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
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
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
Z N and we
i =
D ( P H T * ) *
- 1
- P^"
n T .
Since P
and T are Sylow p-subgroups of G, and their intersection is
contained in D, it follows by the minimality of D that
Pi T D. In
particular, we have K C D C
, and thus K
C P. But
= 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|
Proof. By Brodkey's theorem, we can choose S,T £ Sylp(G) such that
SDT = Op{G). Then
This yields |G|/|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\
Then Op(G) I, and thus G is not simple unless
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