Sylow number equals index of Sylow normalizer

Statement
Let $$G$$ be a finite group, $$p$$ be a prime number, and $$P$$ a $$p$$-fact about::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

 * 1) uses::Sylow implies order-conjugate: Any two $$p$$-Sylow subgroups are conjugate.
 * 2) uses::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).