ID 25
Note that in statement (3), a "maximal subgroup" is maximal among
proper subgroups. But in most other situations, the word "maximal" does
not imply proper. If a group G happens to be nilpotent, for example, then
the whole group is a maximal nilpotent subgroup of G.
To help with the proof that (4) implies (5), we establish the following.
1.27. Lemma. Let X he a collection of finite normal subgroups of a group G,
and assume that the orders of the members of X are pairwise coprime. Then
the product H — H X of the members of X is direct. Also, \H\ = H \X\.
xex
Proof. Certainly \H\ 111^1 Also, by Lagrange's theorem, \X\ divides
\H\, for every member X of X, and since the orders of the members of X
are pairwise coprime, it follows that H \X\ divides \H\. We conclude that
\H\ = Y[ \X\, as wanted.
Now to see that U X is direct, it suffices to show that
x
n
]J{Y e x

Y
+
X}
=
l
for every member X E X. This follows since by the previous paragraph, the
order of H F for Y ^ X is equal to H \Y\, and this is coprime to \X\. •
Proof of Theorem 1.26. We saw that (1) implies (2) in Theorem 1.22.
That (2) implies (3) is clear, since if M G is a maximal subgroup, then
Ncf(M) M, and so we must have N G ( M ) = G.
Now assume (3), and let P G Sylp(G) for some prime p. If N G ( P )
is proper in G, it is contained in some maximal subgroup M, and we have
M G. Since P G Sylp(M), it follows by Lemma 1.13, the Prattini argument,
that G =
NQ(P)M
C M, and this is a contradiction. Thus P G, and this
proves (4).
That (4) implies (5) is immediate from Lemma 1.27. Now assume (5). It
is clear that (4) holds, and it follows that (4) also holds for every homomor
phic image of G. (See Problem IB.5.) Since we know that (4) implies (5),
we see that every homomorphic image of G is a direct product of pgroups
for various primes p. The center of a direct product, however, is the di
rect product of the centers of the factors, and since nontrivial pgroups have
nontrivial centers, it follows that every nonidentity homomorphic image of
G has a nontrivial center. We conclude by Lemma 1.20 that G is nilpotent,
thereby establishing (1). •
Recall that Op(G) is the unique largest normal psubgroup of G, by
which we mean that it contains every normal psubgroup. We define the
Fitting subgroup of G, denoted F(G), to be the product of the subgroups