann (1967).

Lemma 1.2.7. The wreath product A\B contains an isomorphic

copy of every group t h a t is an extension of A by B .

Proof. See Neumann (1967) .

3 . Hall Subgroups.

Definit ion 1.3.1. Let IT denote a s e t of primes and T i t s

complement i n the s e t of a l l primes. An in teger n i s ca l led a IT-number

whenever every prime d iv i sor of n l i e s i n IT . A subgroup H of a

group G i s ca l led a IT-subgroup of G whenever I H I is a IT-number, and

it i s c a l l e d a Hall T-subgroup of G when H i s a T-subgroup and [G:H]

is ITq-number. Note t h a t when IT cons i s t s of exactly one prime p , a

Hall IT-subgroup of G is a Sylow p-subgroup of G . A Hall IT'-subgroup

of the group G i s a l so ca l led a Hall IT-complement, provided G has a

Hall IT-subgroup.

Lemma 1.3.2. Let H be a Hall IT-subgroup of G . Then:

(i) I f H -( K 5 G , then H is a Hall T-subgroup of K .

(ii) If M Q G , then HM/M i s a Hall T-subgroup of G/M.

Proof. (i) H i s c l ea r ly a IT-group. Now IK:H 1 , d ivides I G : H ~ , but J G : H J i s prime t o any p i n , s o I K : H J is prime t o any p i n

IT . Hence H i s a Hall T-subgroup of K .

(ii) HM i s a group because M - Q G . Also I G / M : H M / M I = I G : H M ~ , and

as H 5 HM , I G : H M I d ivides I G : H I s ince I G : H I / I G : H M I = I H M I / I H I . Hence I G : H M ~ i s prime t o any prime i n IT .

Then IG/M:HM/H I is prime t o any prime i n IT . Also HM/M HPH n M ,

and H i s IT-group, then HM/M i s IT-group. Hence HM/M i s a Hall

bmma 1.3.3. L e t H and K be subgroups of G whose indices

are r e l a t i ve ly prime. Then HK 5 G and i n f a c t G = HK and

I G : H n K I = I G : H I I G : K ~ . Moreover i f H , K are Hall subgroups of G ,

then H n K i s a Hall-subgroup of G and i f H i s a Hall IT-subgroup

of G and M 5 G, then H fl M i s a Hall T-subgroup of M .

Proof. Let I G ~ = n , I G : H I = n I G : K I = n and I H n K I = a , 1 ' 2

then / H I = ah, I K I = ak f o r some i n t e g e r s h,k . I G 1 Since - - - n s o n = nlah = n2ak, s o n h = n k , bu t I H I l' 1 2

(nl,n2) = 1 , s o h = n r , k = n r f o r some i n t e g e r r . Then 2 1

n = a n n r . On t h e o t h e r hand n = I G I > I H K ~ = I H I I K I - - (ah) (ak) - -

1 2 I H " K I a

2 ahk = an n r , it follows t h a t r = 1. Then

2 1 an n

2 1 I H K ~ = an n = n , and hence G = HK . Also I G : H n K I = ------- 2 1 - n n = a 2 1

Let p be any prime dividing I H n K I . Now H n K 5 H ,

H n K 5 K and H , K a re Ha l l subgroups, then p does not d iv ide

IG:HI , I G : K I s o does no t d iv ide I G : H I I G : K I , s o does not divide

I G : H I I G : K I = IG :H n K I . ~ h u s H n K is a Hall =-subgroup of G s ince

H n K i s IT-group because H fI K 5 H .

Fina l ly w e show H n M is a Hall IT-subgroup of M . A s

H fl M 5 H , s o H fl M i s IT-group. ~ l s o I M : H fl M I = I M H : H ~ s ince MH

i s a subgroup of G , then IMHI divides I G l , - s o 1MH:HI d iv ides

1G:HI.. Then 1HM:HI i s r e l a t i v e l y prime t o any prime p i n IT , hence

H fl M is a Hal l IT-subgroup of M .

Remark 1.3.4. A f i n i t e group G need no t contain any Ha l l

IT-subgroup ,

Example. L e t G = A and n = {3,5}, = 23.3.5, a n d a n y Hal l 5

(3,s)-subgroup of % would have index 4 . Suppose H 5 A5 and

[ A ~ : H ] = 4. Take t h e regu la r r epresen ta t ion of A5 as a permutation

group on t h e four cose t s of H i n A5 . Then t h e r e exists a homomorphism

f : A5 + S4. Also Ker f # { l ) f o r i f Ker f = 11) then As would be

isomorphic t o a subgroup of S4 . But I A ~ ~ > IS41 . Therefore Ker Q %,

which i s a contradiction because A i s simple. Thus A5 has no sub- 5

g r o q of index 4 .

Theorem 1.3.5. Let G be a solvable group and suppose

IT - c IT (G) . Then

(1) G has a t l e a s t one Hall IT-subgroup.

( 2 ) Any two Hall IT-subgroups are conjugate.

(3) Any IT-subgroup of G l i e s i n some Hall IT-subgroup of G .

Proof. We consider two cases :

Case (a ) : Suppose G contains a minimal normal subgroup M ,

such t h a t I G / M I i s not a IT-number.

(1) By induction on I G I , G/M contains a Hall IT-subgroup H/M say.

Since G/M i s not IT-group, H/M < G/M, s o H < G . Hence again by

induction H must

I G : K I = I G : H ] - I H : K of G/M , s o I G : H I

I H : K ~ . Then I G : K

contain a Hall IT-subgroup, say K . Now

a l so IG/M:H/M I = IG:H I and H/M is Hall a IT-subgroup

is not d iv i s ib l e by any prime i n IT , and a l s o

is not d iv i s ib l e by any prime i n IT . Since K i s

a l so a IT-group K i s a Hall IT-subgroup of G .

(2) L e t K , K2 be any two Hall IT-subgroup of G , then by (1.3.2) (ii)

KIM/M , K2M/M are Hall IT-subgroups of G/M, and by induction they are

conjugate i n G/M. So we have ( K ~ M ) ' = K M f o r S O W g i n G , s o 2

g K~ 5 K ~ M = 1

K 2 M . Now G/M i s not IT-group, SO K2M/M < G/H , s o

K2M < G . But K' , K2 are Hall IT-subgroups of G , so by (1.3.2) (i),

ICg K are Hall IT-subgroup; of 1, 2

K2M . Again by induction K:,K~ are

con jugate i n K M I s o K K are con jugate i n 2 1' 2 K2M . Hence K K are

1' 2

conjugate i n G .

( 3 ) L e t J be any IT-subgroup of G . Then JM/M i s a T-subgroup of G/H.

Hence JM/M l ies i n some Hall IT-subgroup H/M of G/M . But G/M is

not IT-group , s o H/M 2 G/M, s o H < G . Then, a s i n (1) , H contains

a Hall IT-subgroup K , which is a l s o a Hall IT-subgroup of G , and s o KM/M i s

a Hall IT-subgroup of G/M. But KM/M 5 H/M . Then KM = H , hence

J 5 J M 5 KM since J M 5 M . Hence J 5 KM. And then by induction

J l i e s i n Hall IT-subgroup of KM . But K i s a Hall IT-subgroup of KM.

Thus J I K .

Case (b) . Suppose G has no minimal normal subgroup M , such

t h a t I G / M I is d iv i s ib l e by some p i n IT' . Let MI, M2 be two

d i s t i n c t minimal normal subgroups of G , then M n M 5 G , S O 1 2

M n M 2 = < 1 > . Hence I G I = I G : M n M I = I G : M 118 -M n M 2 1 = 1 1 2 1 1 - 1

I G : M ~ I I M M 1 2' - M 2 I . Since ne i t he r I G : M ~ I nor / G : % I a re d iv i s ib l e by p

i n IT' , so I M M :M I is not d iv i s ib l e by p € . Thus G i s T-group, 1 2 2

a contradict ion, because IT # IT ( G ) . Therefore, G must have a unique

minimal normal subgroup say M , and G/M i s n-group. Now i f ] M I i s d iv i s ib l e by p i n IT , then G i s IT-group, since G/M is IT-group,

so 1 ~ 1 i s IT' -group. But M i s a minimal normal subgroup of a solvable

group G I then M i s p-group, s o M i s a Sylow p-subgroup of G .

(1) Let L/M be a minimal normal subgroup of G/M. Then

where p i s an in t ege r , and q i s a prime

i n IT , s ince G / M i s IT-group. Clearly

a p f q , where ] M I = q , since M i s

IT'-group. Let L be a Sylow q-subgroup 1

of L . ~ e t N = N ( L ) , N = N ( L ) = N 1 n M - 1 G 1 M 1

W e claim N = < 1 >. F i r s t we show N 5 Z ( L ) . Ll

1 N1 as M 9 G , and a l s o Now N = N n M -

M n L, = < 1 >, s o N fl L~ = < 1 >, since < 1 >

N = N n M . B U ~ N , L ~ 4 N~ , SO [ N , L ~ J 5 N n L, = < 1 > . Hence 1

N 5 C (L ) . Also, M i s a minimal normal subgroup of G , and SO M G 1

is abelian. As N 9 M we see t h a t N cen t ra l izes 5 M = L , s o

N I Z (L) , s ince N < L . However Z ( L ) i s a cha rac t e r i s t i c subgroup of

L , and L P G , so Z ( L ) 5 G . Now since M is the unique minimal

normal subgroup of G . Then e i t h e r z (L) = < 1 > o r M 5 z (L ) ,

Z (L) n M # < 1 > which contradicts the minimality of M . If

M 5 Z(L), then M cen t ra l izes L , and s o L 5 L . But L1 is a 1 1

S y l m q-subgroup of L , so ILL I , I G : L ~ I a re coprime, so Ll i s a

cha rac t e r i s t i c i n L , and a l s o L -Q G . Hence L 9 G . Thus we have 1

a normal q-subgroup of G , s o M i s not the unique minimal normal

subgroup of G , a contradiction. Thus Z ( L ) = < 1 > . But

N 5 Z(L) = < 1 > . Therefore N = < 1 > .

We now show N (L ) i s a Hall 7T-subgroup of G . L1 is a G 1

S y l w q-subgroup of L , and L 5 G , s o by the F r a t t i n i argument

Now complement

i n G . Then s ince M is a TI-group, and G/M i s a T-group, it follows

t h a t , N (L i s a Hall IT-subgroup of G . So i n t h i s case G has a G 1

Hall IT-subgroup.

(2) Let H be a Hall T-subgroup of G . Then I G : H I and I G : L J a re

coprime, so by (1.3.31,G = HL and thus I L : L fl H I = I H L : H ~ = I G : H I =

s o every Hall IT-subgroup of G contains a Sylow q-subgroup of L . W e

may assume w.1.o.g. t h a t L 5 H , then L = H fI L 5 H , as L 2 G , s o 1 1

H 5 N (L ) . However H I NG(L1) are both Hal l IT-subgroups of G , so G 1

H = N (L ) . Thus every Hall IT-subgroup of G i s the normalizer of some G 1

conjugate of L , andhence is conjugate t o NG(L1). 1

( 3 ) I f K is a IT-subgroup of G , and not a Hall IT-subgroup of G ,

then KM < G , s o G = HKM. Then I G : H I = IHKM:H I = IKM:H n K M I .

a a But I G : H I = p , s o IKM: H n KMI = p . On the other hand, K n M = < 1 > ,

a because M is a TI-group, and K i s a T-group, s o I KM/KI = / M I = p . Hence H n KM i s a subgroup of H and i s of the same order as K . Now I K M : K ] = M and M i s a IT1-group, s o K is a Hall IT-subgroup of

KM, therefore K i s conjugate t o H n KM i n KM. Hence K i s

conjugate t o a subgroup of H .

4. Primitive Groups.

Definition 1.4.1. I f M 5 G , then we define Cor M = n M~ , G

g CG

s o Cor (MI is the la rges t normal subgroup of G lying i n M . G

Definition 1.4.2. A group G # < 1 > i s c a l l ed primitive i f

there e x i s t s a maximal subgroup M of G such t h a t Cor (M) = < 1 > . G

Remarks 1.4.3.

(1) I f G i s a group, and M is a maximal subgroup of G , K = Cor M , G

then G/K i s pr imi t ive , and a l l the pr imit ive f ac to r groups are

obtained i n t h i s way.

(2) It follows e a s i l y t h a t 4(G) = fI { C O ~ ( M ) : M i s maximal i n G},

because 4 ( G ) = fI IM: M is maximal i n G), but any subgroup

conjugate t o the maximal subgroup i s maximal, s o 4 (GI = fI {Mg: M is

maximal). Then 4 ( G I = n {COX (MI : M i s maximal i n G}. We note

t h a t d(G) = < 1 > when G i s pr imit ive .

( 3 ) I f H 5 G and G = HK fo r a l l a subgroups K such t h a t G/K i s

primitive then G = H by (2) . (4) I f IGI = p when p i s prime, then G i s pr imit ive , s ince < 1 >

i s a maximal subgroup.

Lemma 1.4.4. A n i lpo ten t group i s primitive i f f it has a prime - order.

Proof. I f G i s n i lpo ten t and pr imi t ive , then there e x i s t s a

maximal subgroup M of G such t h a t n M~ = < 1 > . But M -4 G s ince G

g CG

i s n i l p o t e n t , s o fI M~ = M . Hence M = < 1 > , s o G i s a c y c l i c of g EG

prime o rde r p . The converse is obvious.

Lemma 1.4.5. Let G be a p r imi t ive group and M a maximal subgroup

of G such t h a t Cor ($1) = < 1 2. I f H # < 1 > i s any n i l p o t e n t normal sub- G

group of G , then M i s a complement of H i n G , i - e . , M n H = < 1 >

and MH = G .

Proof. F i r s t , H $ M s i n c e H a G and Cor (MI = < 1 > . Thus

s ince M i s maximal and H i s normal, G = MH. Suppose M n H # < 1 >.

Also H a G s o M n H d M and s i n c e H i s n i l p o t e n t NH(M H ) > M n H .

Hence M n H a MNH(M n H) = G by maximality of M . This con t rad ic t ion

shows H n M = < 1 > , s o M i s complement of H i n G .

Lemma 1.4.6. Let G be a p r imi t ive so lvab le group. Then G

has a unique n o n - t r i v i a l n i l p o t e n t normal subgroup M . In p a r t i c u l a r '

M i s t h e unique minimal normal subgroup of G .

Proof. Let M be a maximal subgroup of G , with

Cor (M) = < 1 > , and l e t N be a minimal normal subgroup of G , H a G

n i l p o t e n t normal subgroup of G . Since N , H a r e n i l p o t e n t normal

subgroups, s o NH i s a n i l p o t e n t normal subgroup of G , and s o by (1.4.5),

G = NHM. Now lNHl = I G : M I = I N I , s o N = NH. Then H = < o r H = N .

So i n both cases G has a unique n o n - t r i v i a l n i l p o t e n t normal subgroup

N , and s o i n p a r t i c u l a r G has unique minimal normal subgroup of G .

Defini t ion 1.4.7. I f G has unique minimal normal subgroup N ,

we say G i s monol i th ic with monolith N .

Theorem 1.4.8. Let G be a p r imi t ive group, S < G , and M be

minimal normal subgroup of G . Then t h e fol lowing a r e equivalent .

(1) G = SM. (2) S i s amaximal subgroup of G with Cor (S) = < 1 > . G

Proof. 1 + 2 . S fI M 5 M s ince M is a b e l i a n and S fI M S S

as M 3 G , s o S n M -d G and by (1.4.6) , M i s a unique minimal normal

subgroup of G , s o S conta ins no minimal normal subgroup of G . Hence Cor s = < 1 > . NOW l e t s 5 S* < G and S* n M = < 1 > .

G

G = SM, s o G = S*M, b u t S* n M 2 M as M i s abe l i an , S* fI M Q S*

a s M I1 G , s o S* M 51 G. Hence S* fl M = < 1 > by the minimality

of M , now IS*^ = I G / M I = I S [ and S = S* . Therefore S i s a

maximal subgroup of G with Cor S = < 1 > . G

2 + 1 follows immediately from the maximality of S and M ) S.

Theorem 1.4.9. Let G be a group, M be a minimal normal sub- .

group of G and S1, S2 be maximal subgroups of G such t h a t MS1 = MS2 = G

and Cor S = < 1 > = Cor S G 1 G 2 Then S1 and S2 a r e conjugate

under M ' .

Proof. If G = M, then

S1 = S2 done. Otherwise, l e t L/M

a be a chief f a c t o r of G , now ] M I = p

s ince M i s a minimal normal subgroup

of G , I L / M I = qB s i n c e L/M is a < 1 >

minimal normal subgroup of G/M. Now b ~ ( 1 . 4 . 6 1 , ~ i s not n i l p o t e n t , s o

p # q . Since S1, S2 a r e complements of M i n G by (1.4.51, it

follows t h a t S fI L i s a Sylow q-subgroup of L and S2 fl L i s a 1

X Sylow q-subgroup of L . Therefore S1 n L = (~S2 n L) f o r some

X X X x i n L , s o S1 n L = s2 fl L 5 s2 . NOW i f s1 # S2 , then

G = < S ,sX > and then S n L would be a normal subgroup of G I s o 1 2 1

S n L would be a non-trivial n i lpo ten t normal subgroup of G , which 1

X contradicts (1.4.6) , since G i s pr imit ive . Thus S1 = S2 f o r some

Theorem 1.4.10. Let G be a group. Then the following are

equivalent :

1. There e x i s t s a minimal normal subgroup M with C ( M ) = M . G

2. There e x i s t s a maximal subgroup S with Cor S = < 1 > and G = SM . G

Proof. (See the Figures a t t he end of the proof ) .

(1) implies ( 2 ) . I f M = G , then G i s cyc l ic

of prime order and t h e theorem holds , since i n t h i s case

S = < 1 > . Otherwise l e t L/M be a chief f ac to r of G ,

I L / M ) = qa and I M I = pB. Now i f p = q, then L i s a p-group, so

M n Z ( L ) f < 1 > but M f l z (L) 3 G , s o M 5 z (L) by minimality of M

and s o CG ( M ) 1 L , a contradiction. Therefore p # q . Let T be a

Sylcw q-subgroup of L and S = N (T) . By the F r a t t i n i argument G -

G = SL = STM = SM. I f T were normal i n G , then [T,M] 5 T n M = < 1 >,

s o T 5 C (M) = M , a contradiction. Therefore S < G , a l s o S M 3 S G

as M 2 G I S n M fl M s ince M i s abel ian, s o S n M 3 SM = G . Hence S n M = < 1 > by minimality of M . Now

s o Cor S 5 C (M) = M. Therefore Cor S = < 1 > . G G G

2 + 1. L e t M be a minimal normal subgroup of G and S a

maximal subgroup of G with K = C o r S = < 1 > and G = S M . Now G

S n C G ( ~ ) 2 S as C (M) 2 G I and it is c l ea r tha t M nomalizes G

S n C G ( M ) = < 1 > , a l so M C C (M) since M i s abelian. Therefore G

S ~ M C S ~ C (MI = < w . T ~ U S c (MI = M . G G

CHAPTER 2

In t h i s Chapter we w i l l i nves t iga te the important ideas of

closure operators, &covering subgroups and &projectors.

Any c l a s s 5 of groups i s supposed t o be non-empty and contain,

with any group G , a l l groups isomorphic t o G . We a l so follow the

convention t h a t every c l a s s contains a l l groups of order 1. An

X-group i s a group belonging t o A . - The following a re some p a r t i c u l a r c lasses of groups, together

with notation t h a t s h a l l be used f o r them throughout t h i s t he s i s .

A : the c l a s s of f i n i t e abelian groups -

N : the c l a s s of f i n i t e n i lpo ten t groups -

7 : the c l a s s of f i n i t e super solvable groups -

P : the c l a s s of f i n i t e p-groups -

: the c l a s s of f i n i t e solvable T-groups -

U : the c l a s s of f i n i t e solvable groups -

1. Closure Operators.

Closure operators on c lasses of groups, introduced by

P. Hall (1963) , have proved t o be a convenient way of describing the

most useful types of c lasses , such a s formations (c lasses t h a t a re

Q-closed and Ro-closed) and F i t t i n g c lasses (c lasses t h a t are

S -closed, N -closed). n 0

A comprehensive study of c losure opera tors was made by

Maclean (1969) i n a work which i s no t ava i l ab le t o m e .

Def in i t ion 2.1.1. A c losure opera to r i s a map C a s s ign ing

t o each c l a s s of groups Y a c l a s s of groups CY such t h a t :

(1) Y c C y i . e . C i s expanding -

2 ( 2 ) C Y = CCY (which w e w r i t e C Y ) , s o C is idempotent

(3) ~f Y c Z , then C Y c CZ , i . e . , C i s monotonic. - -

~f Y = cy we say y i s C-closed.

