If
pk divides the order of
G, then the number of subgroups of
G of order
pk is congruent to 1 mod
p. (Proofof the Frobenius Theorem?)
The property of having order
pk is isomorphism invariant, so by the theorem that
T is a group property (which is true for a
p group
P whether or not
P is a subgroup of a larger group
G or as a subgroup of a
p Sylow of
G), we can assume
G is a
p group. Let
Gact on the set of all subgroups of order
pk by conjugation. Since
G is a
p group, the orbits of the action have size a power of
p. The fixed points of this action are normal groups of
G, so it suffices to prove that the number of normal subgroups of
G of order
pk is congruent to 1 mod
p. We proceed by induction on the order of
G. The base case, where
G is cyclic of order
p is trivial.
Let
n be the size of the set of pairs of the form (
A, N) where
A is a group of order
p contained in the center
Z of
G,(
Z is nontrivial since
G is a
p group), and
N is a normal subgroup of order
pk of
G containing
A. We then count
n in two different ways:
For the first count, fix
N. Then the number of pairs of the form (
A, N) is simply the number of groups of order
p in
Zn N. (Since
G is a
p group, any normal subgroup
H of
G has nontrivial intersection with
Z. This can be seen by having
G act on the non-identity elements of
H by conjugation, and noting that this action must have fixed points). Now since
Z n N is abelian, by the lemma(Let G be an abelian group; then the number of subgroups of order
p is congruent to 1 mod
p.) the number of subgroups of order
p of
Z n N is congruent to 1 mod
p. Thus,
n is equal to the number of normal subgroups
N of order
pk , modulo
p.
For the second count, fix
A of order
p in
Z. We want to count the number of normal subgroups
N of order
pk containing
A. Note that the homomorphism
G?G/A preserves normality, so that in
G/A we want the number of normal subgroups of order
pk-1. By the induction hypothesis, this is congruent to 1 mod
p. Also, since
Z is abelian, by the lemma(stated above in the first count),the number of
A of order
p in
Z is congruent to 1 mod
p. Thus
n = 1(
p).