If p k divides the order of G, then the number of subgroups of G of order p k is congruent to 1 mod p. (Proofof the Frobenius Theorem?) The property of having order p k is isomorphism invariant, so by...

Ifp^k

divides the order of
G,
then the number of subgroups of
G
of orderp^k

is congruent to 1 mod
p.
(Proofof the Frobenius Theorem?)
The property of having order
p^k

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
p^k

by conjugation. Since
G
is ap
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
p^k
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
p^k

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
p^k

, 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
p^k

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
p^k-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).