Def in i t ion 2.1.2. I f 3 i s any c l a s s of groups, then t h e

c l a s s e s S ~ I sn& S N b Q ~ I R ~ X . N ~ Y , E &, E X, D ~ & and P& def ine as follows: P 4-

1. G € SL i f f G i s a subgroup of an X-group. -

2. G € snL i f f t h e r e e x i s t s H Cd such t h a t G 4 H .

3. G € sN& i f f G is a subnormal subgroup of an A %group.

4. G C QX i f f G i s an epimorphic image of an - %group.

5. G € R X i f f G has normal subgroups N - Jk s uch t h a t 0-

6 . G € N X i f f G has normal subgroups NI I . ,Nk such t h a t 0-

G = NIN Z... Nk and N 6 X , i = 1,2 ,... ,k. i -

7. G € E X i f f G has normal p-subgroup M such t h a t G/M 2 . P'

8. G € E X i f f G has normal subgroup M I 4 ( G I , G/M 6 Y . 4-

9. G C PX - i f f G has - %subgroups N ..., N such t h a t G = N 1 ' k

"Nk.

10. G € D X i f f G i s t h e d i r e c t product of groups i n X . 0- -

Lemrna2.1.3. The opera tors S, Q, Ro t D E and E a r e 0 ' 4 P

closure opera tors .

Proof. The e n t r i e s marked x on t h e c h a r t below a re elementary.

W e w i l l prove t h e remaining.

Ex Mon Id

To show R is idempotent l e t G € R ~ X . Then G has normal 0 0- L

subgroups N1 ,N2 , . . . tNk s u c h t h a t n N . = < I > and G C R X 1

i=l Pi 0-

i = 2 . . . , k . Therefore each G/Ni has normal subgroups

i j = 1, ... ,L. such t h a t n M~~ = N~ 1

and ~ / k ~ ~ f 5 . Now j=1

k 2 n Mij =

". = < 1 > . Then w e have G C R ~ X and s o R X = R ~ X . 1 0- i j i= 1

W e show E is idempotent. I f G € E 2 x then G has normal 4 Q '

subgroup K 5 4 (GI . G/K C E z . Therefore G/K has normal subgroup

H/K 5 4(G/K) such t h a t G/H C - X . Now H/K 5 4(G/K) = d(G)/K, s o

H 5 4 (G) . Therefore G f ~8 . Thus E i s idempotent. 4

Fina l ly t o show E i s idenpotent l e t G C E (E X I , s o G has P P P-

normal p-subgroup M such tha t G/M C E X . Therefore G/M has normal P

p-subgroup H/M

p-group. Thus

Note. -

such t h a t G/H C X . Then G 6 E X s ince H i s a - J?-

Sn is not a closure operator because it i s not

idempotent. This i s because a normal subgroup of a normal subgroup i s

not necessari ly normal i n G , otherwise every subnormal s e r i e s would be

normal s e r i e s .

Remarks 2.1.4. (i) I f C is a closure operator then f o r any

c l a s s - X, CX - i s the smallest C-closed c l a s s containing - X . (ii) I f A and B a re any two closure operators and we denote by

(A,B)X - the smallest c l a s s containing - X which i s both A-closed and

B-closed, then {A,B) i s a l s o closure operator.

(iii) A p a r t i a l ordering of closure operators i s defined as follows

A I B means t h a t 0 5 - ( i v ) The product AB

by ABX - = A ( B X ) , i s i n -

The following

closure operators t o be

BX - f o r any c l a s s X . -

of two closure operators A and B , defined

general not a closure operator.

lemma i s useful c r i t e r i o n f o r the product of two

a closure operator.

Lemma 2.1.5. L e t A and B be closure operators. Then AB i s

a closure operator i f f BA 5 AB; moreover, i f t h i s is the case,

AB = < A,B > .

Proof. Let BA 5 N3. Since (AB) = A (BA) B 5 AABB = AB , the

condition i s su f f i c i en t . On the o ther hand, i f AB i s a closure operator,

A 5 AB, B 5 AB and AB 5 < A , B > , s.o AB = < A,B > and BA 5 AB.

Lemma 2.1.6. The re l a t i o n BA 5 AB i s v a l i d in t h e following

cases :

( i v ) A = E 4

B = D , Q 0

Proof. (i) F i r s t w e show SE 5 E S . Let G C SE X . Then P P P-

G 5 H C E X and hence H has a normal p-subgroup K such t h a t H/K C X. P- -

NOW G/G n K % GK K 5 H/K c X , SO G/G n K c sX, bu t G n K i s a normal / - -

p-subgroup of G , the re fo re G C E SX . Hence SE 5 E S . P - P P

TO show SR 5 RoS. Let G f S R ~ X . Then G 5 H C R ~ & . 0

Hence H has a normal subgroups Ml ,M2 . . . . ,Mn such t h a t n M . = < 1 > 1

i= 1

n V i = 1,2 ,... ,n and n ( G ~ M . ) = < 1 > s o G C R ~ S ~ . Hence SR 99,s.

1 i=l

0

For t h e remainder of (i) t h e v e r i f i c a t i o n , is s t ra ight forward.

W e t u r n now t o (ii) . (ii) L e t G f E N X . Then G has normal p-subgroup M such t h a t

P 0-

G/M C NO? . Hence G has normal subgroups N1, N2 containing M such

. Since Ni t h a t N ~ / M C - X , i = 1,2. and G = N N has normal

p-subgroup M such t h a t C E X , S O N1.N2 p-

G C N E . Therefore E N 5 N E . 0 P- P O O P

To show E S 5 SE . Let G C E SX . Then G has normal P P P -

p-subgroup M such t h a t G/M C sX, - SO G/M 2 H 5 K C - X . NW

M\K C E X, where M\K denotes t h e wreath product of M and K s i n c e P-

M'\K has normal p-subgroup such t h a t M\K I[M S K C X . By (1.2.6). / - k CK

MIK has a subgroup isomorphic t o M\H s i n c e H 5 K . By (1.2.7) M\H has

a subgroup isomorphic t o G s ince G i s an extens ion of M by H .

Therefore G C SE X . P-

(iii) ~ e t G C R ~ Q X - . Then t h e r e a r e subgroups N. 4 G where 1

i = 1 ,2 ,..., n and n N . = < 1 >, and groups Gi

C X with M. -9 Gi such 1 1

i=l

t h a t G M i L e t Oi be t h e induced homomorphism Oi : G + G ~ / M ~ . L

Fix a set , inc lud ing t h e i d e n t i t y , of cose t r ep resen ta t ives of Mi

i n

represen ta t ive . The group H = <(.q 1 (g)ml,...,qn(g)mn): g C G, m i C M

f o r a l l i > i s contained i n t h e d i r e c t product G1 x G2 x . . . x Gn ,

and t h e subgroups L i = < (ql (.g)ml,.. . .qi-l (g)mi-l ,1 Icpi+l ( g ) m i + . . . : g C Ni,

n f o r a l l j # i m C M . > a r e normal i n H . W e have n Li = < 1 )

j I i=l

The preceding i s from P a t r i c k DIArcy's t h e s i s (1971),

" f-abnormality and t h e Theory of F i n i t e Solvable Groups " .

To show Suppose G E E . Then t h e r e i s an P Q E ~ - P -

F E X - and N 5 F such t h a t (G) 2 C F/N , where 0 (G) i s t h e l a r g e s t P

normal p-subgroup of G . Let 9 be the induced homomorphism 0: G-+ F/N

def ined by 8 (g) = Q (g) N where Q (g). i s one element of a given set ,

inc luding the i d e n t i t y , of cose t r e p r e s e n t a t i v e s of N i n F . The

group A = <(g,cp(g)n) : g E G , n E N > i s contained i n G x F . Now

A 0 (GI x < l > - P' P-

A C Q E X . Y F E X , s o A E E X a n d h e n c e

P

But G % A

< I > X N r S O G E QE X . Thus

P' P _C QEP

( i v ) Let G E D E X s o G = G~ x G~ x . . . x G~ where € E X , 0 Q- Gi Q-

i = 1 2 , . . . , . Hence f o r each i , Gi has normal subgroup Mi

such

t h a t Mi 5 Q ( G i ) , G/Mi 6 X - . Let M = M x M2 x ... 1 X M n s o M 4 G

and G/M = G Z ( G ~ / M ~ ) x . . . x (G /M I C D ~ X , a l s o

M1 X ... X M n n n

M = M 1

x M2 x . . . x Mn 5 Q(G1) X 4(G2) X . . . x Q(Gn) 5 + ( G I , t he re fo re

G C E D ~ X . Hence D E 5 E D 0 4 4 0 '

To show 0E4 C E4Q. Let G/K E QE X where G E E& . Then 9-

G has normal subgroup N 5 4 (G) , G/N E - X . Now G/K NK/K G/NK E QX / - s i n c e G/N E - X . On t h e o t h e r hand NK/K 5 6 (G/K) s ince N 5 4 (G)

t h e r e f o r e G/K f E ~ Q X - . Thus QE E Q . 6 - 6

34

Lemma 2.1.7. Let X and y be C-closed c lasses of groups. - - Then X n Y is C-closed. - -

Proof. C i s a closure operation. Then c ( X - n - Y) 5 @ ,

c(X - n - Y) r cx, SO c(X - n - Y ) r cX - n cY - = - x n - Y a l so - X n - Y r c(X - n - Y ) .

Therefore C ( X n Y ) = X n y . - - - -

Remarks 2.1.8.

1. I t i s c l ea r t h a t the c lasses A - , U are Q , S and Do-closed. + -

2 . The c l a s s of cycl ic groups is Q , E closed but not Do-closed. 6

Proof. It i s c l ea r ly Q-closed. To show it is E closed, 4-

l e t - X denote the c l a s s of cyc l ic groups. Let G E E X . Then G has d-

normal subgroup M 5 9 ( ~ ) , G/M E X - . Choose x i n G such t h a t xM

generating G/M, s o G = < x,M > = < x > since M 5 + ( G I . Then G i s L

cyc l ic and - X i s E -closed. But X i s not closed since the d i r ec t 9 -

product of cyc l ic groups i s not a cyc l ic group i n general.

3. The classes U, N, 7 , 1I and P are R and E closed. - - - - - 0 4-

TO show - U i s R -closed suppose G C R& . Then G has 0

normal subgroups Nl , . . . ,Nk such t h a t fI N = < 1 > and G/N~ E & i

i=l

f o r a l l i = 1.2. ..., k . We define a map 8: G - + G / N ~ X G/N2 X ... X G/Nk

by 8 (g) = (gN1,. . . ,gt\) . It i s c l e a r t h a t 8 is well defined. Now

we w i l l show 9 i s a homomorphism:

g-g' C Ni -1 f o r a l l i , s o f o r a l l i , g ( g ' ) C n Ni = < 1 > .

Hence g = g ' , and s o 8 is 1-1 . Therefore G i s isomorphic t o a

subgroup of G / N ~ x G/N~ x ... x ~ / i ~ , but G/N1 x ... x G / N ~ is

solvable, s o G C U , Thus U i s R -closed. In the same way we can - - 0

show , , and - P are R -closed. 0

To show - U i s E -closed, l e t G € E.U . Then G has normal d @-

subgroup K 5 d ( G ) : G/K C - U . Now ~ ( G I i s solvable, s o K i s

solvable. Also G/K i s solvable , so G C - U . Therefore 9 is E -closed. 9

TO show is E -closed l e t G C E$ . Then G has a normal 6 .

subgroup K 5 6 (GI , G/K f - N , Now G/$ ( G ) Y c g since E i s 4 (GI /K

Q-closed. Then by (1.1.71, G i s n i l po t en t , s o - N i s E -closed. d

It is

not d i f f i c u l t t o show and a re a l s o E -closed. 4

4. - A is R -closed but not E -closed. We can show A is Ro-closed 0 d

by the same argument as i n ( - 3 ) .

To show 4 is not E -closed. Let G be any f i n i t e p-group, 9

so G i s not necessar i ly abel ian, f o r example, Q8 . Now G / d ( G ) i s

abel ian, s ince G is p-group, so G C E A , but G f A . Then 4- -

E A f A and A is not E -closed. d- - - 4

3 6

5. A is not N -closed, because i f w e take - 0

4 2 2 -1 G = p 8 = < a , b : a = e , b = a , b a b = a ' l > , then H = < a > , K = < b >

are both normal i n G and H , K c A b u t HK = G Q A . - -

2. Covering Subgroups.

In 1963 W - ~ a s c h u t z publ ished a f a r - reaching genera l i za t ion

of H a l l and Car te r subgroups.

Def in i t ion 2.2.1. Let - X be a c l a s s of groups. Then a sub-

group E of a group G i s c a l l e d an - %covering subgroup of G i f

2. I f E 5 H 5 G and K d H, H/K € then H = EK. -

Remarks 2.2.2. 1. I f E i s an - X-covering subgroup of G

with E 5 H 5 G and H € A, then H = E. This i s because H/< 1 > 'S H E X - -

and E i s an - X-covering subgroup, s o E = H. Therefore i f E i s an

%covering subgroup of G then E is a maximal X-subgroup of G . - -

2. I f G € - X then G i s i t s own - X-covering.

Def in i t ion 2.2.3. Let - X be a p c l o s e d c l a s s . Then we c a l l - X

a homomorph .

Lemma 2.2.4. I f E i s an - X-covering subgroup of G and i f

E 5 H 5 G , then E i s a l s o an - %covering subgroup of H .

Proof. Follows from the d e f i n i t i o n .

Lemma 2.2.5. Let - X be a homomorphand l e t E be an

X-covering subgroup of G . I f N 5 G then EN/N is an X-covering sub- - - group of G/N .

Proof. E n N 5 E a s N 5 G and EN/N % E/E n N € - X s ince

X i s a homomorph - Now suppose t h a t EN/N 5 H/N 5 G/N and

. Then H/H € X and s o , a s E I H I G and E is an 0 -

%covering subgroup of G we g e t H = EH Hence it follows t h a t - 0 '

H/R = (EN/N) (H /N) and consequen%ly EN/N is an &-covering subgroup of G/N. 0

Examples 2.2 -6. 1. Sylow p-subgroups a r e P-covering sub- -

groups of G where P is t h e c l a s s of f i n i t e p-groups . I f E is -

Sylow p-subgroup of G , then E E P. L e t E 5 H 5 G , H 9, - 0

H /H~ C - P . Now IH:E] is prime t o p s ince E is Sylow p-subgroup

of H , and I H : H ~ ] is d i v i s i b l e by p only , s o [H:E], [H:HO] a r e

coprime. Then by (1.3.31, H = EH. Thus Sylow p-subgroups a r e P-covering -

subgroups. S imi lar ly Hal l IT-subgroups a re 11- - covering subgroups.

Conversely, i f E i s - P-covering subgroup of G , then E i s

Sylow p-subgroup of G ; f o r , i f E is - ?-covering subgroup of G , then

E i s a p-group and by (2.2.2 1 (i) , E i s a maximal P-subgroup of G . -

Then E i s Sylow p-subgroup of G . A n i l p o t e n t se l f -normal iz ing subgroup of a so lvab le group is

a a l l e d a Car te r subgroup.

2. E i s a Car te r Subgroup of G i f f E i s an - N-covering

subgroup of G . Proof. I f E i s a Car te r subgroup of G then

E C N . Let E 5 H 9 G , H 4 H and H/HO C N . - 0 - i s a

n i l p o t e n t se l f -normal iz ing subgroup of t h e n i l p o t e n t group H / H ~ , b u t

every proper subgroup of a r l i lpotent group must be proper subgroup of

i t s normalizer , s o H / k 0 = E H ~ / H ~ . Hence H = EH and consequently 0 '

Car te r subgroups a r e N- cove r i n g subgroups. -

Conversely, i f E i s an - N-covering subgroup of G , then E

i s a Car te r subgroup of G . Since E i s an - N-covering subgroup o f G ,

so E i s ni lpotent . Now l e t x € N ( E ) so E 5 < x > E f G and G

< x > E/E r < x >/E n < x > c ; but , E i s an N-covering subgroup

of G SO < x > E = E(E n < x > ) = E . Then x € E and E = N ( E ) , so G

E is a Carter subgroup of G .

Remark 2.2.7. Not every group has an 5-covering subgroup.

Example. I f G is a non-abelian p-group, then G has no

A-covering subgroup. This is because: ~f s were an A-covering sub- - - group of G , then S would be abelian. Also S G ' /GI would be an

A-covering subgroup of G/Gt , but G/G' € A s o S G ' /G ' = G/Gt , t h a t -

i s G = SG'. But G ' 5 d ( ~ ) f o r every p-group, s o it would follow

t h a t G = S , a contradiction because G i s non-abelian.

The following r e s u l t allows us t o p u l l an - X-covering subgroup

of a given f ac to r group of a group G back t o an 5-covering subgroup

Lemma2.2.8. Let - beahomornorph and M S G . I f H/M

i s an - %covering subgroup of G/M and E is an - X-covering subgroup of

HI then E i s an 5-covering subgroup of G .

Proof. By (2 -2 - 5 ) , EM/M i s an - %covering subgroup of

H/M, H/M € - X s ince H/M is an X-covering - subgroup of G/M, s o

H/M = EM/M, t h a t i s H = EM. ~t E 5 F 5 G , F/F~ c 3 . We claim

F = F E. Now F 5 N (F ) and F 5 N ~ ( M ) = G s o F normalizes FOM, 0 G 0

and a l so M normalizes F M s ince M 5 F M. Then FOM 5 FM- We show 0 0

M F ~ F F/FO FM E covers - FM Now - - - - & - e

F ~ M ' F ~ M F ~ M

F ~ M n F - ( F ~ M n 6% . B U ~ F/F~ c 5 ,

40

and X - i s Q-closed, so MF/MFO C - X . Now EM/M = H/M , s o EM/M i s

an &covering - of G/M . But E 5 F so EM 5 FM, t h a t i s EM/M 5 FM/M.

- Then EM/M i s an - X-covering subgroup of FM/M, s o allrK C X . "o/" "0

-

Therefore MF/M = (EM/M) (MF = EMF /M , t h a t i s MF = EMF ?' 0 -

We next show E covers ME n F . A s F 4 F, EM n FO 5 EM n F, ME n F, o

E M n F - (EM n FIF, M E F ~ n F - , E M ~ F ,, - - - (by t he modular law)

E M ~ F , E M ~ F ~ F , F~ F0

= F/F, C - X .

EM C X . ~ u t E i s an %covering subgroup of H = EM Therefore EM n Fo - -

and E 5 EM n F since E 5 F , s o E i s an - %covering sbbgroup of EM n F.

Then EM n F = E(EM n F,) ........ (*) .

F = F fl a = F fl EMF, = (ME n F)F, (by t he modular law)

Thus E i s an - X-covering subgroup of G .

The above r e s u l t i s found i n ~ a s c h k [1963], although a s

C- Godsil po in t s out t h a t it is unnecessary t o assume X is R -closed. - 0

To end t h i s sec t ion , we introduce another important c l a s s of

con jugate subgroups ca l led - X-projectors . This notion i s another

general izat ion of Carter and Hall IT-subgroups .

Definit ion 2.2.9. Let X be a c l a s s of groups and G a -

group. A subgroup H of G i s ca l led an 5-maximal i f :

(i) H E - X and (ii) H 5 H* 5 G I H* E X - , implies H = H* .

Definit ion 2.2.10. Let - be a c l a s s of groups and G a

group. A subgroup E of G i s ca l l ed an - %projector of G i f EN/N

i s a maximal - %subgroup of G/N whenever N 2 G .

Remark 2.2.11, Taking N = < 1 > we see t h a t an - X-projector

i s - X maximal. It is a l so c l ea r t h a t a homomorphic image of an ?-projector

i s an - X-projector i n the homomorph. Thus i f E i s an - X-projector than E

i s - %maximal and EM/M i s an - of G/M , when M i s minimal

normal. The following lemma a s se r t s t h a t the converse i s t rue .

Lemma 2 -2.12. E i s an - %projector of G i f f E i s

%maximal subgroup of G and EM/M is an %projector of G/M f o r any - -

minimal normal subgroup M of G .

Proof. Let N 5 G . We may assume t h a t M 5 N . Now

iS an )(-maximal subgroup of /M % G/N. But

(EM/M) (N,!M) (NjrM) = EN/M /M Y EN/N, so EN/N i s 5-maximal Subgroup of / P G/N. Hence E i s an X-projector of G . -

Remarks 2.2.13. (a ) I f - i s Q-closed and G i s a f i n i t e

group, then every &-covering subgroup of G i s an - X-projector of G .

(b) Let - X be any c l a s s and G any f i n i t e group. I f a subgroup E of

G is an - X-projector i n every subgroup U containing E , then E

is an - X-covering subgroup of G .

Proof. (a) Let E be an %covering - subgroup of G and

N 2 G , s o EN/N is an %covering - subgroup of G/N. Then EN/N is

X-maximal subgroup of G/N. Therefore E i s an X-projector of G . - -

(b) L e t E I U 5 G and U/V € X - . Since E i s an X-projector - of U

EV/V is - X-maximal i n U/V. Hence U = EV, s o E i s an ?-covering

subgroup of G .

