26 1. Sylow Theory

Op(G) as p runs over the prime divisors of G. Of course, F(G) is character-

istic in G, and in particular, it is normal.

1.28. Corollary. Let G be a finite group. Then F(G) is a normal nilpotent

subgroup of G. It contains every normal nilpotent subgroup of G, and so it

is the unique largest such subgroup.

Proof. By Lemma 1.27, we know that |F(G)| is the product of the orders of

the subgroups Op(G) as p runs over the prime divisors of G. Then Op(G) G

Sylp(F(G)), and thus F(G) has a normal Sylow subgroup for each prime. It

follows by Theorem 1.26 that F(G) is nilpotent.

Now let N G be nilpotent. If P G Sylp(iV), then P N by Theorem 1.26,

and thus P is characteristic in N and hence is normal in G. It follows that

P Q Op(G) C F(G). Since N is the product of its Sylow subgroups and

each of these is contained in F(G), it follows that N C F(G), and the proof

is complete. •

1.29. Corollary. Let K and L be nilpotent normal subgroups of a finite

group G. Then KL is nilpotent.

Proof. We have K C F(G) and L C F(G), and thus KL C F(G). Since

F(G) is nilpotent, it follows that KL is nilpotent. •

Problems I D

1D.1. Let P G

Sylp(jHr),

where H C G, and suppose that N

G

(P) C i7.

Show that p does not divide |G : H\.

Note. In these problems, we are, as usual, dealing with finite groups.

ID.2. Fix a prime p, and suppose that a subgroup H C G has the property

that CG(X) C JH" for every element x £ H having order p. Show that p

cannot divide both \H\ and \G : H\.

ID.3. Let H C G have the property that iJ H

if5'

= 1 for all elements

geG~H.

(a) Show that NG(K) C # for all subgroups X with 1 K CH.

(b) Show that H is a Hall subgroup of G. (Recall that this means that

|if | and \G : H\ are coprime.)

Note. A subgroup H G that satisfies the hypothesis of this problem is

usually referred to as a Frobenius complement in G. In Chapter 6, we

give a different definition of a "Frobenius complement", which, in fact, is

equivalent to this one. It is easy to see (as we will show) that a Frobenius

complement in the sense of Chapter 6 satisfies the hypothesis of this problem.