Sylow number equals index of Sylow normalizer

Statement

Let $G$ be a finite group, $p$ be a prime number, and $P$ a $p$ -Sylow subgroup of $G$ . Then, if $n_{p}$ denotes the number of $p$ -Sylow subgroups, we have:

$[G : N_{G} (P)] = n_{p}$

Facts used

Sylow implies order-conjugate: Any two $p$ -Sylow subgroups are conjugate.
Group acts on set of subgroups by conjugation: Under this action, the isotropy subgroup for any subgroup is its normalizer, and the index of the normalizer equals the number of conjugate subgroups to it.

Proof

The proof follows directly by piecing together facts (1) and (2).