Number of $p$-groups of a given size in a Sylow subgroup

Question

Problem: Let $p$ be a fixed but arbitrary prime number and $a$ a non-negative integer. Let $P$ be a Sylow $p$-subgroup of $G$.

Then prove that the number of subgroups of $G$ of order $p^a$ which are not contained in $P$ is divisible by $p$.

I am able to do this if all the Sylow $p$-subgroups are disjoint by using the fact that they are all isomorphic. Someone please help.

Answer 1

Consider the action by conjugation of $P$ on the set of subgroups of order $p^a$ that are not contained in $P$. There are no fixed points under this action (because if one of these subgroups was normalized by $P$ then it would be contained in $P$) so all orbits have length divisible by $p$.

Answer 2

$P$ acts on $\Omega=\{\text{subgroups of }G\text{ of order } p^ a\}$ by conjugation. Let $\Omega_{0}$ be the set of fixed points. Let $Q \in \Omega_{0}$. Then $P$ fixes $Q$ by conjugation, that is $P \subseteq N_G(Q)$.

So $PQ$ is a subgroup, but since $P$ is Sylow thus maximal, we must have $PQ=P$, so $Q \subseteq P$.

We conclude that $\Omega_{0}=\{Q \unlhd P: |Q|=p^a \}$. But $P$ is a $p$-group, whence $|\Omega| \equiv |\Omega_{0}|$ (mod $p$).

$P$ also acts on $\Lambda =\{\text{subgroups of }P \text{ of order } p^ a\}$ by conjugation. Note that the set of fixed points under this action is also $\Omega_{0}$. Therefore $|\Lambda| \equiv |\Omega_{0}|$ (mod $p$) and we get $|\Omega| \equiv |\Lambda|$ (mod $p$) as desired, since $\Omega=\Omega_{0} \cup \{Q \leq G: |Q|=p^a, Q \nsubseteq P \}$, a disjoint union.