IF 39

1.38. Theorem. Fix a prime p, and let G be any finite group. Choose

S,T G Sylp(G) such that D — S C\T is minimal in the set of intersections

of two Sylow p-subgroups of G. Then Op(G) is the unique largest subgroup

of D that is normal in both S and T.

If we assume that S and T are abelian in Theorem 1.38, we see that D S

and D T, and so in this case, the theorem guarantees that D — Op(G). In

other words, Brodkey's theorem is an immediate corollary of Theorem 1.38.

Proof of Theorem 1.38. Let K CD, where K S and K T. We must

show that K C Op{G), or equivalently, that K C P for all Sylow p-subgroups

PolG.

Let N = NG(K) and note that S C N, and thus S G Sylp(N). Now let

P G Sylp(G) and observe that P n N is a p-subgroup of N. Then P D N is

contained in some Sylow p-subgroup of TV, and so we can write P D N C

Sx

for some element x G N. (We are, of course, using the Sylow D-and C-

theorems in the group N.) Now T C TV, and thus also

Tx

Z N and we

have

PnP^PniVnPcs'nT^D".

Then

i =

(Dx)x~l

D ( P H T * ) *

- 1

- P^"

1

n T .

Since P

x

and T are Sylow p-subgroups of G, and their intersection is

contained in D, it follows by the minimality of D that

Px

Pi T — D. In

particular, we have K C D C

Px

, and thus K

x

C P. But

iTx

= X since

x G iV = No(i^), and thus if C P, as required. •

1.39. Corollary. Let P G Sylp(G), where G is a finite group, and assume

that P is abelian. Then \G : Op(G)| \G : P|

2

.

Proof. By Brodkey's theorem, we can choose S,T £ Sylp(G) such that

SDT = Op{G). Then

This yields |G|/|P|

2

l/|O

p

(G)|, and the result follows by multiplying both

sides of this inequality by |G|. •

We can recast this as a "good" nonsimplicity theorem.

1.40. Corollary. Let P G Sylp(G)7 where G is finite and P is abelian, and

assume that \P\

|G|1//2.

Then Op(G) I, and thus G is not simple unless

\G\=p.