Examples 2.2.14. (i) In any f i n i t e so lvable group G , t h e

?-projectors coincide with t h e B-covering subgroups and a r e j u s t t h e -

p-Sylow subgroups.

(ii) Let G = As. has no Car te r subgroup and no E-covering subgroup,

where - N i s t h e c l a s s of f i n i t e n i l p o t e n t groups. On the o t h e r hand,

As has a subgroup E s V , t h e Klein four group. In f a c t , E -4 A4 5 A~ .

Now E € E l s i n c e V € N . Since q i s simple t h e only normal -

subgroups o f A5 a r e < 1 > and As s o it i s c l e a r t h a t EM/M L

is E-maximal i n A5/M , where M is any normal subgroup of A5 . Thus

Remarks 2.2.15. From t h e previous example w e can see t h e

f ol lcwing :

(i) A f i n i t e group G need no t contain any &-covering subgroup.

(ii) In genera l - X-covering subgroups and &-projectors do no t coincide.

Hawkes [1969, p. 2431 shows t h a t t h e not ions of - X-projectors

arid - X-covering subgroups coincide when X - i s a s a t u r a t e d formation and

G i s a f i n i t e so lvable group.

CHAPTER 3

SCHUNCK CLASSES AND FORMATIONS

A l l groups a r e supposed t o belong t o t h e c l a s s 2 of f i n i t e

so lvable groups unless e x p l i c i t l y s t a t e d otherwise; thus "G i s a group"

and "G C - U" a r e equivalent .

1. Schunck Classes.

W e t u r n t o the ques t ion of which c l a s s e s - X give rise t o

%covering subgroups. -

Defini t ion 3.1.1. Let X be a c l a s s of groups. Then i s - -

c a l l e d a Schunck c l a s s i f t h e fol lowing condi t ions a r e f u l f i l l e d :

(i) - is Q-closed. . .

(ii) I f G i s a group and G/K C 5 f o r a l l p r imi t ive f a c t o r groups G/K . of G , then G € X . -

Schunck (-1967) introduced t h i s concept, and he c a l l e d such a

c l a s s a s a t u r a t e d homomorph b u t now such c l a s s e s o f t e n r e f e r r e d t o a s

Schunck c las ses .

Examples 3.1.2. (i) < 1 > i s a Schunck c l a s s and i s contained

i n every Schunck c l a s s .

(ii) - N , t h e c l a s s of f i n i t e n i l p o t e n t groups i s a Schunck c l a s s .

Proof (ii). A s we know - N is a Q-closed. Suppose G/K C - N

f o r a l l p r imi t ive f a c t o r groups G/K of G . Now i f M is any maximal

subgroup of G l e t Cor M = K . Then M/K is a maximal subgroup of G

the primitive group G/K. But G/K i s n i lpo ten t s o M/K 2 G/K and

M S G , SO G C N . -

(iii) - 1 , the c l a s s of f i n i t e supersolvable groups is a Schunck c l a s s .

Proof (iii) . A s we know - 7 i s Q-closed. Suppose G/K - 7

f o r a l l pr imit ive f ac to r groups G/K of G and l e t M be any maximal

subgroup of G s o M/K i s a maximal subgroup of the primitive super-

solvable group G/K, where K = Cor M . Now I G/M I = IG/K M/K I = p / where p i s prime since G/K C - 7 , s o G € - 7 . Thus - 7 i s a Schunck

c lass .

( i v ) - A , the c l a s s of f i n i t e ahel ian groups i s not a Schunck c l a s s .

Consider G where G be any p-group, s o G i s not necessary abelian.

However, G / + ( G ) i s abel ian, and by (1 .4 .3 ) (2 ) , + ( G I C K . f o r a l l

primitive fac tor groups G/K of G . Now G/K I G/+ (GI/K/Q ( G I c - A

since - A i s Q-closed, Therefore G/K 6 - A f o r a l l primitive f ac to r

groups G/K of G , but G + - A .

Lemma 3.1.3. I f H 9 G and HK = G f o r a l l primitive fac tor

groups G/K of G , then H = G .

Proof. Suppose H < G , s o there e x i s t s a maximal subgroup M

of G with H C M . Then HK 5 MK = M , where K = Cor M , and HK # G

a contradiction. Thus H = G.

Lemma 3.1.4. If X - is a homomorph and every f i n i t e solvable

group G has an - X-covering subgroup, then - X i s a Schunck c lass .

Proof. L e t G/K C - f o r a l l p r imi t ive f a c t o r groups G/K of

G . By hypothesis G has an A-covering subgroup say E , s o G = EK

f o r a l l pr imi t ive f a c t o r groups G/K of G , s o by (3.1.3) , G = E € - X . Hence X i s a Schunck c l a s s . -

Main Theorem 3.1.5. Let - X be a Schunck c l a s s and G a

f i n i t e solvable group. Then :

( a ) There e x i s t s an - X-covering subgroup of G .

(b) A l l - X-covering subgroups of G a r e conjugate.

Proof. (a) I f G € then G i s i t s own X-covering and we - a r e done, s o assume G f 5 . We consider two cases , and the proof is by

induction on I G 1 .

Case (1) . Suppose G has a minimal normal subgroup N such

t h a t G/N f - X. Then by induction G/N has an X-covering - subgroup H/N

say. Now H/N € - while G/N f 5 , s o H < G and by induction again H

a l s o contains an - X-covering subgroup E say. Now H/N i s an - X-covering

subgroup of G/N and E i s - a n - X-covering subgroup of H , then by

(2.2.81, E is an - X-covering subgroup of G and we a re f in ished.

Case (2 ) . Assume G/N C - f o r a l l minimal normal subgroups N . I f G i s no t p r i m i t i v e , then a l l i t s p r imi t ive quo t ien t groups l i e i n

X s o G € s ince X i s a Schunck c l a s s , a contradic t ion. Thus G is - - -

p r i m i t i v e , and by (1.4.6) , G has a unique minimal normal subgroup N ,

and N i s abe l i an s ince G i s solvable . Now G is pr imi t ive s o the re

exists a maximal subgroup M of G wi th CorG(M) = < 1 > and by (1.4.51,

s ince N i s n i l p o t e n t M i s a complement of N i n G . Then G = MN,

M X G/N C - X, obviously M f u l f i l l s t h e requirements f o r t h e d e f i n i t i o n

of an - X-coyering subgroup of G . Hence G has an X-covering - sub-

group .

(b) If G C X t h e r e s u l t i s t r i v i a l . So suppose G { 5 and w e -

consider two cases.

Case (1). Let G/N C X - f o r a l l minimal normal subgroups N

of G , and l e t El, E2 be two d i s t i n c t - %covering subgroups of G . I f N i s a minimal normal subgroup of G , then G = E . N s i n c e

1

G/N - X , i = 1.2. S e t C = CG(N) and our claim i s t o show t h a t C = N.

c = c n G = C n ( E . N ) = N ( E . fl c) s ince N 3 c. Now Ei C 2 Ei a s 1 1

C -9 G s ince N -Q G. Obviously N c e n t r a l i z e s C n Ei , s o

c n E . ? E . N = G . 1f c n E~ # < 1 > , then by hypothesis G/C fl Ei 6 X 1 1 -

s o G = Ei(C n Ei) = E a con t rad ic t ion , t h e r e f o r e C n Ei = < 1 > . i

Thus C = N , s o by (1.4.10), G is p r i m i t i v e , now G = E.N and 1

E. n N = < 1 >, i = 1 ,2 s o each Ei

i s t h e complement of N i n G 1

and by (1.4.9) , a l l t h e complements o f N i n G a r e con jugate s o

El, E2 a re conjugate.

Case (2 . Assume G has a normal subgroup N such t h a t

G/N f - X and the proof is by induct ion on / G 1 . By (-2.2.51 , EIN/N, E2N/N a r e

two - X-covering subgroups of G/N, where E and E2 a r e a s be fo re , 1

and by induct ion E N/N, E N/N a r e conjugate i n G/N, s o t h e r e e x i s t s 1 2

g C G such t h a t ( E ~ N ) ~ = E N , E~ 5 E ~ N = E2N a l s o it i s easy t o show 2 1 1

E: i s an - X-covering subgroup of G , s o E:, E2 a r e two - X-covering

subgroups of G and so by (2.2.4) , E?, E2 are two X-covering sub- -

groups of E N. Also E2N/N f X while G/N f 5 , s o E2N < G and 2 -

again by induction E~ , E2 a re con jugate i n E2N . Thus El , E2 are

conjugate i n G .

Lemma 3.1.6. I f X i s a Schunck c l a s s , then 3 i s E -closed. - d

Proof. Let G be a group N 5 G I N 5 + ( G I and G/N f - X . Since - X i s a Schunck c lass so by (3.1.5 1, G has an - X-covering sub-

group E say, therefore G = EN 5 E$(.G) = E so G f - X . Hence 5 is

Lemma 3.1.7.

In other words Schunck

Proof. Let

c lass . I t su f f i ce s t o

belongs t o X . Let K -

Schunck classes are closed under d i r ec t products.

c lasses (of solvable groups I a re Do-closed.

G = H x L, where H , L € X and X is a Schunck - -

show t h a t every primitive quot ient group of G

2 G such t h a t G/K = H x L/K is primitive

F i r s t we show H 5 K o r L 5 K . Suppose not. Then we have

K n H < H and K n L < L . Project H 3 G/K by h - 3 h K and l e t + ? - H = { h ~ : h f H} 4 G/K and s imi la r ly = {EK: 8 f L} 4 G/K. Then

ii n = < 1 > , therefore G/K i s not pr imit ive , a contradiction.

Hence H C. K o r L 5 K . Now i f H 5 K then G/K G/K H/K. But / G/H 'Z L f - X and - X is Q-closed, therefore G/K C - X and G f - X s ince

X i s a Schunck c lass . -

Remark 3.1.8. Not a l l < Q,E >-closed c lasses a r e Schunck 9

c lasses . For e x w l e , the c lass of cyc l ic groups is Q-closed and

E -closed, bu t not closed under d i r e c t product (see 2.1.8. (2 1 ) , s o by 9

(3.1.7) i s not a Schunck c l a s s .

Lemma 3.1.9. Let X be a Schunck c lass . Let G = AB with -

A 5 G and B Q G . I f V/A i s an - X-projector of G/A and W/B i s

an - X-projector of G/B, then V n W A n B i s an X-projector of / - G/A n B.

Proof. The proof i s by induction on I G 1 . We note t h a t

A/A fl B , B/A n B are normal subgroups of G/A fl B and

G/A fl B = (A/A n B) (B/A n B) a l s o A/A n B , B/A fl B have an

X-pro jectors , then the hypothesis holds i n group G/A fI B , SO by induction -

we may assume A n B = < 1 > . Suppose V < G , now V/V n B S VB/B = G/B

a l s o - v n w V n . V / V f l B and -- V ~ W ( V ~ W ) B - V B ~ W - - - G ~ W

V n B - c v n w n B " B -

V n B B B

V ~ W = W/B , SO -

V n B i s an - X-projector of V / V fI B . Also V/A i s an

X-projector of V/A , s o by induction (V = V n W i s an - V ~ B ~ A

X-projector of V / V n A fI B = V and s o V fI W i s an X-projector of G. - -

Hence we may assume V = G and by symmetry a l s o t h a t W = G . Now

A I G / B = W / B C X - and B I G / A = v / A € X , - s o G = A X B € D X = X 0 - -

s ince - X is a Schunck c l a s s , s o the conclusion follows t r i v i a l l y .

Theorem 3.1.10. Let X be a Schunck c lass . I f A and B - have - %projectors E and F , then E x F i s an - X-projector of A x B.

Proof. Apply the lemma with

V = FA and W = EB. The r e s u l t follaws A B

because v fl w = FA fl EB = E x F ( A fl B)

= E X F . < 1 >

We note i n passing t h a t t he analog of Theorem (3.1. l o ) , f o r

X-covering subgroups is a l s o t rue . -

Lemma 3.1.11. P ( N ) = {G: G has an - X-covering subgroup) is

< Q,DO >-closed whenever - X i s .

Proof. Let G C P ( N ) , s o G has an 5-covering subgroup

say E . I f N 2 G then by (2.2.5 ) , EN/N i s an %covering subgroup -

of G/N , s o P ( H ) i s Q-closed.

To show it i s D -closed, l e t G = A x B where A I B C P(H), 0

so A has an &-covering subgroup say E and B has an*%covering -

subgroup say F . By Theorem (3.1.10) , E x F i s an - %covering subgroup

of A x B . Thus G C P ( N ) , and s o P ( N ) i s D -closed. 0

2. Formations.

We now t u r n our a t t e n t i o n t o t h e r e s u l t s der ived by

Def in i t ion 3.2.1. Let - F be a c l a s s of groups. We c a l l - F

a formation provided:

(i) - F i s Q-closed.

(ii) F i s R -closed. - 0

Remark 3.2.2. Condition (ii) of the Def in i t ion (3.2.1) is

equ iva len t t o t h e fol lowing

(ii)' I f N1,N2 ? G I G/N1, G/N2 c - F , then G/N1 n N 2 € I , f o r i f

w e assume (ii) and N1,N2 -4 G and G&,G/N~ C - F then t h e Q-closure of

F impl ies G/II~ 1 N ; / N i / N l fl N~ E F i = 1.2. - - Also

the o t h e r hand, i f N1, . . . 'Nk a r e normal subgroups of G such t h a t

k n N . = < 1 > G/N~ C F , i = 1 , 2 f . . . I k and i f (ii)' holds then t h e

1 - i=l

hypotheses imply G/hl n N2 € . Again G/N3 - F s o G/ (N l n N2) n N3 ( - F,

k k and continuing i n t h i s way we f i n a l l y g e t Ni c - F ; b u t n N . = < 1 >,

1 i= 1

Remark 3.2.3, Condition (ii) ' c b be replaced by t h e

fol lowing condi t ion :

( 2 ) ' I f N , N 4 G such t h a t G/N1, G/N2 ( F and N n N2 = < 1 >, 1 2 - 1

We want t o show t h a t (2) ' holds i n general i f ( 2 ) ' is only

f u l f i l l e d i n those cases i n which N1 and N2 a r e minimal normal sub-

groups of G : f i r s t of a l l it can

I be assumed N . # < 1 > f o r i = 1 , 2 .

1

Let Mi be a minimal normal subgroup

of G , M . 5 Ni . Then 1

H 2

z G/.,/.,N dN1 c s ince - F

is Q-closed. Now

G ~ ~ ~ N ~ / M ~ ~ pfi2p2 1

Furthermore , ( M ~ N ~ / M ~ ) fI ( N ~ / M ~ ) = < 1 > so by induction G/M2 C - F . Likewise, G/M1 C - F . Now we have one of the r e s t r i c t e d cases and we can

conclude G € F . -

Lemma 3.2 -4. The in te rsec t ion of formations and the union of

ascending chain of formations i s formation.

Proof. Follows eas i ly .

Lemma 3.2.5. I f a c l a s s - F , i s < Q,S,DO >-closed, then - F

is a formation.

Proof. We need t o show - F i s R -closed. Let G be a group 0

with normal subgroups N1, N2 such t h a t G/Nl, G/N2 € - F . &fine

e: n N2- -+ (G/N~) X ( G / N ~ ) by 8Ig(IJ1 n N,, 1 = (gN1,gN2). By

the proof of (2.1.8) p a r t ( 3 ) , e i s a monomorphism, so G h 1 fI N 2 %

t o a subgroup of (G/N1) X ( G / N 2 ) , bu t - F i s D -closed and 0

S-closed, s o G / N ~ fl N~ C - F . Thus - F is R -closed. 0

O w main application of (3.2.5) is the following lemma.

Lemma 3.2.6. The c lasses - A, 5 E, 1, and - U are

formations.

Proof. A l l these c lasses are S, Q and D -closed so by 0

(3.2.5) a r e formations.

Remark 3.2.7. A formation is i n general no t closed under

passing t o subgroups, not even t o normal subgroups. In order t o construct

examples we need the following: I f H/K is a chief f ac to r of G and

I H / K I i s some power of the prime p , then H/K i s ca l led a p-chief

f ac to r of G .

Example 3.2.8. Let - F = {G: G i s solvable and IM/N\ # 2 f o r

a l l 2-chief fac tors M/N of G }. Then - F i s a formation.

Proof. Since a l l 2-chief fac tors of G/N are isomorphic t o

a subset of those of G , - F i s Q-closed. For the proof of the va l i d i t y

of Ro-closed l e t G/N, G/M C - F and l e t H/K be a 2-chief f ac to r of

G/M n N . W.1.o.g. it is between M and N n M . Since

MN/N Y M/M n N, there i s a chief f ac to r ii/i? lying between N and MN such

t h a t H/E Z H/K, hence I H/K I $ 2

and G/M n N - F . Thus - F i s a

formation. M N

For ins tance , A4 C - F s i n c e A4 is solvable and the orders of

t h e chief f a c t o r s of A4 a r e 4 and 3 , However, < A , b u t C 2 - 4

C2 f . Thus - F i s no t c losed under subgroups, n o t even under normal

- C F . subgroups, s i n c e otherwise V C F and then a l s o C2 -

Defini t ion 3.2.9. I f f i s a formation and G a group -

then t h e - F-residual of G is t h e subgroup GF = n {N: N 4 G and G/N ( - F). -

Example. GA = G I where - A i s t h e c l a s s of f i n i t e abe l i an -

groups.

B m a r k 3.2.10. By condi t ion (ii) of the Def in i t ion of a

format ion , G i s t h e uniquely determined smallest normal subgroup of G

whose f a c t o r group belongs t o f .

Lermna3.2.11. L e t - F b e a f o r m a t i o n , H Q G . Then G / H C F -

i f f GF 9 H :In p a r t i c u l a r G C F i f f GF = < 1 >. - -

Proof. I f GF 9 H , then G/H E ( s ince - F is -

Q-closed, and t h e converse i s obvious.

Lemma 3.2.12. If - F i s a formation and M 2 G , then

Proof. I f ( G / M ) ~ = R/M, then G/R Z -

#

/- s o GF 9 R. On t h e o t h e r hand G/M -

( G / M ) ~ 5 GFM/M t h a t i s n / M 5 GFM/M . Thus RIM = G p M . - - - - /

In order t o generalize Car te r ' s r e s u l t s , ~a schGtz needed t o

consider a spec i a l kind of formation which he defined ( i n [ ~ a s c h i t z , 19631)

as follows:

Definition 3.2.13. Let F be a formation. Then F i s s a i d - -

t o be sa tura ted provided the following condition i s s a t i s f i e d : i f G j?

and M is a minimal normal subgroup of G such t h a t G/M € F , then M -

has a complement and a l l such complements are conjugate i n G . In order t o get a b e t t e r understanding of sa tura ted formations

we prove the following character izat ion and then provide some examples.

Theorem 3.2.14. I f F is a formation then the following are - equivalent :

(i) i s a sa tura ted formation;

(ii) f is E -closed; d

(iii) - F i s a Schunck c lass .

i i i Suppose - F i s a sa tura ted formation andG/Q(~) € F . - Assume G i s minimal with t h i s property, subject t o G j! f . Let N be

a minimal normal subgroup of G ; N (. Q (.GI . Now G/Q ( G I 2 G/N Q (GI /N / = (G/N) d(G/N) F and s o by minimality of G , G/N € F s o by (i), N / - - is complemented, a contradiction because N 5 4 ( G I . Thus G/+ (G) € F -

implies G C - F , and s o - F is E -closed. Q

(ii) implies (iii). Suppose - F i s E -closed and G/K € F Q -

f o r a l l primitive fac tor groups G/K of G . By (1.4.3) (21, 4 (G) = f l ~

f o r a l l primitive f ac to r groups G/K of G and by R -closure 0

G/9(G) = G / ~ K € - F , s o G € - F since F i s E closed. Thus i s a - +- Schunck c lass .

(iii) = (i). Suppose i s a Schunck c lass and G has a

normal subgroup M such t h a t G/M € F , while G d F we m u s t show M - -

has a complement and a l l complements a re con jugate. By (3.1.5) I G has

an r-covering subgroup E , s o G = EM. NOW as G f , E E F , E < G . -

Also since M i s abelian and normal E fl M 5 EM = G. By rbinimality

of M I E n M = < 1 > , hence M is complemented. Any complemnt of M

w i l l be isomorphic t o E and s o w i l l l i e i n - F . Therefore a l l

complements of M are - f-covering subgroups of G and again by (3.1.51,

are con jugate. Thus M is complemented i n G and a l l i t s complements

a re conjugate, s o f is a sa tura ted formation. -

To provide some examples I need the following:

Definit ion 3.2.15. Let ll - denote the c l a s s of solvable IT-groups,

IT a f ixed s e t of primes. Let C ( r ) = (G: f o r a l l proper normal subgroups

M of G , G/M n ) . - C ( r ) i s ca l led t he c l a s s of IT-perfect groups.

