L. J. ZENG563

iv) every maximal subgroup [8] of G is normal;

v) G is the direct product of its Sylow subgroups.

Proof: i)→ii) Let G be nilpotent with class c. If

G, then 1ii

GH G

since

1ii i

GG GG

.

Hence

01 c

HGHG HGG

and H is subnormal in G in c steps.

ii)→iii) Let

G. Then H is subnormal in G and

there is a series 01n

HH HG

i

. If i is the

least positive integer such that

H. Then

1ii

HH

and

iG

NH.

iii)→iv) If M is a maximal subgroup of G, then

G

NM, so by maximality

G

NM G

and

G.

iv)→v) Let P be a Sylow subgroup of G. If P is not

normal in G, then is a proper subgroup of G

and hence is contained in a maximal subgroup of G, say

M. Then

G

NP

G; however this contradicts Lemma 4.

Therefore each Sylow subgroup of G is normal and there

is exactly one Sylow p-subgroup for each prime p since

all such are conjugate. The product of all the Sylow sub-

groups is clearly direct and it must equal G.

v)→i) by Lemma 2 and Lemma 3.

Theorem 1. Assume that every maximal subgroup of

a finite group G itself is not nilpotent. Then:

i) G is soluble;

ii) mn

Gpq where p and q are unequal primes;

iii) there is a unique Sylow p-subgroup P and a Sylow

q-subgroup Q is cyclic. Hence and . GQPPG

Proof. i) Let G be a counterexample of least order. If

N is a proper nontrivial normal subgroup, both N and

GNare soluble, whence G is soluble. It follow that G is

a simple group.

Suppose that every pair of distinct maximal subgroups

of G intersects in 1. Let M be any maximal subgroup:

then certainly

G

NM If Gn and

m

,

then M has nm conjugates [10] every pair of which

intersect trivially. Hence the conjugates of M account for

exactly

1mn n

n

mm

nontrivial elements. Since m ≥ 2, we have

1

22

nnn

nm

in addition it is clear that

21.

n

nnn

m

Since each nonidentity element of G belongs to ex-

actly one maximal subgroup, n – 1 is the sum of integers

lying strictly between 1

2

n

and n – 1. This is plainly

impossible.

It follows that there exist distinct maximal subgroups

M1 and M2 whose intersection I is nontrivial. Let M1 and

M2 be chosen so that I has maximum order [8]. Write

G

NNI. Since M is nilpotent,

1

M

NI by

Lemma 5, so that 1

NM

. Now I cannot be normal

in G; thus N is proper and is contained in a maximal

subgroup M. Then 11

NM MM

, which con-

tradicts the maximality of

.

ii) Let 1

1k

e

pe

e

Gpk, where and the i

are distinct primes. Assume that . If M is a maxi-

mal normal subgroup, its index is prime since G is solu-

ble; let us say

0

i

3

p

k

1

:GMp

. Let i be a Sylow

pi-subgroup of G. If , then i and, since M is

nilpotent, it follows that i; also the since .

Hence P1Pi is nilpotent and thus

P

M1iP

G

P3k

1,

i

PP1 (by Lemma

5). It follows that

1

NP

G and 1. This

means that all Sylow subgroup of G are normal, so G is

nilpotent. By this contradiction k = 2 and

GPG

12

2

e

p

1

e

Gp.

We shall write 2

pp

and q.

1

iii) Let there be a maximal normal subgroup M with

index [6] q. Then the Sylow p-subgroup P of M is nor-

mal in G and is evidently also a Sylow p-subgroup of G.

Let Q be a Sylow q-subgroup of G. Then G = QP. Sup-

pose that Q is not cyclic. If

p

Q, then ,

PG

since otherwise QGP, which is cyclic [6]. Hence

,

P is nilpotent and ,1gP. But this means that

,PQ 1

and GPQ

, a nilpotent group. Hence Q

is cyclic.

In an insoluble group [3] Hall π-subgroups, even if

they exist, may not be conjugate: for example, the simple

group PSL (2, 11) of order 660 has subgroups isomor-

phic with D12 and A4: these are nonisomorphic [10] Hall

2,3 -subgroups and they are certainly not conjugate.

However the situation is quite different when a nilpotent

Hall π-subgroup is present.

Theorem 2. Let the finite group G possess a nilpotent

Hall π-subgroup H. Then every π-subgroup of G is con-

tained in a conjugate of H. In particular all Hall

π-subgroups of G are conjugate.

Proof. Let K be a π-subgroup of G. We shall argue by

induction on

, which can be assumed greater than l.

By the induction hypothesis a maximal subgroup of K is

contained in a conjugate of H and is therefore nilpotent.

If K itself is not nilpotent, Theorem 1 may be applied to

produce a prime q in π dividing

and a Sylow

q-subgroup Q which has a normal complement L in K.

Of course, if K is nilpotent, this is still true by Lemma 5.

Now write 12

HH

where H1 is the unique Sy-

low q-subgroup of H. Since , the induction hy-

LK

Copyright © 2011 SciRes. AM