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 $|\Omega| \equiv 1 \bmod p^a$.

  • $\begingroup$ Do you mean the size of the orbit of Q is the index..? $\endgroup$ – yoyostein Sep 19 '15 at 11:48
  • $\begingroup$ Yes, thanks. Corrected! $\endgroup$ – Derek Holt Sep 19 '15 at 13:23
  • $\begingroup$ Thanks, I have understood every part except the part where no element g of P outside Q can normalize Q. Why is that so? $\endgroup$ – yoyostein Sep 21 '15 at 7:41
  • $\begingroup$ 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? $\endgroup$ – Derek Holt Sep 22 '15 at 20:46
  • $\begingroup$ I understand why it would be larger. However, why would <Q,g> be a p-subgroup? $\endgroup$ – yoyostein Sep 23 '15 at 0:13

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