Lemma 3.2.16. C ( T ) i s a Schunck c l a s s and i f GII is the -

n-res idua l of G then Gn is an C(n)-covering subgroup of G . - -

proof. C(n) i s c l ea r ly Q-closed. Put Y = C ( n ) , R = G n and - -

l e t G/K € - Y f o r a l l pr imit ive f ac to r groups G/K of G . Suppose

G € - Y . If R i s a maximal subgroup of G then Cor R = R s ince G

R 3 G s o G/R = G/Cor R y a contradiction. So there e x i s t s a -

maximal subgroup M of G such t h a t R 5 M then R I Cor M s o G

G/Cor M I[ - a contradiction because a l l pr imit ive f ac to r groups of G

l i e i n - Y . Thus G Y and therefore Y i s a ~chunck c lass . - -

To show R i s a Y-covering subgroup of G . Let H be a -

Y-covering subgroup of G . Then HR/R is a !-covering subgroup of -

the 'IT-group G/R so HR/R = < 1 > and H 5 R . Hence H i s a

Y-covering subgroup of R . ~f H < R , then R f y and R has a unique - -

normal subgroup M minimal subject t o R/M 6 fl . Clearly M i s a - cha rac t e r i s t i c i n R 5 G , s o M 5 G . Now I G : M ( = I G : R ( * ( R : M ~ is a

product of primes i n

we must have R = M.

t r ad ic t ion . Hence R

IT t h a t i s G/M € il - and R = G n 5 M , therefore -

Hence R i s a 'ITt-group and s o R € Y a con- -

= H .

Remarks and examples 3.2.17.

(1) Not every formation i s a Schunck class . For

Example - A i s a formation but not a Schunck c lass .

(2) Not every Schunck c lass is a formation. For

Example ~ ( ( 2 ) ) i s a Schunck c l a s s but not a formation. To show it

i s not a formation, l e t G = C3Q8 be the semidirect product of the

Q8 = < a ,b , a 4 2 2 = e , a = b , ba = b - l > with the cyc l ic group c3 = < a >,

where a: Qg ' Q8, a ( a ) = b , ~ ( b ) = ab . Since every element of Q8

i j has a unique representat ion a b , i = 0,1,2 ,3 and j = 0,1 , w e

j extend a t o a l l elements of Q8 by a(aibJ) = [ a ( a ) li [ a (b) ] . This i s con-

s i s t e n t with t he def in i t ion of a ( a ) and a ( b ) . By considering two cases 8 = 0

i j k t i j k t and 8 = 1 one can check a [a b a b ] = ~ ( a b ) a ( a b 1. Also it i s easy t o see

t h a t l a / = 3 i n t he group G = < a , b , a >. Now l e t M P G Suchthat

I G / M I = 2" . Then s ince a i s of order 3 , a: € M . Since M 4 G ,

-1 - -1 l -laa = a- (a(a-') ) * a = a 0 b a. S imi lar ly , aa € M , bu t aa = a a a = aa a

- 1 -1 -1 - 1 I? € M , b u t C? = b OLb = rn b cb = a * ( a ( b ) ) b = a-abb = aa . Since

Cia and a € M it follows t h a t a € M , and, form t h e f i r s t , b € M , s o

M = G . Therefore t h e r e i s no normal subgroup of index 2n . Hence

G = C3Q8 C C((2) ) . Now H = C x G it i s c l e a r t h a t H 'f C({2}), b u t 2

* * i n H the re is another normal subgroup C2 with H = C x G . Therefore 2

From Theorem (3.2.14) and t h e preceding example we can see

t h a t Schunck c l a s s e s a r e more genera l than s a t u r a t e d formations.

I n genera l , t he most d i f f i c u l t p a r t of showing t h a t a c l a s s of

groups i s s a t u r a t e d l i e s i n proving it i s E -closed. ~ a s c h i i t z provided dJ

L

a p a r t i a l s o l u t i o n t o t h i s by descr ib ing an important method of cons t ruc t ing

s a t u r a t e d formations .

F i r s t we need some d e f i n i t i o n s .

Def in i t ion 3.2.18.

(a) I f H/K is a chief f a c t o r of G and I H / K I is d i v i s i b l e by

prime p , then we c a l l H/K a p-chief f a c t o r of G . I n a so lvable

group every ch ie f f a c t o r i s a p-chief f a c t o r . I f H/K i s a ch ie f

f a c t o r of G , then t h e c e n t r a l i z e r of H/K i n G , i s def ined a s

cG (H/K) = {x: ~ l h - l x h € K f o r a l l h € H). ~ u t ~ (H/K) w i l l denote

t h e group automorphisms induced by G on i ts chief f a c t o r H/K , s o

~ u t (H/K) 1 G / C ~ (H/K). We note t h a t i f A (H/K) = 1 > , then H/K i s G G

c en t r a l i n G/K .

(b) A formation function f i s a map from the s e t of prime

numbers t o c lasses of groups such t h a t f ( p ) i s e i t h e r empty o r a

formation f o r a l l p .

(c) I f f i s a formation function, we define an associated

c l a s s - F( f ) as the c l a s s of a l l f i n i t e groups G sa t i s fy ing :

AutG (H/K) 6 f (p) whenever H/K i s a p-chief fac tor of G . I f

f (p) = 4 , we i n t e r p r e t t h i s t o mean G has no p-chief fac tors .

F ( f ) i s ca l l ed the l oca l formation defined by f , and a formation i s -

ca l l ed l oca l i f it has the form F ( f ) . Before going any fur ther we need -

t o r e c a l l the following:

A p a i r ( 5 1 , ~ ) consis t ing of a s e t $2 and a group G i s s a id t o be a

group with operator domain $2 if there i s a function 8 from $2 i n t o

end(^) . A group with an operator domain 9 is ca l l ed an %group. L e t

G and G ' be $2-groups. A homomorphism f from G i n t o G' is s a i d

0 0 t o be an $2-homomorphism i f f o r any 0 € $2 and g € G , f (x ) = f (x) .

Lemma 3.2.19. F ( f ) a s defined above i s a formation. -

Proof. Let G € F ( f ) , M -Q G and l e t H/K be a chief f ac to r -

of G/M s o H/K i s a chief f ac to r of G . Now

/ G /&JHfi) s ince H/K i s a chief G c (H/K) % /M = (G/M)

f a c to r of G/M s o (G/M) C (H/K) C f (p) hence G/M E ( f ) , and s o /

TO show - F(f) i s Ro-closed, l e t G/N, G/M E - F ( f ) . Consider

a chief s e r i e s of G/M fl N through M . I f H/K i s a chief f ac to r i n

t h i s s e r i e s and K 2 M then we may consider it a chief fac tor of G/M

and by assumption the induced automorphism group is i n E (p ) . Hence

w.1.o.g. we may suppose H/K i s a chief f ac to r of G/M f l N ly ing

between M and N fI M . Since MN/N & M/M fl N , there i s a chief

f ac to r / lying between N and MN such t h a t H/z %H/K , and we

can consider H/K as a chief f ac to r of G/N . Then

Aut (H/K) 1 Aut (H/E) s ince the isomorphism between H/K and H/K i s G G

an operator isomorphism, s o t o every chief f a c t o r between M and M fl N

there is an isomorphic chief f ac to r between N and MN whose automorphism

group G/N is isomorphic t o t h a t of H/K from G/M fl N . I f

G/M, G/N C - F ( f ) then A U ~ ~ ( H / K ) C f ( p ) . Thus G / M n N C - F(f ) and

F( f ) is a formation. -

(i > H/K i s a

then H/K

Definit ions 3.2.20.

Let - F(f ) = - F be a formation l oca l ly defined by f (p) . I f

p-chief f ac to r of some group G such t h a t AutG (H/K) E f (p) ,

is. ca l l ed an - F-central chief f ac to r of G and i s ca l led

F-eccentric, i f ~ u t ~ (H/K) f f (p) . ~ h u s G C F i f f every chief f ac to r - -

of G i s - f-central .

(ii) Let < 1 > = Go < G 1

< G2 < ... < G 2 = G b e a

G . Then we define F (GI = fl {C ( G . Gi 1): G & ~ a P G - J - -

chief s e r i e s of

p-chief f ac to r

of G) . From ( l . l . l 5 ) , it is c l e a r t h a t F (GI 5 F (GI. P

We c a l l a group G , p-ni lpotent i f it has a normal p-complement,

t h a t i s t h e Hal l p'-subgroup of G is normal.

(iii) Let IT c IT (G) and l e t o ~ ( G ) denote t h e l a r g e s t normal -

IT-subgroup of G and 0 (G) i s t h e l a r g e s t normal IT'-subgroup of G . IT'

I t i s e a s i l y seen t h a t OIT(G) = Cor (S where ST denotes a Ha l l G T

( i v ) L e t a = G/OIT(G) . Then OIT(G) = i . Now consider t h e unique

maximal normal IT'-subgroup Or, (a) = On, (G/OIT (GI and denote i ts inverse

image i n G by Or ,IT, ( G I . S imi la r ly we def ine OIT,IT' , r ( G I t o be t h e

inver se image i n G of GIT ( . G / O ~ , ~ , (G) . Continuing i n t h i s way, we obta in

a sequence of c h a r a c t e r i s t i c subgroups o f G , 1s 0 (GI < IT

(.GI 5 ... .

Lemma 3.2.21. Let f = F ( f ) be a l o c a l formation. Then - - G C - F i f f G/F (G) C whenever f (p) # 6 and p 1 ( G I when

P

f ( p ) = 6 .

proof; Let p l l ~ I . m e n G € - F only i f A U ~ ~ ( H / K ) € f ( p ) f o r

a l l p-chief f a c t o r s H/K of G . But AutG (H/K) S G/bG (H/K) and by

s o t h e r e s u l t follows.

Lemma 3.2.22. F (G/d (GI ) = F ( G I /dI ( G I . P P

Proof. By t h e same argument a s i n (1.1.121 .

Theorem 3.2.23. I f - f i s def ined l o c a l l y by f (p) then - f

i s a s a t u r a t e d formation.

Proof. By (3.2.191, 1 i s a formation, s o by (3.3.14) it is

s u f f i c i e n t t o show t h a t - F i s E closed. ~f G / & ( G ) C F then by (3.3.211, 6 -

(G/d(G) ) € f (p) then by ( 3 -2.22) t h i s implies

/4 (GI € f (p) , and t h i s impl ies G/ '~(G) C f ( p ) . Then again by

(3.2.21), G € f s o f i s E -closed and F is a s a t u r a t e d formation. - - dl -

Gaschutz and Lubeseder (1963) have proved t h e converse of

Theorem (3.2.23) namely t h a t every s a t u r a t e d formation i s l o c a l l y defined.

However most of t h e proof i s module-theoretic, and w e omit it.

Def in i t ion 3.2.24. L e t - f be a formation l o c a l l y def ined by

a formation func t ion f .

(i) W e say t h a t t h e system ) i s i n t e g r a t e d provided t h a t

f o r every prime p , f (p1 5 .

say t h a t t h e system ) i s - f u l l provided f o r every p ,

w i l l e s t a b l i s h t h e fol lowing ex i s t ence and uniqueness

r e s u l t s , both e s s e n t i a l l y from [ c a r t e r and Hawkes, 19671. The f i r s t of

t h e s e r e s u l t s says t h a t every l o c a l formation can be expressed a s an

i n t e g r a t e d l o c a l formation.

Lemma 3.2.25. Let F be a formation l o c a l l y defined by f (p) . -

Then - F = F ( f where F ( f i s a formation l o c a l l y def ined by f l (p) 1 - 1

Proof. By (-3 -2.4) , f (p) n F i s a formation and s o by (3.2.19), - F ( f i s a formation. - 1

Clear ly - F(f 1 ) - - c F s ince f o r any p-chief f a c t o r of G , i f Aut (H/K) l i e i n G

f ( p ) n - F then l i e i n f ( p ) . To show F C F(f l ) l e t G C F and l e t - - - -

H/K be any p-chief f a c t o r of G , s o AutG (H/K) C f (p) . On the o t h e r

and s o by t h e d e f i n i t i o n of a l o c a l fonnation we see t h a t G C - F ( f l ) .

F ina l ly we no te t h a t f (p) = 4 i f f f l ( p ) = 9 and s o i f

G C - F implies p 1 I G J then G C F( f ) impl ies p 1 I G ] and conversely. - 1 So - F = - F(f l ) and w e have f ( p ) - - - F c F ( f l ) .

Before g iv ing some examples w e need the fol lowing lemma.

Lemma 3.2.26. G i s supersolvable i f f Aut (H/K) is abel ian G

of exponent d iv id ing p-1 f o r a l l p-chief f a c t o r s of G .

Proof. This i s because a l l p-chief f a c t o r s of G a r e c y c l i c

of o rde r p . For d e t a i l s see [ ~ e s k i n s and Venzke, 1982, p. 41 .

Examples 3.2.27.

(i) Let f ( p ) = 1 f o r a l l p , where 1 denotes t h e c l a s s con-

s i s t i n g of only t h e t r i v i a l group. A chief f a c t o r H/K of G i s

F-central iff G C G ( H / ~ ) = < 1 > i f f G = CG(H/K) , s o G C F i f f every - / -

chief f ac to r i s cen t r a l , then G f - F i f f G i s ni lpotent . Hence

F = N , and we note t h a t {f (p) ) is integrated. - -

@ i f p ,f! IT, IT i s a f ixed s e t of primes (ii) f ( p ) =

U i f p € n, where U i s the c l a s s of f i n i t e solvable groups. - -

Now G C - F( f ) i f f IT(G) cn so G C - F(f ) i f f G is an-group.

Hence - F( f ) = c lass of IT-solvable groups. We note t h a t i f (p) ) i s not

in tegra ted i n f a c t f o r each p € IT, - F( f ) = 3 c U = f (p) . - - - n

(iii) For a l l p , f (p) = - N where - hj i s the c l a s s of abelian

groups of exponent dividing p-1.

G C - F(.f) i f f A u ~ ~ ( H / K ) i s abelian of exponent dividing p-1, SO by

(3.2.26), G € - f ( f ) i f f G i s supersolvable. Thus - F ( f ) = L the

c l a s s of supersolvable groups, and therefore - 1 i s a formation loca l ly

A defined by f ( p ) = .

Lemma 3.2.29. Let C be cyc l ic group of prime order p. P

I f - Y i s a c l a s s of groups we c a l l the s e t of primes IT a cha rac t e r i s t i c '

of - Y i f f I T implies C f Y . P -

Hence f o r a loca l format ion - F , IT = {p : f (p) # 4 ) .

Lemma 3.2.29. I f F is a sa tura ted formation of - cha rac t e r i s t i c IT , then every group G f i s IT-group and a l l n i lpotent

IT-groups belong t o - F .

Proof. F i s a sa tura ted formation sof i s loca l ly defined. - - Now IT is a c h a r a c t e r i s t i c of - F so n = ip: f (p) # $ 1 . If p I IG I

where G f - F then f ( p ) # 4 implies C F implies p f IT s o G i s P -

a IT-group.

I f G is a n i l p o t e n t IT-group, then G / + ( G ) i s a d i r e c t

product of c y c l i c groups of order p f o r some C 7T . Therefore i

by (3.1.7) , G / ~ I ( G ) C , s i n c e - f i s a s a t u r a t e d formation s o G C f . -

Therefore a l l n i l p o t e n t IT-groups belong t o - f . Lemma (3.2 -25) shows t h a t every l o c a l formation can be

expressed as an in tegra ted l o c a l formation. However t h i s expression i s

not unique. For example, i f w e l e t

c l a s s of a l l 2-groups i f p = 2 f = {$

i f p f 2

and

c l a s s of elementary abelian..2-groups i f p = 2 f 2 (p) =

i f p f 2

then c l e a r l y f l (p) and f 2 (p) both l o c a l l y def ine t h e c l a s s of a l l

Before continuing w e need some t e c h n i c a l lemmas f i r s t .

Lemma 3.2.30. I f H/K i s a p - c h i e f f a c t o r of G , then

G / C ~ (H/K) has non- t r iv ia l p-subgroups .

Proof. W e f i r s t note t h a t G/C (H/K) a G

(H/K) a n d s o w e may assume w.1.o.g. t h a t K = < 1 > , and

t h a t H i s a minimal normal p-subgroup of G . Let A S G/H and put

G* = HA, t h e semidirec t product of H by A . L e t P be a normal

p-subgroup of A. Now A, normalizes H by const ruct ion and A,normalizes P s o

A , normalizes H P . On t h e o t h e r hand, H normalizes H P s i n c e H 5 HP,

t he re fo re HP 5 HA = G* . Now H 5 HP and H i s a p-group s o

H n Z ( H P ) f < 1 > . H is a minimal normal p-subgroup of G , s o H is

a minimal normal p-subgroup of G* , b u t HP 5 G* s o Z ( H P ) 9 G* . Hence H 5 Z (HP) and the re fo re P I- HP 9 CG, (H) . But CG, ( H ) = H by

t h e cons t ruct ion s o P 5 HP 5 CG,(H) = H. Thus P = < 1 > s o G / c ~ ( H )

has no normal p-subgroup.

Def in i t ion 3.2.31, Let - X and Y be a c l a s s of groups. -

Then we def ine X * Y = {G - - : t h e r e e x i s t N E - X , N P G , G/N y}. W e -

w i l l prove i n t h e next chapter t h a t - X * - y is a formation when - Y is

a formation and - X i s a formation which is S -closed. n

We know t h a t E X = (G: G has a normal p-subgroup M, G/M C A}. P- -

So E X = P * X where P i s t h e c l a s s of f i n i t e p-groups. I ? - - - -

Lermna 3.2.32. E X i s a formation when X i s - a formation. P- -

Proof. Since E X = P * X and P i s S --closed s o t h e P- - - - n

lemma fol lows.

Theorem 3 . 2 . 3 3 . Let F be a formation def ined l o c a l l y by

t h e two d i s t i n c t i n t e g r a t e d formation funct ions f l , f 2 . Denote t h e

value of i

a t p by Fi(p) whenever f i ( p ) # 4 . Then

F = F(f3) where

Proof. (i) f l ( p ) = Q i f f f (p) = 4 s i n c e i f t h e r e e x i s t s 2

Po s u c h t h a t f l (po) = c $ and f 2 ( p o ) f Q . then C C F ( f 2 ) and

Po c f F(.fl) a con t rad ic t ion s i n c e F ( 4 = F = F ( f Z ) . Po

1

(ii) We show t h a t F = F ( f 3 ) By (3.2.32) , E Fl(p) i s a formation P

s o F ( f j ) is a formation. L e t G C F( f3) and H/K be a p - c h i e f

f a c t o r of G s o A u ~ ~ ( H / K ) C E F ( p ) . By (3 .2 .30) . 0 ( A U ~ ~ ( H / K ) ) = < 1 >, P I P

A U ~ ~ ( H / K ) C Fl(p) SO G t F . Also i f A U ~ ~ ( H / K ) € F1(p) C E F (p) P l

then G C F( f3 ) . Thus F = F ( f 1. Hence we may assume w.1.o.g. t h a t 3

Fl (p) and F2 (p) a r e E -closed. P

(iii) Suppose then t h a t f o r some p , F ip) # F2(P) . -t 1

G C F ~ ( ~ ) \ F ~ ( P ) and assume G i s minimal i n t h i s c l a s s . W e show

no such G e x i s t s .

(a) W e w i l l show , f o r any < 1 > f K 9 G , G/K C F2 (p) by minimality

of G and by R -c losure 0 n K C F2(p) . Since G 1 F2(p) .

K 5 G

n K f < 1 > s o fl K i s t h e unique minimal normal subgroup of G , K 2 G K Q G

and ~ / f i K c F~ (p) impl ies 5 n K . SO n K = G F. (p) and

G ~ i ( ~ ) K 2 G K Q G 3.

G ~ . (p) i s t h e unique minimal normal subgroup of G .

1

W e p u t H = G . I f H i s a p-group, G/H C F2 (p) = E F2 (p) F2 (P) F

s i n c e F2 (p) i s E -closed. But G/H C E F 2 ( ~ ) where H i s a normal P P

p-group impl ies G C FZ (p) a con t rad ic t ion . Therefore H i s not a

p-group .

(b) Let G * = Z \ G , t hus G * = ( z x Z x ... x Z ) r G t h a t i s G* P P P P

i s t h e semidi rec t product of A by G where A i s t h e d i r e c t product

of I G 1 copies of Z . We note C ~ *

(A) = A. W e show t h a t H does P

not c e n t r a l i z e a l l t h e chief f a c t o r s of G* t h a t l i e i n A , f o r i f

it d id , then AH would be n i l p o t e n t s ince AH/H % H is abe l i an , and

a l l t h e chief f a c t o r s below A would be c e n t r a l . N o w H i s q-group,

and A i s a p-group s o H i s Sylow q-subgroup of AH and then H 4 AH.

Then [ A , H ] 5 H and a l s o A 5 AH by const ruct ion s o [A,H] 5 H n A = < 1 >

and H 5 CG, (A) = A a con t rad ic t ion . Thus H does no t c e n t r a l i z e a l l

t h e ch ie f f a c t o r s of G* i n A .

