Normalizer of Sylow p-Subgroup

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

Then any $p$-subgroup of $$N_G \left({P}\right)$$, the normalizer of $$P$$, 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 Largest Subgroup, $$P \triangleleft N_G \left({P}\right)$$, thus by Subgroup 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 $$Q \subseteq P$$.

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