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

xn

lie in if, where neJV , observe that

x~1xn

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

H9

for g G G. If this

intersection is nontrivial, consider a nontrivial Sylow subgroup Q of H fi

H9

and show that Q and

Q9

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)\

p2,

and deduce

that a group of order

p2

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.