( c ) The claim i s t h a t G* C F . G*/A X G C F1 (p) and A i s a p-group

s i n c e f l i s i n t e g r a t e d , s o . G* C F .

(d) Now H i s t h e unique minimal normal subgroup of G and SO

G n CG, (n / s ) 2 H o r G n cG* (R/s) = < 1 > , bu t by (b) , H does not

c e n t r a l i z e a l l t h e ch ie f f a c t o r s of G* i n A , s o t h e r e e x i s t s a chief

f a c t o r K/L of G* i n A such t h a t H $ CG, (K/L) . For t h i s chief

f a c t o r w e have G n C *(K/L) = < 1 > and a s A i s abe l i an A 5 CG,(K/L) G

s o A = c (K/L) s i n c e G n c,, (K/L) = < 1 > . Hence G*

G*/CG, (K/L) a G 1 F2 (p) s o G* has an F-accentric p-chief f a c t o r

and G* 1 f c o n t r a d i c t i n g (iii). Hence G can n o t e x i s t and s o

F1(p) = F2(p) . Thus every l o c a l formation i s def ined by a unique

i n t e g r a t e d formation funct ion .

P. Schmid c i t e d i n 1131 has shown t h a t the previous r e s u l t

i s s t i l l t r u e i n a f i n i t e group.

3. Normalizer and Closure Operators on Sch.unck Classes.

The f i r s t lemma genera l i zes a c l a s s i c a l r e s u l t on normalizers

of Sylow subgroups.

Lemma 3.3.1. Let - be a Schunck c l a s s , G a group, E an

%covering subgroup of G and N (El 5 T I G . Then N (TI = T . - G G

Proof, Its proof is j u s t t h e s tandard F r a t t i n i argument,

namely:

h Let h f N G ( T ) s o T h = T and E h 5 T = T . Then E and

h E a r e 5-covering subgroups of T , hence they a r e conjugate under T .

h t Since E = E f o r some t c TI h t f N ( E ) 5 T . But h t f T implies

G

Lemma 3.3.2. Let - be a Schunck c l a s s . Then N (E l = E f o r G

a l l f i n i t e so lvab le groups G , E an - %covering subgroup of G , i f f

N c X . - - -

Proof. Suppose N (E) = E and w e w i l l show t h a t N c X . G - - -

L e t G t - N . I f G is p r imi t ive then by ( 1 . 4 . 4 ) , G Z Z f o r some prime P

p s o G = E f - X . So we may assume G i s not p r i m i t i v e . But again by

(1.4.41, every p r imi t ive f a c t o r group G/K i s isomorphic t o Z f o r P

some prime p s o G/K f 5 f o r a l l p r imi t ive f a c t o r groups G/K of G . Hence G f - X s i n c e - X is a Schunck c l a s s .

Conversely, i f - N - c - X and E 5 < x > E I NG(E) where x i s

any element i n N (E) then < x > E/E %'< x >/E n < x > f N s o G -

< x > E/E f - X . But E i s an - X-covering subgroup of G s o

< x > E = EE = E and E = N (E) a s required. G

Lemma 3.3.3. Let - X be a Schunck c l a s s , - N - C X , - G* a f i n i t e

not necessa r i ly so lvable group, G 3 G*, G so lvab le , H 5 G , H 5 4 (G*)

and G/H C X . Then G C X . - -

Proof. Let E be an - %covering subgroup of G . Then

it now fol lows t h a t E =

= G * ~ G = G , ~ ~ G C X -

Theorem 3.3.4.

G = EH s ince G/H C X . By F r a t t i n i - a G * argument G* = N (El G = NG, (El EH = N (E)H, G* G*

+(G*) G

b u t H 5 +(G*) s o G* = N (El . By (3.3.21, G*

H

Let be a Schunck c l a s s , G a f i n i t e not -

necessa r i ly so lvable group, G 2 G , Go so lvab le , E an %covering 0 -

subgroup of Go

and H Z G . Then N (EH/H) = NG(E)H/H. (Thus t h e G/H

proper ty of being a normalizer of an - X-covering subgroup of a so lvable

normal subgroup is preserved under epimorphisms of f i n i t e groups) .

Proof. L e t gH C N (EH/H) s o (EH/H) gH = EH/H , t h a t i s G/H

( E H ) ~ = E ~ H = EH, a l s o E~ 5 E ~ ( H n G ~ ) = [E(H n co) lg = (EH O G o l g =

E ~ H n co = EH n co = E (H n G ~ ) s o E and ~g are ?-covering subgroup

of E ( H n Go) , S O there e x i s t s x C E , h C H n G I such t h a t

Eh = EXh = Eg . T h u s gh-l C N G ( E ) and g H C N G ( E ) ~ / ~ , so

N (EH/H) ~ N ~ ( E ) H / H . G/H

let x c N ( E ) s o x~ c N ~ ( E ) H / H . NOW ( E H ) ~ = E ~ H = EH , s o G

(EH/H) = EH/H. H e n c e XH C N (EH/H) and N (EH/H) = N G ( E ) H/H . G/H G/H

D e f i n i t i o n 3 . 3 . 5 . L e t - X be a Schunck class.

Then l e t N ( X ) - = {G: G has a normal X-covering - subgroup). N ( X ) -

is called the n o r m a l i z e r of - .

T h e o r e m 3.3.6, Let - X be a Schunck class, G a group and E

an - X-covering subgroup of G . T h e n the f o l l o w i n g hold.

( i ) N G ( E ) is an N ( X ) - c o v e r i n g - subgroup of G .

( i i ) N ( X ) - i s a Schunck class.

Proof. (i) E 5 N ~ ( E ) SO E i s an - X-covering subgroups of

N G ( E ) . A l s o E 9 N ~ ( E ) therefore N G ( E ) C ~ ( 5 ) . L e t N ( E ) 5 H 5 G , G

H/K C N ( X ) - since E 5 N ( E ) 5 H and E i s an %covering subgroup of H , G -

EK/K i s an - X-covering subgroup of H/K. B u t H/K C N ( 5 ) so

EK/K 4 H/K, t h a t i s , N (EK/K) = H/K. B u t by ( 3 . 3 . 4 ) , N (EK/K) = NG ( E ) K/K G/K G/K

so H/K = N ( E ) K / K and H = N ( E ) K . T h u s N G ( E ) i s an N ( X ) - c o v e r i n g G G -

subgroup of G .

( i i ) F o l l o w s i m m e d i a t e l y f r o m ( i ) and f r o m (3.1.41, since N ( X ) -

is Q - c l o s e d .

Lemma 3.3.7. Let X be a Schunck c l a s s . Then N ( X ) = X - - -

i f f N c X . - - -

Proof. Suppose E 5 and l e t G € N ( X ) - s o G has a normal

X-covering subgroup E t h a t i s N (E) = G . However N c X , s o by - G - - -

(3 .3 .2 ) , E = N ( E ) = G € X and N(X) c X . Thus N ( X ) = 5 . And G - - - - -

t h e converse follows e a s i l y from (3.3.2) .

Lemma 3.3.8. N c N ( X ) f o r a l l Schunck c l a s s e s X . - - - -

Proof. For any N(X)-covering - subgroup NG(E) of G ,

N (El 3 N ( E ) 9 G s o by ( 3 . 3 1 , N (N (El = N ~ ( E ) . Then by (3.3.2) . G G G G

Remark 3.3.9. Let G be any f i n i t e so lvable group. Then by

(2.2.161,that % is a C(r ) -cover ingsubgroup of G .. But G n " s o - -

G E N ( C ( r ) ) . Therefore N( .C( IT) ) i s t h e c l a s s of f i n i t e so lvab le groups . .

As w e know Schunck c l a s s e s of f i n i t e so lvable groups a r e

{Q,E ,D }-closed, b u t n o t n e c e s s a r i l y R -closed as w e showed i n example 0 0

(3.2.17) ( 2 ) . ~ a s c h G t z (1969) asked whether every {Q,E D )-closed c l a s s & 0

i s a Schunck c l a s s . Hawkes (1973)gives a negat ive answer, as we can s e e

i n t h e fo l lowing example.

Example 3.3.10. Le t - Y denote t h e c l a s s comprising groups

of o rde r 1, groups of order 2 and non-abelian groups of o rde r 6 .

S e t X = E D,!. By (2.1.6) and (2.1.5), E D i s a c losure opera to r s o - 6 d 0

by (2.1.6) and (2.1.5) , EmQ i s closure operator s o

QX - = QE D Y 5- E QD Y = E D Y = X s ince D y i s Q-closed. Hence & 0- 0- 4 0- - 0-

X = { Q , E ~ , D ~ } ? . Let G denote t he extension of an elementary abelian -

group K = < a > x < b > of order 9 by an inver t ing involution a .

Then among the maximal subgroups of G are < a,b > , < a , a > and

< b , a > s o & ( G ) = < 1 > . A t yp i ca l element of D Y i s e i t h e r < 1 > 0-

o r a d i r e c t product of some cyc l i c groups of order 2 and some copies

of S3 . Thus it is not hard t o see t h a t G f - X and every primitive

fac tor of G does belong t o 5 , even t o Y . Therefore -

X = {Q , E D }A but X is not a Schunck class . - d ' o - I -

In order t o construct a closure operator Ro s a t is f ying

- - Do 5 R 5- Ro such t h a t X i s a Schunck c l a s s i f f X = { Q , E ~ , R ~ } ~ we

0 - - need the following:

Definit ions 3.3.11. (i) Let L 5 G and K , H 5 G . We say

L covers H/K if H 5 KL which i s equivalent t o saying H 5 K(H n L) . Notice t h a t ' i f L 2 H then ' L covers H .

(ii) A subgroup M i s ca l l ed a complement of a chief f ac to r

L/J i n G i f G = ML and M n L = J .

G

Lermna 3.3.12. L e t L/J be a complemented chief fac tor of a

group G and M one of t he maximal subgroups of G complementing it.

Writing C = CG (L/J) and K = Cor (M) . Then G

( a ) K = M ~ C .

( b ) C/K i s a ch ie f f a c t o r which is G-isomorphic with L / J .

Proof. ( a ) M n C 5J G ,follows e a s i l y . Now t o show M n C i s

t h e l a r g e s t normal subgroup of G contained i n M , l e t M fI C 5 H P M

and H 9 G , SO [L,H] 5 L n H 5 L fI M = J s o H 5 c (L/J) = c , bu t G

H 5 M s o H ~ C ~ M . Thus H = M f l C and K = M f l C .

( b ) M complements L/J s o G = LM, M f l L = J . Assume without l o s s o f

g e n e r a l i t y t h a t J = < 1 > , s o L i s a minimal normal subgroup of G .

By ( a ) , K = M n c (I,) = C M ( ~ ) . NOW L is a minimal normal subgroup of M

G s o L i s a b e l i a n and t h e r e f o r e L 5 C G ( L ) But C ( L ) = C (L) n G G G

= c (L) n MI, = L (M fl c (L) ) by t h e modular law. Therefore c ~ ( L ) = L C ~ ( L ) G G -

Def in i t ion 3.3.13. A crown of G is a f a c t o r of t h e form C/R

where C is t h e c e n t r a l i z e r o f a complemented ch ie f f a c t o r H/K and

The fo l lowing lemma shows t h a t a normal subgroup of G e i t h e r

covers a crown C/R o r i s proper ly contained i n C .

Lemma 3.3.14. Let C/R be a crown of G and N 3 G . Then

t h e fol lowing a r e equ iva len t :

(i) The image of C/R under t h e n a t u r a l homomorphism G + G/N is

a c o r n of G / N ;

(ii) N does not cover t he fac tor C/R ;

(iii) RN < C .

Proof. By (i) CN/N/ RN/N 1 CN/RN is a crown of G/N and

since by def in i t ion a crown is a non- t r iv ia l normal f ac to r it i s c l e a r

t h a t (i) implies (ii) .

Assume (ii) holds, and l e t

is a normal subgroup of G properly

[N,c] 5 c n N 5 L , and since C i s

t r i v i a l normal fac tor of G between

L = cfl N R = (c f l N ) R . Then L

contained i n C . Now

the cen t r a l i ze r i n G of any non-

C and R , we have N I CG(C/L) = C.

Therefore NR = NR n C = L < C , so (ii) implies (iii) .

Final ly f o r (iii) implies (i) t he proof involves the theory of

groups modules and we omit it.

Definit ion 3.3.15. If - X i s a c lass of groups, define

- of normal subgroups R X a s follows : G f Kz i f f G has a s e t i ~ ~ } ~ = ~

0-

Ni sa t i s fy ing

(a) G/Ni f - X I i = 1 ,2 , ..., t . t

(8 ) n N~ = < 1 ) . and i=l

(y) For each crown C/R of G , there e x i s t s i

t h a t Ni does not cover C/R . -

Lemma 3.3.16. R i s closure operatox. 0

Proof. Evidently - - c R X and, i f c 0- - -

- To prove R i s a closure operator , it remains t o

0

C 1 2 , t such

- Furthermore Do 5 R 5 Ro.

0

y , t h e n E X C R Y . - 0- - 0-

show it ' is idempotent.

t G C E ~ X = E (k A) . Then G has. normal subgroup. iNi with 0- 0 0-

t G/Ni C E , n N . = < 1 > and f o r each crown C/R of G , t h e r e

0- 1 i= 1

e x i s t s i € 1 I 2 , . . . t } such t h a t N does not cover C/R . Thus each i

G/Ni has normal subgroups iNi j / ~ i ti j=1 such t h a t

(c) Each crown of G/N i s not covered by a t l e a s t one N . ./N The i 11 i '

c l e a r l y s a t i s f i e s condi t ions (a) and ( 6 ) of t h e Def in i t ion (3.3.15) .

Let C/R be . a crown of G . There e x i s t s i 1 , . . t ] such t h a t

Ni does no t cover C / R . By (3.3.14), c . . 1 1 i s a crown

of G/;U~ and by condit ion (.c) above t h e r e e x i s t s a j C f1 ,2 , . . . .t 1 . }

such that N i j does not cover it. Again by (3.3.14) w e have RNi j < C

and a l s o by (3 -3.14) , N does not cover t h e f a c t o r C/R . Therefore i j

t h e s e t {Nij} a l s o s a t i s f i e s (y) . Thus G C E . It fol lows 0-

- - 2 - R = R X and t h a t R i s a c losure operator . 0- 0- 0

- - It i s obvious t h a t R 5 Ro . To show t h a t Do 5 Ro , l e t

0

G = G x G2 x ... x Gr with < 1 > # G. ex f o r i = l , . . , r S e t 1 1 -

N = IT G . Then t h e normal subgroups c l e a r l y s a t i s f y i

j Z i j

condi t ion (a) and ( B ) . It fol lows e a s i l y from the p r o p e r t i e s of

d i r ec t product t h a t each chief fac tor is central ized by a t l e a s t one N i

so C = N f o r some i and the f ac to r of the form N . /R i s never a i 1

crown since f o r i # K , N 5 G and N does not cover N . /R and is not k k 1

properly contained i n Ni .

Theorem 3.3.17. The condition X = {Q,E , R }X i s both necessary - 4 0 -

and su f f i c i en t fo r X t o be a Schunck c lass . -

Proof. Let X be a Schunck c lass and l e t G C R X . G has a - 0-

family {N it of normal subgroups sa t i s fy ing condition ( a ) , (.@I and (y) . i i=l

Let G/K be any primitive fac tor of G . Let C/K denote the unique

minimal normal subgroup of G/K and C/R the crown of G associated

with C/K. It follows from the hypothesis t h a t there e x i t s

i C 1 2 . . . t } such t h a t N does not cover C/R and by (3.3.14) , i

NiR < C . Let C/T be a chief f ac to r of G with T 1 NiR . Then . G/K%G/T Q G / N . R / T / N . R 1 1 C Q X = - - X SO G/K C - X . G/K i s an a rb i t r a ry

primitive f ac to r of G , so G/K C - X f o r a l l primitive f ac to r groups

G/K of G . But X i s a Schunck c l a s s s o G C and R X = . Since - - 0- -

a Schunck c lass is Q-closed and E -closed, the necessity of the condition 4

i s es tabl ished.

We prove the suff ic iency by contradiction. Suppose there

e x i s t s a {Q,E , R )-closed c l a s s X which i s not a Schunck class . Let 4 0

-

G be a group of minimal order subject t o G/K C - X f o r a l l primitive

fac tor groups G/K of G and G f X . I f < 1 > # N 4 G , then a l l -

the primitive fac tors of G/N belong t o - and by minimality of G ,

G/N € - X . Hence by E -closure of X , + ( G I = < 1 > . Let N denote t h e 4 -

s e t of a l l minimal normal subgroups of G . I f IN1 = 1 , then G is

p r imi t ive and s o G € - X a con t rad ic t ion . I f I N I > 1 then

fI IN: N € N } = < 1 > . If C/R i s a crawn of G , e i t h e r R > < 1 > o r

R = < 1 > and c f 1 . In e i t h e r c a s e , t h e r e i s an N N such t h a t

RN < C . Thus t h e set N of a l l minimal normal subgroups o f G

s a t i s f i e s condi t ions ( a ) , (81 and (y) of Def in i t ion (3.3.15). Hence

G C %? = - X . This f i n a l con t rad ic t ion completes t-he proof .

CHAPTER 4

A COMPOSITION AND A LATTICE OF SCHUNCK CLASSES

1. Composition of Classes.

Most of t h e fol lowing m a t e r i a l i s from ~ a s c h & z ' s Notes on

Pure Mathematics (1979) and Er ickson 's paper, "Products of Satura ted

Formations" (1982 ) .

Defini t ion 4.1.1. Let - X be a c l a s s of groups and Y a

formation. Then XY -- = {G; Gy E 5 ) i s c a l l e d t h e formation product

of X with Y . -

Remarks 4.1.2. (i) I f - X i s a l s o a formation then G C Xy - - i f f

1 > . This i s because ( G ~ ) ~ = < 1 > i m p l i e s G C X and - - Y - -

Also i f G C XY - - then Gy C - X . But - X i s a formation so -

1 >.

XY . But X C XY does no t hold i n genera l . -- - - --

Example. Let 5 = b: G i s so lvab le , [M:N] f 2 t/ 2-chief f a c t o r s

Lemma 4.1.3. Let X - and - Y be a formation. Then XY -- i s a

formation.

Pxoof. Let G C Xb' - - , M 3 G . BY (3.2.12). ,

I ( G / M ) ~ = G M ~ M '% fI M C - X since

Y / Gy C and X i s Q-closed. - - - - -

Thus G/M C Xy and Xy i s Q-closed. - - --

To show XY - - i s R -closed, l e t G/hl,G/N2 C XY . Then jus t as 0 -

above (G/N~) ^ . G ~ G ~ fl N. C X for i = 1 . 2 . By R -closure of X , 1 - 0 - - -

h 2 I y . Hence n N~ f XY -- . -

Theorem 4.1.4. Let X be a Schunck class and Y a sa tura ted - - formation. Then XY i s a Schunck class . --

Proof. (See the figure i n the end of the proof . By the same

argument as i n (4.1.3) ,we can show Xy is Q-closed. --

Let G/N C XY f o r a l l primitive factor groups G/N of G - - and l e t S be an - X-covering subgroup of GY . by the F r a t t i n i argument

- G = N (S)Gy . I f no M < G w i t h . S 5 M and G =

G MGY e x i s t s , then

- max -

N (S) = G . t ha t i s S -4 G . and Gy /S 5 q5(G/S). Since Y i s G - -

saturated formation. it follows tha t G/S C - Y and Gy 5 S . Therefore -

G ~ = s C X - and G C X Y . -- I f M < G with S 5 M and G = M G ex i s t s . - max

Y -

Then l e t N = Cor M . Now G Y - -

G/N C - by hypothesis , Since S i s an - A-covering suhgroup of

G ~ ' G~ = s k y fl N ) 5 SM = M . B U ~ G = G M = M a con t rad ic t ion t o t h e

- - - Y -

assumption t h a t M < G . G

Theorem 4.1.5. Let - X be a Schunck c l a s s , 5 5 , and - Y i s

a formation. Then Xy - - i s a Schunck c l a s s and - N - C XY -- .

Proof. Let G/N C XY - - f o r a l l p r i m i t i v e f a c t o r groups G/N

of G and l e t S be an ?-covering subgroup of Gy . By (3.3.21, -

N ( S ) = S, and from the proof of (-4.1.4) , S I? GY , s o we observe

Y -

t h a t now Gy = S C - X implies t h a t G ( XY - - . Thus - XY - i s a Schunck -

Example 4.1.6. AN i s a formation by (4.1.31, b u t no t a

, t h e semidi rec t product Schunck c l a s s : Consider t h e group G = C Q

of Q8 with a subgroup C3 of ~ u t Q~ . G has t h e subgroup C2 of

Q8 as i t s c e n t e r and C2 5 4 ( ~ ~ 1 5 6 ( G ) s ince Q8 2 G . Now

C &f , b u t G { AN s i n c e ( G ) ~ = Q~ f A . ~ h u s AN i s no t - - - - -

a Schunck c l a s s .

