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.
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