## A sufficient condition for Sylow intersections of bounded index

Prove that if $n_p \not\equiv 1$ mod $p^k$, then there are distinct Sylow $p$-subgroups $P$ and $Q$ in $G$ such that $P \cap Q$ has index at most $p^{k-1}$ in both $P$ and $Q$.

(We will follow the strategy of this previous exercise.)

Let $P \leq G$ be a Sylow $p$-subgroup. Now $P$ acts on $\mathsf{Syl}_p(G)$ by conjugation; note that if $Q \in \mathsf{Syl}_p(G)$ is distinct from $P$, then using the Orbit-Stabilizer theorem and Lemma 4.19, we have $|P \cdot Q| = [P : N_P(Q)]$ $= [P : P \cap N_G(Q)]$ $= [P : Q \cap P]$. Since the orbit containing $P$ itself has order 1, then $n_p(G) = 1 + \sum [P : Q_i \cap P]$, where $Q_i$ ranges over a set of orbit representatives. If every $[P : Q_i \cap P]$ is divisible by $p^k$, then $n_p(G) \equiv 1$ mod $p^k$, a contradiction. Thus some $[P : Q_i \cap P]$ is a power of $p$ and is at most $p^{k-1}$.

We can see then that $Q_i \cap P$ has index at most $p^{k-1}$ in both $P$ and $Q_i$.