Theorem 4.1.7. Let - X, - Y and Z be formations. Then -

Proof. (i) G/H C Xy - - i s , hy ( 3 . 2 . l l ) , e q u i v a l e n t t o

This i s equivalent t o G H /H r~~ H fI G C X and t o ( G y I X 9 H n G y 9 H, Y - - / . - - - -

t h a t i s G/H C XY - - i f f ( G ~ ) ~ 9 H . But GXy i s t h e smal l e s t normal sub- - - --

group such t h a t Thus

The preceding theorems o f f e r far - reaching p o s s i b i l i t i e s f o r

t h e cons t ruct ion of ~ c h & c k c l a s s e s .

Examples 4.1.8. The fol lowing Schunck c l a s s e s a r e obtained.

(i) NN. - - .. . . . N - , t h e c l a s s of groups of n i l p o t e n t length I r . Y

r

(ii) ' S -P 1 S -P-P S 1 S -P . . . . .S S ., t h e c l a s s of so lvab le groups of p-length - P I -P

5 r where S and S a r e t h e c l a s s . of so lvab le IT' and IT-groups -P ' -P

respec t ive ly , where 7-r = ip).

(iii) S ,S X where 5 i s a formation. This follows by (4 .1 .5) . -P -P-

Another method of producing new c l a s s e s from o l d is t h e

extens ion of one c l a s s by the o the r .

Def in i t ion 4.1.9. Let and Y be a c l a s s of groups. Then - -

w e def ine - * - Y a s fol lows * Y = (G: t h e r e e x i s t s N C A, N -4 G , G/N € Y}. - - - -

Theorem 4.1.10. I f - Y i s a formation and - i s a formation

which i s c losed under normal subgroups. then - * - y i s a formation.

Proof. ( G / N I / (MN/N) Z G/MN G / M ~ C s ince G/M C - Y . Since MN/N - ,y t h e fol lowing implies G/N C - * - Y and s o - * - y i s

Q-closed.

L e t G/N, G/M C - * x , s o t h e r e e x i s t J and K , subgroups

of G , such that J /M 3 G / M , G / J € - Y , J /M C - and

K/N 3 G/N, G/K C - Y , K/N C - X . Thus G/K n J C s ince - Y i s a

formation. As ( J fI K ) J n K = - and (J n K )

J ~ K ~ M K ~ M -

' C X s i n c e X i s S -closed. S imi lar ly it fol lows t h a t - K n M - - n

' C . By R - c losure of X it folows t h a t J f l N - 0 -

J f l K -- c x , b u t (G/M n n K/M n N) c Y and M ~ N - K ~ M ~ J ~ ~ N - -

J n € x s o G/M fl N C X * Y . T ~ U S X * Y is a formation. M n N - - - - -

Lemma 4.1.11. Let X and Y be a formation. Then - -

(ii) I f X i s S -closed, then Xy = X * Y . - n -- - -

Proof. (i) Is obvious.

G/N C - Y and thus Gy Ir N . But X i s S -c losed s o - n Gy C - X and - -

G C XY . Thus XY = X * y . - - -- - -

However unl ike t h e formation product XY, X * Y need not be -- - -

assoc ia t ive . For example, l e t G = A4 and l e t C be t h e c l a s s of

c y c l i c groups. Then G C (C*C)*C. However, G has no n o n - t r i v i a l normal 6

c y c l i c subgroups, s o G 4 C* (.C*C) .

Lemma 4.1.12. I f - X is a formation, E is an - %covering

subgroup of G , and M,N 5 G , then EM n EN = E ( M n N ) .

Proof. Let K = EM n EN, now M n N 5 EM fl E N , E 5 EM fl EN = K.

so J@ fl N = =1)/M fl N fl K 6 - X s i n c e 1( i s a formation. But E(M fl N) M fl N / /

I

i s an X-covering - subgroup of G/M f l N s o E ( M fl N) M n N i s - X-maximal

i n G/M n N , s o t h a t K = E ( M n N ) .

Theorem 4.1.13. L e t X - and - Y be s a t u r a t e d formations, and

suppose t h a t t h e order of each group i n - X i s coprime t o t h e o rde r of

each group i n - y . Then * y i s a formation. I f E i s an 7 -

%covering subgroup of G and F i s a Y-covering subgroup of NG (E) -

then EF i s - X * - Y-covering subgroup of G . In p a r t i c u l a r - X * - Y i s

a Schunck c l a s s .

Proof. 5 * Y i s Q-clos-ed by t h e same argument a s i n (4.1.10).

To show it is R -closed, l e t G/M,G/N c X * y , s o t h e r e 0 - -

e x i s t normal subgroups J , K of G such t h a t

J/M ZJ G/M, G / J C - Y , J / M C 5 and K/N L' G/N,G/K C - y , K/N € - X . Choose

a Hal l T-subgroup H of G where i s t h e s e t of primes d iv id ing

I J / M I o r I K / N I . Then by t h e hypothes is on 5 and - Y , H 5 J n K ,

and HM = J, HN = K. Thus H/H n M % HM/M = J / M E - X and

R -closed. By (4.1.12) , H ( M n N ) = HM n HN s i n c e H i s z-covering sub- 0

J n K HM n HN - H ( M n N ) , - - H group of G . SO - - € A .

M n N - M n N n N - Now M n N M n N

We consider t h e second conclusion. Cer ta in ly EF - X * - Y . Now suppose t h a t EF 5 V , and V/A c & * - Y . We w a n t t o show t h a t

EFA = V . Since V/A c & * , t h e r e i s a normal subgroup K of

V , V/K € Y and K/A c & . Since ( _ I V / K I , I E I ) = 1, E 5 K - , and

t h e r e f o r e E i s an - X-covering subgroup of K . Thus K = EA s i n c e

K/A E - X . Since any two V-conjugates of E a r e n e c e s s a r i l y K-conjugate,

k f o r any x E V we have E ~ = E f o r some k E K, a n d s o

xk" r NV(E) . Hence x C N (El-K and V = N (E) K. Then v v

V/K = N ~ ( E ) K / ~ C N ~ ( E ) / ~ ( E I fl K SO N ~ ( E ) N ~ ( E ) fl K t Y s i n c e / -

V/K C - Y . Now F I v n N (E) = N (3) and F i s a Y-covering subgroup G V

of NG (El . It follows t h a t

i s an 5 * Y-covering subgroup of G and by (3.1.41, - X * - Y i s a

Schunck c l a s s .

The previous r e s u l t appears i n I ~ a l e , M. P. , 1976 1.

There is a l s o a cons t ruct ion f o r Schunck c las ses - corresponding

t o subgroup generat ion.

Def in i t ion 4.1.14. Let - and - Y be Schunck c las ses . Then l e t

<X, - - Y > = { G : G = < S,T > f o r every - %covering subgroup S and every

Y-covering subgroup T of G ) -

The fol lowing theorem i s due t o D. Blessenohl and appears i n

t h e Notes on Pure Mathematics of ~ a s c h k z (1979) .

Theorem 4.1.15. L e t - X and - Y be Schunck c l a s s e s , G any

f i n i t e so lvable group. Then

(i) < &,Y_ > i s a Schunck c l a s s .

(ii) The subgroup H i s an < - X,Y - >-covering subgroup of G i f

H i s minimal sub jec t t o H = < S,T > where S i s an - X-covering

subgroup of G and T i s a - Y-covering subgroup of G .

Proof. (i Is a consequence of (i i , s o we need t o prove (ii 2 .

(ii) Cer ta in ly H € < - X , Y - > . Se t < - X,Y - > = L and suppose H 5 L,

K 3 L and L/K € - Z . A s H 5 L t h e - X-covering subgroups of L a r e

p r e c i s e l y t h e - %covering subgroups of G t h a t a r e contained i n L and

s i m i l a r l y f o r - Y-covering subgroups of L . Hence t h e subgroup H

s a t i s f i e s t h e hypothesis of (ii) wi th re spec t t o L . I f L < G then

H i s - Z-covering subgroup of L by induct ion and L = HK. I f L = G ,

then K 3 G , G/K € - Z . Now SK/K i s an - X-covering subgroup of G/K

and TK/K i s a - Y-covering subgroup of G/K, bu t G/K € L and s o

G/K = < SK/K,TK/K > = < S,T > K K = HK/K t h a t i s G = HK. Hence H i s /

The fol lowing approach y i e l d s a more r e s t r i c t e d way t o combine

two Schunck c l a s s e s t o form a new one. And we s e e how we can cons t ruc t

new formations by comparing covering subgroups.

Theorem 4.1.16. Let X - be a Schunck c l a s s , - Y a. formation and

X = IG: an %covering subgroup E of G belongs t o Y } . Then X i s -Y - - - -Y - a formation.

Proof. Let G 'C 5 , N 2 G . L e t E be an :-covering sub- -

group of G , s o E C - y . Then EN/N € - Y s ince - Y i s Q-closed. But

EN/N i s an - %covering subgroup of G/N s o G/N € . Hence ZY i s Q-closed. -Y - -

Now l e t G/N1,G/N2 C 5 . I f E i s an - X-covering subgroup -

of G , then EN ji is an - %covering subgroup of fii , i = 1,2 i

and EN^/^ E A i fI E C - Y , i = 1,2 . By R -closure of Y , 0 -

EJNl n N, n E c - Y so E ( N ~ n N,I (N, n N,) c Y . B U ~ E ( N ~ n / - i s an - X-covering subgroup of pl n a, so G N~ n N, c x / -Y - -

Remark 4.1.17. ZY need not be a s a t u r a t e d formation even i f -

X and Y a r e . - -

Example. Let X = k and Y , t h e c l a s s of 3-groups. Then - - -

X and Y a r e s a t u r a t e d formations. Now consider t h e group H = C Q - - 3 8

a s i n example (4.1.61, Z ( H 1 5 $(H) , H / Z ( H ) ' Z A 4 Now t h e Car te r

subgroup of A4 i s of o rde r 3 , s o an - N-covering subgroup of A i s 4

i n - Y and hence A4 C _Xy . On t h e o t h e r hand t h e Car te r subgroup of H -

must conta in Z ( H ) , s o H t Xy . Therefore Xy i s not a s a t u r a t e d - - - formation.

It i s remarkable t h a t a t l e a s t t h e fol lowing holds :

Theorem 4.1.18. Let p C T , and - X a formation. Then

( 9 s S X i s a s a t u r a t e d formation, where S is t h e c l a s s of f i n i t e -p ' -p- - 'lT

T-solvable groups.

The theorem i s proved wi th t h e a i d of the fol lowing lemma:

Lemma 4.1.19. Let G be a so lvable group, T a f i x e d s e t of

primes and S a Hal l T-subgroup of G . Then i f p C T t h e maximal

normal p'-subgroup of S is contained i n t h e maximal normal p'-subgroup of G .

Proof. (See t h e Figure below). Let H and T be t h e maximal

normal p ' -subgroups of G and S , r e spec t ive ly . I f H # < 1 > , t hen

the a s s e r t i o n follows by induct ion by consider ing G/H i n whose maximal,

but t r i v i a l normal p ' -subgroup t h e group TH/H must be contained;

t h e r e f o r e T 5 H i n t h i s case. So assume H = < 1 > and l e t

L = F i t ( G ) a s S is a Hal l IT-subgroup and H = < 1 > , L 5 S . By

l . 6 , L conta ins i t s c e n t r a l i z e r s o L commutes elementwise with

T . Hence ~ L , T ] = < 1 > and T 5 C (I,) 5 L . Then T = < 1 > . G

G

Proof of theorem 4.1.18. F i r s t of a l l , by (4.1.16). ,

S - = s i s a formation s ince both and S -P 1 S -P- are -p ' -p-

formations. W e show t h a t - Y i s a s a t u r a t e d formation. For t h i s purpose

l e t G b e any so lvab le group, M 5 G , M 5 d(G) and G/M - Y . min

Furthermore, l e t E be an S -covering subgroup of G ; t hus E i s a - IT

Hall IT-subgroup of G , a l s o EM/M is an S -covering subgroup of G/M -IT

and EM/M C S ,S X s ince G/M C Y . I f G has a minimal normal subgroup -P -P- -

K d i f f e r e n t from M , then MK/K 5 4 (G/K) . Now

E G/MK r i s i n c e - Y i s p s l o s e d . Hence by

induct ion G/K E - Y and by R -c losure of Y , G/M n K E y . But 0 -

M n K = < 1 > s o G C - Y . Therefore it may be assumed t h a t M i s t h e

a, unique minimal normal subgroup of G . I f [ M I = q and q # p , s o

q f n then E C S ,S X s ince E M / M C E ~ n M = E and EM^ C S S X . -p -P- -p ' -p-

I f q = p C IT and F/M i s a n i l p o t e n t normal subgroup of G/M , then

F i s n i l p o t e n t s ince - N i s E -clos.ed. By t h e hypothesis G has no d

normal pl-subgroup # < 1 > s ince M i s t h e unique minimal normal sub-

group of G and I M I = qa r q = p C n .- By (.4.1.18), E/M has a t r i v i a l

(E lS = < 1 > ; hence E C S c S ,S and G C Y . Thus Y i s a --P- - -P -P- - - -P-

s a t u r a t e d formation.

Remarks 4.1.20.

(i Obviously $y = nx - f o r a family Y of a formations. Y .

-i - i - ., - J.

(ii) - X - c (SIT)s ,S , t h i s is because i f G € and E i s

-P -p-

an ST-covering - subgroup of G , then E - N s ince - N i s S-closed.

As. we have seen i n t h e beginning of t h i s s e c t i o n , ~ a s c h k z

proved by means of covering subgroups t h a t XY -- i s a s a t u r a t e d formation

whenever - X and - Y a r e s a t u r a t e d formations of f i n i t e so lvable groups.

Erickson i n (1982) extends t h i s r e s u l t t o f i n i t e groups i n genera l .

D e f i n i t i o n 4 . 1 . 2 1 . I f - X i s a c l a s s o f f i n i t e g r o u p s , t h e

support of X - i s t h e s e t of primes d iv id ing t h e o rde r s of groups i n - X

and we denote it by n(X1. -

Note: The support of - X conta ins t h e c h a r a c t e r i s t i c of - X . When X - i s

a s a t u r a t e d formation t h e two not ions a re t h e same.

Lemma 4.1.22 . Let X and Y be formation of f i n i t e groups. - -

Then

Proof. (i) This is immediate s i n c e - X conta ins a l l groups

of order 1 .

I T ( Y ) c IT(%'), s o it s u f f i c e s t o show t h a t ( 1 C . I f - - -- - - --

p E 'irr(X), - then C 6 X s ince X i s a s a t u r a t e d formation. But P - -

(C ) = < 1 > o r C ; i n e i t h e r case C XY , s o t h a t p 6 T(XY) . P V P P -- - -

Proposi t ion 4.1.23. L e t - F be a formation of f i n i t e groups.

Then t h e fol lowing a r e equ iva len t .

(i). I i s a s a t u r a t e d formation.

(ii) I f G i s a f i n i t e group, if N G , and B 5 G wi th

B 5 d ( ~ ) n N and N/B C - f , then N = N X N2 , where

Proof. (i) implies (ii). by Lemma (3.2) of Semetkov (1974) .

(ii) impl ies (i) : Set G = N , ~ ( G S = B s o G / ~ ( G ) C f and

G = N 1

x N2 where N~ C - F ana N ~ Q ( G ) . ~ h u s G = N C F . 2 1 -

Theorem 4.1.24. Let - Y and X - be formations of f i n i t e

groups with - X s a t u r a t e d . Assume t h a t - Y is s a t u r a t e d o r t h a t

IT (Y) c IT (X) . Then XY is a s a t u r a t e d formation. - - - - -

Proof. By t h e same argument a s i n (4.1.13), w e can show t h a t

XY is a formation. --

To show XY - - is s a t u r a t e d . Let G be a f i n i t e group, and

kt A 3 G wi th A 5 + ( G I and G/A C XY - - . W e show t h a t G C Xy -- . min

Let H = GyA , SO t h a t H 9 G and H/A = GyA A = ( .G/A)~ C & s ince - - / -

/

G/A C XY , s o H X s ince X is a s a t u r a t e d . Then H H n $(.GI. C X . - - - - / -

H C 4 T I ( H ) fl r ( X ) = $ and H , S A . I H~ = < 1 ) , then 1 - 2 -

A s s u m e then , t h a t H = A and Y is. sa tu ra ted . Then 2 -

A L a t t i c e of Schunck C1ass.e~.

In t h i s sec t ion we a r e fol lowing t h e paper of Wood (1973 ) .

F i r s t w e need t o r e c a l l t h e fol lowing d e f i n i t i o n s :

Def in i t ions . A p a r t i a l l y ordered set i n which every p a i r of

elements has a jo in and a meet i s c a l l e d a l a t t i c e . A p a r t i a l l y ordered

s e t i n which every subse t has a jo in and a meet i s c a l l e d a complete

l a t t i c e . When a l a t t i c e L h a s a l e a s t element 0 and g r e a t e s t

element 1, w e can speak of complements i n L . S p e c i f i c a l l y , by a

complement of an element b i n t h e l a t t i c e L we mean an element

b ' C L such t h a t b A b ' = 0 and b V b ' = 1. W e say t h a t a l a t t i c e

i s complemented i f it has a l e a s t element and a g r e a t e s t element and

each of i t s elements has a complement.

Def in i t ion 4.2.2. L e t - X be a Schunck c l a s s . I f f o r any

f i n i t e so lvab le group G , every &subgroup of G is contained i n

an - %covering subgroup of G then - X w i l l be s a i d t o have the

Examples. (i) The c l a s s of f i n i t e so lvable IT-groups has t h e

D-property; t h i s is because t h e Ha l l IT-subgroups of G a r e t h e

11-covering - subgroups of G and by (1.3.51, any 'IT-subgroup of G l i e s

i n some Hal l IT-subgroups of G .

(ii) f , The c l a s s cons i s t ing of t h e i d e n t i t y group and - U , t h e c l a s s

of f i n i t e so lvable groups have the D-property. The main r e s u l t i s t h e

following.

Main Theorem 4.2.3. Let p be t h e s e t of a l l Schunck c l a s s e s

wi th D-property. Then D , p a r t i a l l y ordered by inc lus ion , forms under

composite and i n t e r s e c t i o n opera t ions a complete, complemented, l a t t i c e .

We reduce tlie proof of t h i s theorem t o t h e s e r i e s of

p ropos i t ions and lemmas.

Proposi t ion 4.2.4. If X, Y C D , then < X,Y > C 0 . - - - -

Proof. By induct ion on t h e order of G w e show t h a t every

< - X, - Y >-maximal subgroup of G i s conjugate t o an < X - , - y >-covering suh-

group of G .

I f G C < X , Y > then t h e r e i s nothing t o prove. So assume - -

G f < X , Y > , - - H an <X,Y>--maximalsubgroupof - - G and P an

< - X,Y - >-covering subgroup of G . If N i s a maximal normal subgroup

of G , then PN/N is an < - - X,Y >-covering subgroup of G/N s o by

induct ion i n G/N, HN/N 5 (PN)'/N f o r some g E G and HN 5 P ~ N , t h a t

i s H 5 P'N . Now pg is an < - X,Y - >-.covering subgroup of P ~ N . If

P'N < G , then again by induct ion H 5 (pg) g 1

f o r some 91 C G and

s ince H is an < X,Y >-maximal subgroup of G, H = P 991

- -

Now assume G = P'N = PN. Induction al lows us t o assume t h a t

Cor P = < 1 > . This i s because i f we p u t K = Cor P , then G G

HK/K 5 (PK) '/K = pg/K t h a t is HK 5 pg s o H 5 pg ; bu t H i s an

9 < X,Y >-maximal subgroup of G s o H = P . Thus induct ion al lows - - us t o assume Cor P = < 1 > . ~ h u s G i s p r imi t ive with a unique

G

minimal normal subgroup N . Let M = HN. By way of a con t rad ic t ion

assume M < G . Now s i n c e N 5 M , M = M fl G = M fl (PN) = N(P fl M)

by t h e modular law. Thus M/N = N ( P n MI ?, P n M = p n ~ . B U ~ N - P ~ M ~ N

M/N = HN/N S H / H n N c < X , Y > . T ~ U S P n M E < X , Y > . AS M < G , - - - -

t h e subgroup P n M i s contained i n an < X, - - y >-covering subgroup of M .

The subgroup H i s < - X , Y - >-maximal i n G and s o c e r t a i n l y is

< - X, - Y >-maximal i n M ; hence H i s an < - X,Y - >-covering subgroup of

M . Thus t h e r e e x i s t s m E M such t h a t ( P n M I m 5 H . So without

l o s s of g e n e r a l i t y we assume t h a t P n M I H .

