Problems ID 27
What is much more difficult is the theorem of Frobenius, which asserts that
if H is a Frobenius complement in G in the sense defined here, then it is,
in fact, a Frobenius complement in the sense of Chapter 6. Unfortunately,
all known proofs of Frobenius' theorem require character theory, and so we
cannot present a proof in this book.
1D.4. Let G = NH, where 1 N G and N H H = 1. Show that H is a
Frobenius complement in G (as defined above) if and only if CN(H) 1 for
all nonidentity elements h G H.
Hint. If x and
lie in if, where neJV , observe that
G N.
ID.5. Let H G, and suppose that N G ( P ) C H for all p-subgroups P C
iJ, for all primes p. Show that H is a Frobenius complement in G.
Hint. Observe that the hypothesis is satisfied by H D
for g G G. If this
intersection is nontrivial, consider a nontrivial Sylow subgroup Q of H fi
and show that Q and
are conjugate in H.
ID.6. Show that a subgroup of a nilpotent group is maximal if and only if
it has prime index.
ID.7. For any finite group G, the Frattini subgroup ^(G) is the inter-
section of all maximal subgroups. Show that $(G) is exactly the set of
"useless" elements of G, by which we mean the elements g G G such that if
{X U {g}) = G for some subset X of G, then (X) = G.
ID.8. A finite p-group is elementary abelian if it is abelian and every
nonidentity element has order p. If G is nilpotent, show that G/Q{G) is
abelian, and that if G is a p-group, then G/Q(G) is elementary abelian.
Note. It is not hard to see that if P is a p-group, then &(P) is the unique
normal subgroup of P minimal with the property that the factor group is
elementary abelian.
ID.9. If P is a noncyclic p-group, show that \P : &(P)\
and deduce
that a group of order
must be either cyclic or elementary abelian.
ID.10. Let A be maximal among the abelian normal subgroups of a p-group
P. Show that A = CP{A), and deduce that \P : A\ divides (\A\ - 1)!.
Hint. Let G - CP(A). If G A, apply Lemma 1.23.
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