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 $N \lhd G$, we see that:
 * $\gen {P, N} = P N$

from Subset Product with Normal Subgroup as Generator.

Since $P \cap N \le P$, it follows that:
 * $\order {P \cap N} = p^k$

where $k > 0$.

By Order of Subgroup Product:
 * $\order {P N} \order {P \cap N} = \\order P \order N$

Hence from Lagrange's Theorem:
 * $\index N {P \cap N} = \index {P N} P$

By Tower Law for Subgroups:
 * $\index G P = \index G {P N} \index {P N} P$

Thus:
 * $\index {P N} P \divides \index G P$

where $\divides$ denotes divisibility.

Thus:
 * $p \nmid \index {P N} P$

so:
 * $p \nmid \index N {P \cap N}$

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