TITLE: Normal Subgroups in Finite p-groups
QUESTION [8 upvotes]: In finite $p$-groups, the number of subgroups of order $p^k$ is congruent to $1 \mod p$.
Is it true that the number of normal subgroups of order $p^k$ is congruent to $1 \mod p$?

REPLY [3 votes]: I am sorry in advance for posting the answer since I could not add comment.
So as in the proof of Arturo Magidin, $|S_0|$ is congruent to $|S|$ modulo $p$. But then how can we get that $|S_0|$ is congruent to $1$ modulo $p$? I tried to take $N\in S_0$ and let $N$ act on $S_0$ by conjugation, but then all orbits are of size $1$?
And in fact, if $G$ is a finite group, then the number of subgroups of order $p^k$ in a Sylow $p$-subgroup is congruent to the number of subgroups of order  $p^k$ in $G$.