Conjugation action on Sylow subgroups

A group, or one of its Sylow subgroups, acts on the set of Sylow subgroups by conjugation.

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.

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.