Intersection of p-subgroup with Sylow subgroup equals intersection with normalizer

Statement

Suppose $G$ is a finite group and $p$ is a prime number. Suppose $P$ is a $p$-subgroup of $G$ and $Q$ is a $p$-Sylow subgroup of $G$. Then:

$P\cap Q=P\cap N_{G}(Q)$.

In other words, $P\cap Q=N_{P}(Q)$.

Here, $N_{G}(Q)$ is the normalizer of $Q$ in $G$.

Facts used

1. Product formula

Proof

Given: A $p$-subgroup $P$ and a $p$-Sylow subgroup $Q$ of a finite group $G$.

To prove: $P\cap Q=P\cap N_{G}(Q)$.

Proof: Suppose $H=P\cap N_{G}(Q)$. Then, $H$ normalizes $Q$, so $QH$ is a subgroup. By fact (1), we have:

$|QH|={\frac {|Q||H|}{|Q\cap H|}}$.

If $|H|>|Q\cap H|$, then the largest power of $p$ dividing $|QH|$ is bigger than the largest power of $p$ dividing $|Q|$, a contradiction to $Q$ being $p$-Sylow.