$Q \cap N_G(P)=Q\cap P$ where $P,Q$ are both Sylow $p-$groups

Question

We just need to prove that $Q \cap N_G(P) \subset P$. This problem becomes to show $gPg^{-1}=P, g \notin P, g\in Q \implies g \in P$. So how exactly $g$ being a member of another $p$-Sylow group helps him to be a member of $P$?

Answer 1

Apply Sylow Theory in $N_G(P)$: $Q \cap N_G(P)$ is a $p$-group in $N_G(P)$, and hence must be contained in some Sylow $p$-subgroup of $N_G(P)$. Since $P \unlhd N_G(P)$, $P$ is the unique one and so $Q \cap N_G(P) \subseteq P$. (Note that $Q$ does not even have to be a Sylow $p$-group here, just being a $p$-subgroup suffices.)

Answer 2

If $x\in Q\cap N_G(P)$ then this means, $x$ is a $p$-element (of $Q$) and it normalizes a maximal $p$-subgroup (namely $P$); this forces that $x$ should be in $P$ (otherwise, $\langle P,x\rangle$ is a subgroup where $x$ is $p$-element and normalizes $P$; so this group is also a $p$-group, and contains $P$ properly.)

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We do not need that $Q$ is Sylow; we need it to be $p$-group.

Note that $x\in Q$ i.e. $x$ is $p$-element i.e. $\langle x\rangle$ is a $p$-group.

Then $x\in Q\cap N_G(P)$ means $\langle x\rangle$ normalizes $P$, hence their product $P\langle x\rangle$ is a subgroup of $G$; by product formula, its order is power of $p$, and so it is a $p$-group containing $P$. Due to maximality of $P$ as Sylow-$p$, this product should be $P$. Q.E.D.