Normalizer of Sylow p-Subgroup

Theorem
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.

Let $N_G \left({P}\right)$ be the normalizer of $P$.

Then any $p$-subgroup of $N_G \left({P}\right)$ is contained in $P$.

In particular, $P$ is the unique Sylow $p$-subgroup of $N_G \left({P}\right)$.

Proof
Let $Q$ be a $p$-subgroup of $N = N_G \left({P}\right)$.

Let $\left|{Q}\right| = p^m, \left|{P}\right| = p^n$.

By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
 * $P \triangleleft N_G \left({P}\right)$

thus by Subset Product with Normal Subgroup as Generator:
 * $\left \langle {P, Q} \right \rangle = P Q$

Thus by Order of Subgroup Product:
 * $P Q \le G: \left|{P Q}\right| = p^{n+m-s}$

where $\left|{P \cap Q}\right| = p^s$.

Since $n$ is the highest power of $p$ dividing $\left|{G}\right|$, this is possible only when $m \le s$.

Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.

Thus from Intersection with Subset is Subset‎:
 * $Q \subseteq P$

In particular, if $Q$ is a Sylow $p$-subgroup of $N_G \left({P}\right)$, then $Q = P$.