 Strengthened Sylow congruence from intersections

For a finite group and a Sylow subgroup $P$, conjugation by $P$ on all Sylow subgroups has stabilizer $P\cap Q$ at $Q$: the normal Sylow subgroup $Q$ of $N_G(Q)$ contains every $p$-subgroup of that normalizer. Thus the orbit length is $[P:P\cap Q]$. If every $Q\ne P$ has this index at least $p^a$, each such orbit length is divisible by $p^a$, whereas $P$ is a singleton orbit. Summing orbit lengths proves the displayed stronger integer congruence.