A Lemma about Sylow Subgroups

Question

Can you help me prove a lemma, please? It would be very useful for several problems in Sylow theory. I tried a couple of ways to prove it but unsuccessfully.

Let $G$ be a group such that $|G|=p^nq$, where $p$ is prime, and $(p,q)=1$. Let $n_p$ be the number of Sylow $p$-subgroups. Prove that either $n_p\equiv 1\pmod {p^2}$ or there exist two Sylow $p$-subgroups $P_1$ and $P_2$ such that $|P_1\cap P_2|=p$.

Answer 1

Hint 1: Let a given Sylow subgroup act by conjugation on the set of all Sylow subgroups. What are the sizes of the orbits?

Hint 2: Here's a similar question. It's not quite exactly the same question, but it's close. See how you can use it with your situation.

Answer:

Using the second hint, if $p^a = \text{min}([P:P\cap Q] \ | \ P\neq Q\in\text{Syl}_p(G))$, then $n_p\equiv 1\bmod p^a$. Now $a=1$ if and only if $[P: P\cap Q]=p$ for some $Q\in\text{Syl}_p(G)$, and if $a>1$ then $p^2$ divides $p^a$ and so $n_p\equiv 1\bmod p^2$. This gives your desired result.