# Sylow Theorem Question ($n_p \equiv 1 \pmod{p^a}$)

I am stuck at the following Sylow Theorem question:

Let $p$ be a prime divisor of $G$, and suppose that whenever $P_1$ and $P_2$ are two distinct Sylow $p$-subgroups of $G$, $P_1\cap P_2$ is a subgroup of $P_1$ of index at least $p^a$. Show that the number $n_p$ of Sylow $p$-subgroups of $G$ satisfies $n_p\equiv 1 \pmod {p^a}$.

I have reached a dead end and can't think of any ways to proceed. I managed to estimate the size of the group $G$:

$|G|=[G:Q_1][Q_1:Q_1\cap Q_2]|Q_1\cap Q_2|\geq [G:Q_1]q^a|Q_1\cap Q_2|\geq q^a$

Let $\Omega$ be the set of Sylow $p$-subgroups of $G$, and consider the action of $P$ by conjugation on $\Omega$ for some fixed $P \in \Omega$.
Then, for $Q \in \Omega$, the size of the orbit of $Q$ is the index $|P:N_P(Q)|$. But no element $g$ of $P$ outside of $Q$ can normalise $Q$, since otherwise $\langle Q,g \rangle$ would be a larger $p$-subgroup of $G$ than $Q$, so in fact $N_P(Q) = P \cap Q$.
So there is one orbit $\{P\}$ of length $1$, and the remaining orbits, for $Q \ne P$, have size $|P|/|P \cap Q|$, which is divisible by $p^a$. So $|\Omega| \equiv 1 \bmod p^a$.
• I explained that in the answer: "since otherwise $\langle Q, g \rangle$ would be a larger $p$-subgroup of $G$ than $Q$". What is it exactly that you do not understand? – Derek Holt Sep 22 '15 at 20:46