Now H = < A,B > where A i s an ?-covering subgroup of M

and B i s a - Y-covering subgroup of M , and H i s minimal of t h i s

form, by Theorem (4.1.15) . Since P n M z M/N t h e 5-covering subgroups

of P n M and M/N a r e isomorphic. Let Q be an - X-covering subgroup

of P fl M . Then Q A N / N E A/A n N s i n c e AN/N i s an - X-covering

subgroup of M/N . The < - X , Y - >-covering subgroup P of G i s of t h e

form < SIT > where S i s some - X-covering subgroup of G and T i s

some - Y-covering subgroup of G . AS X - has t h e D-property , t h e r e

X e x i s t s x E G such t h a t A 5 S . Hence

X ( A ~ N ) ~ = A ~ N ~ s ~ N S P ~ N = < ~ > s i n c e s ~ < s , T > = P .

Thus A n N = < 1 > and AN/N A/A n N = A . SO t h e - X-covering

subgroups of M n P a r e isomorphic t o ?-covering subgroups of M and

hence a r e - X-covering subgroups of M by t h e D-property of - . Simi la r ly t h e - Y-covering subgroups of P n M a r e - Y-covering subgroups

of M . I f R i s any - X-covering subgroups of P n M then

P n M = < Q,R > 5 H = < A,B > . ow Q i s a l s o an &covering subgroup

of M I R i s a - Y-covering subgroup of M and H i s an < X,Y >-covering

subgroup of M ; s o , by minimality of H , < Q , R > = < A,B > . Thus

P n M = H , which implies H I P , AS H i s < - X,Y - > -maximal i n G ,

H = P. So then G = PN = HN = M which c o n t r a d i c t s t h e f a c t t h a t M < G.

Thus, we have G = HN, H < G s ince G f < - X , y > - . By (1.4.8) and (1.4.91,

H i s conjugate t o P s-ince G is p r imi t ive . Thus < - X.Y - > C D .

Corollary 4.2.5. ~ f X. C 0 , i C 1 . then <z i : i C I > C -1

Proof. This follows by induct ion from t h e Proposi t ion simply

because i n each group G only f i n i t e l y many of t h e _Xi are involved.

Lemma 4.2.6. L e t zl, &. , be Schunck c l a s s e s with

1 1 ~ 5 ~ 1 Y c ! ~ and Zl, - Y1 have t h e D-property. Then

- - X -covering subgroup of Proof. Let G C < X,Y > , Sl any

G and T1 any 11-covering subgroup of G . For any &-covering

subgroup S of G and any - Y-covering subgroup T of -G we have

G = < S I T > . Now s C X c z l , - - T C - Y - cYl and X Y C D s o t h e r e -1 ' -1 a b

e x i s t a , b C G such t h a t S 5 S1, T 5 T1 then

a b G = < S ,T > 5 < S1,T1 > , b u t S ,T 5 G s o G = < S1,T1 > and

1 1

G C < %,Yl > . Hence < X,Y - - > - c < X Y > . -1 '-1

Remark 4.2.7. For X and Y a r b i t r a r y Schunck c l a s s e s -1 -1

the lemma i s f a l s e i n general .

Example. Put X - = - N t h e c l a s s of f i n i t e n i l p o t e n t groups.

Y = X = Y = 1 t h e c l a s s of supersolvable groups. - -1 -1 -

Then S4 C < - N.1 - > b u t S4 f < 1.1 > , even though N C 1 . - - - - -

Then - Z = < - X , y - > . ~ h u s < - X,Y - > i s t h e smal l e s t Schunck c l a s s with

D-property containing both - X and - Y .

Proposi t ion 4.2.9. then

Proof. We s h a l l prove by induct ion on the order of G ,

t h a t i f H 5 G and H is - X n - Y-maximal i n G , then H i s an

X n Y-covering subgroup of G . Suppose H I K 5 G , L 5 K and - - 0

K / L ~ C - X n - Y . NOW H is - X n - Y-maximal subgroup of G s o c e r t a i n l y

H i s - X fl - Y-maximal subgroup of K . I f K < G , then by induct ion H

i s an - X fl - Y-covering subgroup of K and s o K = L H . 0

Now l e t K = G and G/L C n Y. I f G C n y then H = G , and t h e r e i s o - - - -

L

nothing t o prove. Suppose G # - X n - Y . Then t h e r e e x i s t s a ch ie f

f a c t o r L/J with L 9 Lo and s a t i s f y i n g G/L ( - X n - Y b u t G / J 1 - X n - Y .

A s G/L C - X n - Y , t h i s impl ies t h a t G = SL = TL where S is an

%covering subgroup of G , T i s a Y-covering subgroup of G . W e - -

have G/J # - X n - Y s o e i t h e r JS < G o r J T < G . Suppose JS < G . g By t h e D-property of - X t h e r e e x i s t s g C G such t h a t H 5 S . W e

9 have S/S n L i SL/L = G/L C - X n - Y , a l s o H~ 5 JS < G s o H i s an

g X n Y-covering subgroup of JS. Thus S = H (S n L). Hence - - g G = S L = H (3 ~ L ) L = H ~ L = H L . But L C Lo s o G = H L z H L o and

G = HLo . Thus H is an - X fl - Y-covering subgroup of G .

Corollary 4.2.10. I•’ X. € D , i C I , then n A. € . -1 i c I-'

This shows t h a t i s a complete l a t t i c e under t h e composite

and i n t e r s e c t i o n opera t ions . To show t h a t t h e l a t t i c e i s complemented

we need t h e following lemma.

Lemma 4.2.11. Let be a Schunck c l a s s with t h e D-property. -

I f IT i s t h e c h a r a c t e r i s t i c of - X , then SIT fs , where IT

i s t h e

c l a s s of f i n i t e IT-groups.

Proof. Assume t h e lemma is f a l s e and suppose G E SIT i s of

l e a s t o rde r such t h a t G f X . Let N be a minimal normal subgroup o f -

G . By t h e minimality of G , G/N € - , which implies G = EN where E

i s an - %covering subgroup of G . If N < G , then N - X by the

minimality of G . Now N i s an X-subgroup - of G , and - X has t h e

g D-property s o the re e x i s t s g € G such t h a t N 5 S , and N 3 G

impl ies N 5 S . Then G = NS = S € 5 , a con t rad ic t ion . Thus N = G , . s o G i s c y c l i c of order p where p C ?r and hence G 5 , a

con t rad ic t ion . Theref ore ST - C x . -

Proposi t ion 4.2 -12. The l a t t i c e i s complemented.

Proof. Let - F be the u n i t c l a s s and - U t h e c l a s s of f i n i t e

so lvable groups, both c l a s s e s having t h e D-property. L e t - X € D , with Tr

t h e c h a r a c t e r i s t i c of - X and I T ' , t h e complementary set of primes. Let

Y be the c l a s s of f i n i t e 'i~' -groups. Then X n Y = F and < X, y > = U - - - - - - -

by t h e previous lemma.

We a r e going t o end t h i s s e c t i o n by some charac te r i za t ions of

Schunck c l a s s e s with D-property. For th i s . we now s t a t e two condi t ions

t h a t we w i l l need.

The Schunck c l a s s - X s a t i s f i e s condi t ion A i f f o r any group

G and any subgroup H of G with t h e Cor H # < 1 > , then an G

%covering subgroup of H i s contained i n an X-covering subgroup of G . - -

The Schunck c l a s s X s a t i s f i e s condi t ion B i f , whenever an -

X-covering subgroup of G avoids a minimal normal subgroup N of G , -

a l l - %maximal subgroups of G avoid N . usual a subgroup A avoids a ch ie f f a c t o r

Theorem 4.2.12. For t h e Schunck

equivalent :

1. - X has t h e D-property ,

2. - X s a t i s f i e s condi t ion A ,

3 . - X s a t i s f i e s condi t ion B .

And we mean by avoids a s

H/K i f A n H r K .

c l a s s - X t h e fol lowing a r e

Proof. (1) impl ies ( 2 ) , namely: Suppose H 5 G and S i s an - %covering

subgroup of H. Then S i s an - X-subgroup of G . Since has t h e -

D-property S i s contained i n an - %covering subgroup of G . Thus (2)

holds. (Note we d i d n o t need t h e hypothesis t h a t Cor H f < 1 > ) .

(2) impl ies (1) : Let - X s a t i s f y condi t ion A and G be any group. W e

s h a l l a r g ~ i e , by induct ion , on t h e o rde r of G . So l e t H be - X-maximal

i n G and S an &-covering subgroup of G . I f G € - X then

G = S = H and t h e r e i s nothing t o prove.

A s s u m e G { 5 . Let N be a minimal normal subgroup of G .

Now SN/N i s an - X-covering subgroup of G/N and HN/N i s an - X-subgroup

9 1 of G/N. Therefore by induct ion NH/N 5 (SN) /N f o r some

91 C G .

91 Thus H < NH I s N . Now S g1 is an &-covering subgroup of G s o S g1

1 1 is an - &covering subgroup of S N. I f S N < G then by induct ion H

i s conjugate t o S g1 91 1 i n G and i n S N s o H i s conjugate t o S

1 S O H is an - X-covering subgroup of G . So le t G = S N = SN,

Cor S = < 1 > and G p r imi t ive . Let M be a maximal subgroup of G

with H 5 M . By induction H i s an - X-covering subgroup of M . I f

Cor M # < 1 > then H < st f o r some t C G by condi t ion A . But H G

i s an - X-maximal subgroup of G , s o H = st and H i s an - %covering

u subgroup of G . I f Cor M = < 1 > then by (1 .4 .8) , G = MN and M = S

G

f o r some u € G and the re fo re M € - X . But H i s an &-maximal subgroup

u of G . Then H = M = S and H i s an - X-cover ingsubgroupof G .

Therefore - X has the D-property . L

(1) impl ies ( 3 ) : Suppose - X has t h e D-property and l e t N be a minimal

normal subgroup of G avoided by t h e - X-covering subgroup S of G . Let H be - X-maximal i n G . Now 5 has t h e D-property s o H = sg f o r

some g € G and H fI N = sg n N = (S n N ) ~ = < 1 > s i n c e S avoids N . Thus X s a t i s f i e s condi t ion B . -

( 3 ) impl ies (2 ) : Assume t h i s i s f a l s e and l e t G be a counter-example

of l e a s t order . Hence t h e r e e x i s t s H , a subgroup of G , with

CorGH # < 1 > , T an - %covering subgroup of H , and T $ S f o r

any 5-covering subgroup S of G . Choose H of l e a s t order i n G

wi th t h i s proper ty . Obviously G 4 5 , because i f G C - X then G = S

and s o T 5 S a con t rad ic t ion , and H < G . Let N be a minimal

normal subgroup of G with. N I Cor H . I f TN < H , then CorG(TN) f 1 G

and s o by minimality of H , T 5 S f o r some - X-covering subgroup S of

G . Hence H = TN.

Consider G/N. By t h e minimality of G , & s a t i s f i e s condit ion

A i n G/N and a l l subgroups of G/N . Thus by t h e proof of (1) o ( 2 )

every - X-subgroup of G/N is contained i n an - %covering subgroup of

G/N . Hence TN/N 5 ( s N ) ~ / N f o r some g G , and s o T 5 s g N . X s a t i s f i e s condi t ion A i n If s g N < G , than by minimality of G , -

g SgN and then T i s contained i n some conjugate of S . Hence G = S N = SN

and S i s maximal i n G as G f . Now a s N 3 G , S n N 5 S and

S n N 5 N a s N i s abe l i an and s o by minimality of N , S n N = < 1 > . ~ l s o

H = H n G = H n (SN) = (H n S I N so

H/N = ( H n S ) N / N z H ~ S --- - H n s - ~ n ~ C X s i n c e H C X s o a s H ~ S ~ N S ~ N - -

L

h H < G , S n H i T f o r some h € H . But S avoids N s o by condi t ion

B , T avoids N . Now H n S 2 H/N = TN/N . Hence

I H fl 5 ) = I T N / N ~ = JT/T fI N ) = I T ) s i n c e T avoids N . Thus S n H = # h

and T 5 S . This con t rad ic t ion e s t a b l i s h e s t h e theorem.

CHAPTER 5

EXTREME CLASSES OF FINITE SOLVABLE GROUPS

In t h i s Chapter w e cons ider a paper by R.W. Car te r , B. F ischer

and T. 0. Hawkes (.1968) .

1. Extreme and S k e l e t a l Classes.

W e w i l l s t a r t by the following well-known r e s u l t :

Lemma 5.1.1. Suppose G has a unique minimal normal subgroup

N . I f N ) 4 ( G ) then

(i) N = C (N) = it(^), t h e F i t t i n g subgroup of G , and G

(ii) A l l complements of N i n G a r e conjugate.

Proof. (i) If N ) d ( ~ ) , then by (1.1.2 ) (ii)., N has a p a r t i a l

complement M say , i . e . , G = MN. NOW M n N - Q M as N - Q G ,

M n n Q N a s N i s abe l i an , s o M n N -d G and by minimality of N ,

M n N = < 1 > s o N i s a complemnt. Let C = C ( N ) . A s C 3 G , G

C n M 3 M and s ince N normalizes C n M I C n M 3 MN = G . Thus, by

t h e uniqueness of N , C n M = < 1 > . But

c = c n G = c n (MN) = N ( C fl M ) = N , SO N = c ~ ( N ) . ~ l s o s i n c e N i s

abe l i an N is n i l p o t e n t s o N 5 F i t ( G ) and hence < 1 > # N 4 it (G) .

Since F i t (G) i s n i l p o t e n t N n Z ( F i t ( G ) ) # < 1 > , and by minimality

of N , N I Z(Fit(.G)). Then N I Fi t (G) I CG(N) = N , s o

(ii) Follows from (1.4.9), s i n c e G i s p r imi t ive .

Def in i t ion 5.1.2. We say a c l a s s of groups i s extreme i f

(a). - X = IQ,E }X , a n d dJ -

(b) Whenever G has a unique minimal normal subgroup N

and G/N C X . Then G € . - -

Lemma 5.1.3. The c l a s s X i s extreme i f f = IQ,E )X and - - 0 -

X conta ins a l l groups which possess X-covering subgroups. - -

Proof. Suppose - X is extreme, and - X does no t conta in a l l

groups with 5-covering subgroups. L e t G be a group of minimal o rde r

such t h a t G has an - X-covering subgroup H , and G f 5 . N m i f N

is a minimal normal subgroup of G , then HN/N i s an - X-covering sub-

group of G/N, s o G/N € - X by minimality of G . Then G = HN s ince H

i s an - X-covering subgroup of G .

W e c la im t h a t N i s a unique minimal normal subgroup of G . To show t h a t it s u f f i c e s t o show t h a t C ( N ) = N, S O p u t C = CG ( N ) ,

G

now c = c n G = c n (HN) = N(.C n H ) , s i n c e N 5 c . NOW c f l H 5 H a s

G/C n H C - x , s o G = H ( C n H) = H , a con t rad ic t ion . ~ h u s c n H = < 1 >,

and s o C = N. Thus N is unique, s o by (5.1.2 ) ( b ) , G C - X and t h i s

con t rad ic t ion proves t h e n e c e s s i t y . Conversely, suppose - X = IQ,E~}?

and t h a t - X ccjntains a l l groups which possess ?-covering subgroups. We

w i l l show - X s a t i s f i e s condi t ion (b). Suppose N i s a unique minimal

normal subgroups of G and G/N € - X . I f N 5 6 (GI , then G C - X s ince

X i s E -closed. So assume N $: +(G), s o by t h e proof of (5.1.U (i) t h a t - 4

N i s complemented i n G , by H say , and G = HN, H %G/N - X s o H

i s an - %covering subgroup of G. Then G € - X by hypothesis .

Before giving some examples of extreme c l a s s e s we s p e c i a l i z e

the concept s l i g h t l y .

Def in i t ion 5.1.4. We c a l l a c l a s s - X of groups s k e l e t a l i f

(i) 5 = { Q , E ~ } ~ , and

(ii) Whenever G has a minimal normal subgroup N such t h a t

G/N 6 - X and N i s complemented i n G with a l l i t s complements conjugate

then G € X . -

Lemma 5.1.5. Ske le ta l c l a s s e s a re extreme.

Proof. Suppose - X i s s k e l e t a l and le t N be a unique minimal

normal subgroup of G and G/N 6 - X . Since - X: i s E -closed w e may Q

assume t h a t N ) Q (GI , s o by (5.1.1) (i) , N i s complemented i n G by M

say , t h a t i s G = MN, M n N = < 1 > and l e t C = C ( N ) . Now G

c n M Ir MN = G , b u t N i s a unique minimal normal subgrdup of G , s o

i f C n M # < 1 > then N 5 C n M I M a contradic t ion. Thus

c n ~ = < 1 > a n d s o C = C ~ G = C ~ (MN) = N ( c ~ M ) = N , s o

N = C (N) and by (5.1.1) (ii) a l l t h e complements of N a r e conjugate, G

hence by (5.1.4) (ii) , G € - X . Thus - X i s extreme.

Examples 5.1.6. (.l) Let - X be t h e c l a s s of f i n i t e groups

which a r e generated by a t most r elements, then - X i s s k e l e t a l

c l a s s where r 1 2 .

Proof. It is easy t o show t h a t - X i s Q-closed and E -closed, d

s o it remains t o show - X s a t i s f y condit ion (ii) of Def in i t ion (5 -1.4) .

To do t h a t , l e t N be a minimal normal subgroup of G , G/N 6 3 , and

suppose t h a t N i s complemented by H i n G and t h a t a l l complements

a r e con jugate. I f H = < h 1,-

,h > we w i l l prove t h a t t h e r e e x i s t s r

- n € N such t h a t G = < hl,h2 ,..., h n > . I f hl = h2 - ... = h = 1, r r

then G = N and s o G i s c y c l i c group s o t h e r e s u l t i s c l e a r l y t r u e .

Therefore suppose h # 1 and suppose the r e s u l t i s t r u e f o r a l l groups 1

of order l e s s than ] G I . Let K be t h e l a r g e s t normal subgroup of G

contained i n H and f i r s t suppose K f < 1 > . Then by induct ion we

have G/K = < hl Y/K,hZ/K, ..., h nl//;( > = i h K,...,hrnK>/K f o r some r 1

n C N . Let H* = < h . . . ,h n > . I f H* # G , then H* complements 1 ' r

N , s o H* conjuage t o H , bu t K - Q H s o w e have K S H * . Thus

H* = < h K,h2K,. . . 1

K,h K > = G , a con t rad ic t ion . Hence w e may lhr-1 r

assume K = < 1 > . Now N n N (H) 4 NG(.H) a s N 3 G and G

N n N ( H I 3 N a s N is abe l i an , s o N fI N ( H I I! G and, by t h e G G

Then H i s self-normalizing i n G . Enumerate the elements of N

thus N = {nl = l , n 2 , ... ,n ) where S = I N ! , and def ine S

H = < h , h ,h n . > , i = 1 , 2 , . . . , S . I f a l l t h e Hi complement i 1 2'" ' 'hr-1 r 1

N i n G , then e i t h e r they form a complete conjugacy c l a s s o r H = H , i j

f o r some i # j . I f t h e former, w e have hl € K , a con t rad ic t ion

because K = < 1 > . If t h e l a t t e r , then h n C Hi , s o n r i -%I-I- C Hi i r

a l s o h n € Hi s ince Hi

= H , t h e r e f ore Hi conta ins r j j

- 1 -1 ( n ~ l h - l ) . (h n . = n n . On t h e o t h e r hand, n n # 1 otherwise 1 r r j i j i j

-I n = n s o i = j a con t rad ic t ion . Therefore H conta ins n n # 1 i j i i j

a con t rad ic t ion . Thus H = G f o r some i and G € X s o X i s s k e l e t a l i - -

c l a s s .

( 2 ) Let - X be t h e c l a s s of groups which a r e generated by a conjugacy

c l a s s of elements. Then - X i s a s k e l e t a l c l a s s .

Proof. It is easy t o see t h a t - X i s both Q-closed and

E -closed. To show X s a t i s f y condi t ion (ii) of Def in i t ion (5.1.4) , 9 -

l e t N be a minimal normal s u b g r ~ u p of G , G/N C - X and suppose t h a t

N i s complemented by H i n G and t h a t a l l complements a r e conjugate.

By hypothesis H = < h = h h 1, 2 ' " '

,h. , r 2_ 1 > , where t h e h f o r m a r i

complete s e t of conjugates of h i n H . I f H is not normal i n G ,

then t h e r e e x i s t s g € G such t h a t hg f H . Then as H i s a maximal

subgroup of G we have < H ,hg > = G , s o G generated by t h e

conjugates of h i n G . Thus G C X - . I f H i s a normal subgroup

of G , our hypothesis implies t h a t H i s t h e unique complement of N

i n G . Let 1 f n € N . Since G = H x N , w e have N 5 Z ( G ) , and

the re fo re h n,h n , . . . ,h n a r e a l l t h e conjugates of hn i n G . 1 2 r

Let H* = < h . n , i = 1 , 2 , . . . ,r > . Since H*N = G we must have e i t h e r 1

-1 H* = H o r H* = G . I f H* = H I then h h n F H and hl (hln) = n € H

1' 1

a con t rad ic t ion s ince H n N = < 1 > and N # 1 , s o H* = G , G € - X

and X i s a s k e l e t a l c l a s s . -

Before going any f u r t h e r w e need t o r e c a l l t h e following

d e f i n i t i o n :

A chief f a c t o r R/S of G i s c a l l e d a F r a t t i n i f a c t o r i f

R/S 5 ~ ( G / S ) . Obviously any chief f a c t o r i s e i t h e r complemented o r

a F r a t t i n i f a c t o r .

