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) uses::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.