I have been working on the following lemma for a few days. Any hints would be appreciated.
Lemma
Let $G$ be a finite group, $H \le G$ and $P \in Syl_{p}(H)$. If $N_{G}(P) \subseteq H$ then $P$ is a Sylow $p$-subgroup of $G$.
Partial Proof (Contrapostive).
First assume that $P$ is not a Sylow $p$-subgroup of $G$. Therefore $P$ is properly contained in some Sylow $p$-subgroup of $G$, say $Q$. Next let $x \in Q \setminus P$. Therefore $P^{x} \subseteq Q^{x}=Q.$
Now I would like to show that infact $P^{x} = P$ so that $x \in N_{G}(P)$. Therefore $N_{G}(P) \not \subseteq H$. However I am not sure if this is true. One interesting fact is that Q contains only one Sylow $p$-subgroup of $H$. So if $P^{x} \subset H$ then the proof is complete.
I have the feeling that I may be missing something very trivial. Maybe a contradiction proof would work better.