Intersection of Normal Subgroup with Sylow P-Subgroup

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

Let $$N$$ be a normal subgroup of $$G$$.

Then $$P \cap N$$ is a Sylow $p$-subgroup of $$N$$.

Proof
Since $$N \triangleleft G$$, we see that $$\left \langle {P, N} \right \rangle = P N$$ from Subgroup Product with Normal Subgroup as Generator.

Since $$P \cap N \le P$$, $$\left|{P \cap N}\right| = p^k$$ where $$k > 0$$.

By Order of Subgroup Product, $$\left|{P N}\right| \left|{P \cap N}\right| = \left|{P}\right| \left|{N}\right|$$.

Hence from Lagrange's Theorem, $$\left[{N : P \cap N}\right] = \left[{P N : P}\right]$$.

By Index of Subgroup of Subgroup, $$\left[{G : P}\right] = \left[{G : P N}\right] \left[{P N : P}\right]$$.

Thus $$\left[{P N : P}\right] \backslash \left[{G : P}\right]$$.

Thus $$p \nmid \left[{P N : P}\right]$$ so $$p \nmid \left[{N : P \cap N}\right]$$.

Thus $$P \cap N$$ is a Sylow $p$-subgroup of $$N$$.