Sylow's Third Theorem

Sylow’s Third Theorem. Let $G$ be a finite group with $|G| = p^{a}m$ where $p$ is prime and $p\nmid m$. Let $n_p$ denote the number of Sylow p-subgroups of $G$. Then:

1. $n_p \mid m$, and
2. $n_p \equiv 1 \pmod p$.

This theorem is a counting consequence of Sylow's second theorem together with the normalizer and the conjugation action on the set of Sylow $p$-subgroups.

Proof sketch. Fix a Sylow $p$-subgroup $P$. Conjugation gives a transitive action of $G$ on the set of Sylow $p$-subgroups, and the stabilizer of $P$ is $N_G(P)$, so $n_p=[G:N_G(P)]$, which divides $m$ because $P\subseteq N_G(P)$ contributes the entire $p^a$-part to $|N_G(P)|$. For the congruence, let $P$ act by conjugation on the set of Sylow $p$-subgroups; orbit sizes are $1$ or multiples of $p$, and $P$ fixes exactly the Sylow $p$-subgroups equal to itself, yielding $n_p \equiv 1 \pmod p$.