Lemma 5.1.7. Let G be a f i n i t e group and N 4 G . Given

any two ch ie f s e r i e s of G pass ing through N , t h e r e is a 1-1

correspondence between t h e chief f a c t o r s below N i n t h e two s e r i e s

such t h a t corresponding f a c t o r s are p r o j e c t i v e l y r e l a t e d (and hence

G-isomorphic) and such t h a t the F r a t t i n i chief f a c t o r s of one s e r i e s

correspond t o those of the o t h e r s e r i e s .

Proof. W e use induct ion on I N I . I f t h e two s e r i e s have t h e

same minimal subgroup, t h e r e s u l t fol lows by induction. Therefore

assume t h a t t h e minimal normal subgroups a r e N 1 2 1 , N , N f N2 . Again

using induct ion , w e may assume without l o s s of g e n e r a l i t y , t h a t

N = N N 1 2 - I f e i t h e r N 5 + ( G ) o r N fl d(G) = < 1 >, then s i n c e

t h e correspondence N1 - N/N2 I N2 - N/N1

s a t i s f i e s our requirements; f o r t h e chief f a c t o r s concerned a r e e i t h e r a l l

F r a t t i n i o r a l l complemented. I f N n 4 ( ~ ) = Ni , i = 1 , 2 , t h i s

correspondence a l s o works; i n t h i s case Ni

and N/Nj a r e F r a t t i n i and

N and N/Ni a r e complemented, where i f j . F i n a l l y suppose j

N n Q ( G ) = N~ where 1 < N3 < NlN2 and N1 f N3 # N2 . Then

N3, N/N1, N/N2 a r e F r a t t i n i f a c t o r s and N / N ~ . Nl and N a r e no t . 2

In t h i s case w e have N / N 1 l N j N/N2 and N1% N/N3% N2 , hence

N/N1 - N/N2 , N1 * N2 i s t h e requ i red correspondence.

Now f o r convenience w e denote F i t ( G ) by F ( G ) , and w e

def ine induct ive ly t h e terms F i ( . ~ ) of t h e upper n i l p o t e n t series of

a group G by FO(G) = 1, F~ (G)/Fi - ( G I = F ( G F i - l ( G ) ) f o r

i = 2 . - 1 . Thus F ( G I = F (G) . ~ l s o we def ine another s e r i e s 1

Fi(G)/di (G) co incides with t h e s o c l e of G / Q ~ (GI where t h e s o c l e of

G/mi (G) i s defing by M G : M i G i s a minimal normal subgroup

Of G/bi (GI and t h e r e f o r e F . ( G I /Q . (G) i s t h e d i r e c t product of 1 1

complemented minimal normal subgroups of G/Qi (G) .

The n i l p o t e n t length E (G) i s t h e smal l e s t 8 such t h a t

Lemma 5.1.8. I f 0 i s a homomorphism of G then

Proof. The subgroup of G t h a t corresponds t o Q (0 (G) ) i s

t h e i n t e r s e c t i o n P of t h e maximal subgroups of G t h a t conta in t h e

ke rne l of 0 . Since 0 a l s o conta ins Q ( G ) t h e conclusion fol lows,

t h a t is 0 ( @ ( G ) ) - c $(a(G) 1, s ince ~ ( v ) = @ ( G ( G ) 1.

Lema 5.1.9. Let G be a group such t h a t F(G)/Q(G) i s a

chief f a c t o r of G . Then G has a t most one complemented minimal

normal subgroup, and i f N i s t h i s subgroup then 4 (G/N) = Q2 (GI /N .

Proof. We assume Q ( G ) f < 1 > ; otherwise F (GI i s t h e

unique minimal normal subgroup of G and t h e r e s u l t follows e a s i l y .

Since N p Q (G) , t h e hypotheses imply t h a t NQ (G) = F ( G I . Let M be

a maximal subgroup of G . I f MQ2(G) = G , then MF1(G) = G. This i s

s o because (MF (GI ) Q (GI = G implies (MF1 ( G ) ) Q2 (GI /Pl ( G I = G / F ~ ( G ) SO 1 2

( G ) / F ~ (GI 1 (m2 ( G I / F ~ (G) ) = G/F ( G I . ~ u t t h i s means G / F ~ ( G I = 1

G = MF ( G ) = M N d ( G ) = MN. Thus every maximal subgroup of G/N conta ins 1

d2 ( G ) /N , and w r i t i n g K/N = Q (G/N) , s o K/N = ~ ( M / N ) where M / N a r e

maximal subgroups of G/N, then K 1 Q 2 ( G ) s ince every maximal subgroup

of G / N conta ins I $ ~ ( G ) / N . Let 8 be a homomorphism 8: G/N + G / F ( G )

L

def ine by 8 (H/N) = HF ( G ) /F (.GI , s o by Lemma ( 5 - 1 - 8 ) I

s o +2 ( G ) /F ( G ) 1 KF ( G ) /F (G) = K / F (G) , then K = Q2 (G) . Thus

(G/N) = Q~ (GI /N .

NOW suppose N* i s a minimal normal subgroup of G such t h a t

N* f N; then N* Y N*N/N 9 F~ (G) /N = F ( G I N ) . Also N* n Z ( F ~ (GI 1 f < 1 >

and SO N* Z Z ( F ~ ( G I 1 by minimality of N*, s o N* 5 F (GI 5 C G ( N * ) . 2

I f N* $ Q ( G ) then a s above w e have ~ * 6 ( G ) = F ( G ) and hence

N* !E F ( G ) /Q ( G ) because N* fl Q (G) = < 1 > . Now F (GI /Q ( G ) i s a chief

f a c t o r , s o F (GI 5 C (F (GI /4 ( G ) ) and a l s o it is we l l known G

F i t t i n g subgroup conta ins i t s c e n t r a l i z e r , and F (GI /d (GI =

t h a t t h e

F (G/d (GI 1 .

Hence CG (F ( G ) /d ( G ) 1 = F ( G ) . But N* '% F ( G I /4 ( G I , hence F ( G I = CG(N*)

a con t rad ic t ion . Then N* 5 C#I ( G ) s o G has a t most one complemented

minimal normal subgroup.

We now def ine a new c l a s s of groups denoted by G , a c l a s s

which w i l l p lay an important p a r t i n t h e sequel .

Def in i t ion 5.1.10. G C G i f f Fi ( G ) / Q i ( G ) i s a chief f a c t o r

of G f o r i = 2 , . . . , 8 G . We c a l l t h e groups i n 6' extreme groups.

Remark. The i n t e r s e c t i o n of an a r b i t r a r y c o l l e c t i o n of extreme

c l a s s e s i s again extreme. Our nex t r e s u l t shows t h a t G i s t h e i n t e r -

s e c t i o n of a l l such c l a s s e s ; i n o t h e r words t h e smal les t - extreme c l a s s .

Theorem 5.1.11. G i s a subc lass of every extreme c l a s s and L

is i t s e l f an extreme c l a s s .

Proof. F i r s t we show G is extreme. I f G/N € G , N 5 6 ( G ) ,

then by t h e f a c t s F(G/N) = F(G) /N , 4 (G/N) = Q ( G ) / N it i s c l e a r t h a t

G € G . Therefore G i s E -closed. To show it i s Q-closed, suppose 4

G € G and N 0 G . And w e show G/N € G . By induct ion we may assume

t h a t N i s a minimal normal subgroup o f G . I f N 5 + ( G I , then by the

f a c t s j u s t c i t e d t h e upper F i t t i n g and F r a t t i n i of G a r e preserved

under t h e n a t u r a l homomorphism G --t G/N and s o G/N i s extreme. I f

N ) + ( . G ) , then by Lemma ( 5 . 1 . 9 ) , + 0 ~ / ~ ) = 4 ( G ) / N and again w e have 2

G/N € G . Thus G i s Q-closed.

To show G s a t i s f i e s condi t ion (ii) of Def in i t ion (5.1.2) ,

suppose N i s a unique minimal normal subgroup of G such t h a t

G/N € G . I f N 5 + ( G I then G C G s i n c e G i s E -closed. I f 4

N $: 4 ( G I then by (5.1.1) (i) , F ( G ) = N , s o F (GI i s a chief f a c t o r

of G and t h u s G € G . Therefore G i s an extreme c l a s s .

F i n a l l y w e w i l l show t h a t i f F i s any extreme c l a s s , then -

G c f . Suppose not and l e t G be an extreme group of minimal o rde r - -

such t h a t G () - F . I f + ( G I # < 1 > we have G/+(G) € QG = G , and

t h e r e f o r e by minimality of G , G/+(G) € f , s o G € - F s i n c e - F is

E -closed, a con t rad ic t ion . Hence + ( G I = < 1 >. But by d e f i n i t i o n 4

of an extreme group, F (GI i s a minimal normal subgroup of G and

then it is unique. Since F s a t i s f i e s condi t ion (ii) of d e f i n i t i o n -

(5.1.21, w e have G € - F , a con t rad ic t ion . Therefore G - c - F .

The fol lowing r e s u l t i s a c h a r a c t e r i z a t i o n of extreme groups.

Theorem 5.1.12, The fol lowing statements are equivalent :

(i) G € G ;

(ii) C ( G I = m ( G ) , t h e number of conjugacy c l a s s e s of maximal subgroups

(iii) E ( G ) = C ( G ) , t h e number of complemented chief f a c t o r s i n a given

ch ie f s e r i e s of G .

(.iv) I f G* € Q ( G ) , G* has a t most one complemented minimal normal

subgroup .

Remark. Pu t t ing N = G i n Lemma (5.1.7) y i e l d s t h e f a c t t h a t

t h e i n t e g e r C ( G ) def ined i n (iii) above is independent of t h e p a r t i c u l a r

s e r i e s chosen and t h e r e f o r e an i n v a r i a n t of t h e group.

Proof. (i) implies (ii) . L e t G C G and consider t h e

following s e r i e s

By (5.1.1) (ii), t h e r e i s a unique conjugacy c l a s s of maximal subgroups of

G complementing t h e chief f a c t o r F ~ ( G ) / ~ ~ ( G ) f o r a l l i = 1 , 2 , ... , C ( G ) .

But each maximal subgroup of G must complement one of t h e complemented

ch ie f f a c t o r s i n the above chief s e r i e s , and must the re fo re belong t o

one of t h e s e 8 (GI conjugacy c las ses . Thus C ( G ) = m ( G ) .

(ii) implies (iii) . W e prove t h i s by induct ion . Let G be a group

s a t i s f y i n g E ( G ) = m ( G ) , and assume t h a t t h e impl ica t ion holds f o r groups

of o rde r less than / G I . ~f 6 ( G ) # < 1 > , we have

induct ion E ( G ) = E ( G / ~ ( G ) ) = r n ( G / & ( G ) ) = C ( G / @ ( G ) ) = C ( G ) . Therefore we may L

assume @(GI = < 1 > . By (1.1.13), F (GI i s t h e d i r e c t product of r 2 1

complemented minimal normal subgroups of G and we have

8(G) = m(G) 3 ~ ( G / F ( G ) + r Z -8 (G). - 1 + r . Hence r = 1, and again

(iii) impl ies (i). C ( G ) = E ( G ) , so t h e r e i s a t l e a s t one complemented

ch ie f f a c t o r H/K s a t i s f y i n g 4 . ( G ) 5 K < H 5 Fi(G) f o r each 1

i = l , , . . . , 8 . Thus F C G .

(i) impl ies ( i v ) . Let G E G and G* Q ( G I , s o G* € G s i n c e G i s Q-closed

SO Fi (G*) /di (G*) i s a chief f a c t o r of G* . It then follows from Lemma

(5.1.9), t h a t G* has a t most one complemented minimal normal subgroup.

Hence ( i v ) holds.

( i v ) impl ies (i). Suppose G s a t i s f i e s ( i v ) and l e t

G* = G/di (GI E Q ( G I . Since F. ( G I /+. ( G I is t h e d i r e c t product of 1 1

complemented minimal normal subgroups of G / ~ ~ ( G ) = G* , and s i n c e G*

has a t most one complemented minimal normal subgroups of G* it

follows t h a t Fi ( G ) / + . (GI i s a chief f a c t o r of G* . Hence G € G . 1

2 . Main Theorem.

In t h i s Section we are going t o prove t he main r e s u l t i n t h i s

chapter. In order t o do s o we need the following:

I f G i s any group, t he isomorphic images of G and the

uni t groups form a c l a s s of groups which w i l l be denoted by (GI .

I f - and - Y are a rb i t r a ry c lasses of groups, we define a

new c l a s s - Y'X - a s follows.

Definit ion 5.2.1. G f Y'X i f f S (G) fI X c Y . In other words - - - - -

Y'X comprises those groups a l l of whose %subgroups belong t o Y . - - - -

We simply wri te - yS f o r - Y ~ U - where - U is the universal c l a s s ( c l a s s

of f i n i t e solvable groups.). We note t h a t - ySx - i s S-closed and i f

X = U then Y'X i s t he l a rges t S-closed c l a s s contained i n Y . - - - - -

The elementary r e s u l t s contained i n t he following lemma are L

immediate consequences of t he def in i t ion .

which we remarked above;

I

Definit ion 5.2.3. A closure operator C i s s a i d t o be unary

i f f o r every c l a s s X , fl = U C ( G ) . - - G CX -

Note. S, Sn and Q are unary, but N R are not. 0' 0

Definition 5.2.4. I f C i s any unary closure operator, we

define f o r each c l a s s Y a new c l a s s m ( y ) by t h e following: - C -

G € m ( Y ) i f f G = < 1 > o r G f < 1 > and the following two conditions C -

are s a t i s f i e d :

(ii) whenever A € C(.G) and A f G , then A € - Y .

Remarks 5.2.5.

(i) mc (c) = 1 where C i s any unary operator.

(ii) m ( Y ) cons i s t s of groups which do not themselves belong t o y S - - L

but a l l of whose proper subgroups are i n - Y . We c a l l such groups c r i t i c a l

f o r Y . -

Definit ion 5.2.6. We c a l l a c l a s s Y , &-complete whenever

lemma 5.2.7. - Y i s - X-complete i f f ms(Y) - - - c X .

Proof. Let Y be X-complete and suppose t h a t G € m S ( Y ) . < - -

I f G f - X , we have S(G) fI - - - X C Y and therefore G € - Y'X - . But - Y i s

X-complete, so Y'X c Y and G € y , a contradict ion, because - - - - - -

Conversely, suppose ms - c - X and t h a t - Y i s not

X-complete. Let G be a group of minimal order such t h a t G C Y'x\Y . - - - -

Clearly G { X - (.otherwise i f G € - X then G C - Y , a con t rad ic t ion) .

I f H i s a proper subgroup of G then S (HI. fI - X - c S(.G) n - - - c Y ,

t h e l a t t e r inc lus ion follows s ince G € - Y'X - . Hence H € - Y'X - , and

by t h e minimality of G we have H C Y , s o G € m ( Y ) c X , a - s - - -

contradic t ion . Thus - Y i s - X-cmplete.

Lemma 5.2.8. The fol lowing s ta tements a r e equivalent :

Y'X c y - . Then by Proof. (i) impl ies (ii) . Suppose - - - -

Lemma (5.2.2) (i) , - Y'X - i s an S-closed c l a s s contained i n - y ,

and i n f a c t - yS is t h e l a r g e s t closed c l a s s contained i n - Y . Therefore - - - - Y'X c yS . But by (5.2.2) (iv) , - yS = - - - - - Y'U c Y'X .

Thus ySx= ys . - - -

(ii) impl ies (iii) . This fol lows from t h e i d e n t i t i e s

S S (l ) 5 = ( Y ~ X ) ~ X = ysx = vs . - - - - - -

(iii) impl ies (i) . Suppose i s not - X-complete s o ms (Y) 5 . Let -.

G € m s ( ~ ) \ 5 . Then G € - Y'X - = (f)S5 . But - Y' i s X-complete -

s o (f) 'X - - - - - C Y'C Y and G € , a con t rad ic t ion because G C ms (3 .

Thus - Y i s - X-complete.

Lemma 5.2.9. 1f N - Q G and G/N C - X = IQ,E then G has d -

a subgroup G* C X such t h a t G = NG* . -

Proof. Let G* be t h e minimal member of t he s e t of subgroups

which supplement N i n G . Since any supplement of N n G* i n G*

i s a l so supplement of N i n G , G* is a minimal supplement of N n G*

i n G* s o N fl G* 5 4 (G*) . Hence

and therefore G* C E X = X . d- -

Lemma 5 -2.10. I f X - = {Q~E+}? and - Y i s a homomorph then

Proof. Let N Q G C - Y'X - and l e t X/N be an* - X-subgroup . of

G/N. By (5 -2.91, X has a subgroup X* C - X such t h a t X = NX* . By . hypothesis X* C - Y , and therefore X/N = X*N/N s X*/N n X* C - Y since

Y ' i s Q-closed. Hence G/N € Y'X a s required. - - -

Lemma 5.2.11. Let 3 = QX - and l e t - Y be a c l a s s of groups

such t h a t R& n 5 5 - y . Then - Y'X - i s R -closed. 0

Proof. Let G/N1, G/N2 C - Y'X - with Nl n N2 = < 1 > and l e t

X be an - X-subgroup of G . Then X N ~ / N ~ S X/X n N1 C QX - = - X , and

therefore X/X n Nl C - Y by hypothesis. Similarly X/X n N2 C - Y .

Hence X = X / ( X rI N1) fI (X n N2) C RZ. Also X C - X , so

x c R Y n x c Y . T ~ U S G c Y'X and Y'X i s R~-c losed . 0- - - - - - - -

Corollary 5.2.12. If - F i s a formation and - = { Q , E + } ~ ,

then - PX - i s a formation.

Proof. By (5.2. l o ) , E ~ & is Q-closed. By (5.2.11), FX i s

F'X i s a formation. R -closed since R F n = F n c F . Thus - - 0 0- - - - - -

Main Theorem 5 -2.13. Let - F be t h e l oca l formation - f ( f )

where f is a f u l l formation function, and l e t - X be an extreme

c lass . Then - - F'X = - F(f*) where f * (p) = ( f (p) )'x - f o r a l l primes p .

Proof. By (5.2.12), [ ( f ( p ) ) S z = i s a formation s o f*

i s a formation function, therefore - F(.f*) ( .call it - f*) i s loca l ly

defined formation. We f i r s t show - - - - flee f j x . Suppose G € - F* and

l e t X be an - %subgroup of G . We have

XF ( G ) / F (G) 4 X/X n F (G) € QZ= 5 . NOW G/F ( G ) € f * ( ~ ) s ince P P P P

G € - F* , SO it follows t h a t X/X n F ( G I c f ( p ) . Next, P

t rue f o r a l l primes p , it follows t h a t X € - F . Therefore G. € - F'X -

and F* c F A . - - - -

Now suppose, f o r a contradict ion, t h a t t h i s inclusion i s

s t r i c t , and l e t G be a group of minimal order such t h a t G C - f j ~ -

but G j? - F*. For any K Q G , G/K f - F'X - s ince - PX - i s Q-closed,

s o by minimality of G , G/K € - F* . Thus G € !?l (F*) , and there fore , Q -

because i s a sa tura ted formation, G has a minimal normal subgroup

N which i s complemented and w.1.o.g. N i s unique. So by (5.1.11, N

i s se l f -cen t ra l iz ing i n G .

We now show t h a t G/N c f * ( p ) f o r a prime p d iv id ing N .

Every chief f a c t o r of G i s f* -cen t ra l s ince G/N F* , and it -

follows from t h e co ro l l a ry t o Theorem (2.5 ) of [ 71 t h a t G € F* , a -

con t rad ic t ion . L e t X/N C X and suppose I X 1 < I G 1 ; X S ( Px) = ~ S X - - - - -

by (5 -2 .2) (i) and t h e r e f o r e by minimality of G we have X C F*. -

Thus X/F ( X I € Q ( X / N ) 5 QX = X , bu t X C F* s o P - - -

X/F ( x ) c f * (p) = f (p) 'X and hence X/F (x) f (p) . Since P - P

N = C X ( ~ ) , F ( X I i s a p-group and t h e r e f o r e X/N E f (p) = f (p) by P P

hypothesis . On t h e o t h e r hand, i f X = G then by condi t ion (ii) of

Def in i t ion (5.1.2) ,we have G € X s ince N i s unique and X i s - -

extreme. Since G 6 F'X , we have G € F and the re fo re - - -

X/N = G/F (G) c f (p) . Hence we have proved S (G/N) n 5 5 f (p) and P

the re fo re t h a t G/N f (p) '5 = f * ( p ) a s requi red .

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