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I have been working on the following lemma for a few days. Any hints would be appreciated.

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.

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Instead of taking an arbitrary $x\in Q\setminus P$, note that $N_Q(P)$ must properly contain $P$ (since $Q$ is a $p$-group, and you are assuming that $P\neq Q$). So pick $x\in N_Q(P)\setminus P$. It cannot lie in $H$, as $\langle P,x\rangle$ is a subgroup of $Q$.

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It is well-known that for an arbitrary $p$-subgroup $S$ of a group $G$,

$[G:S]\equiv [N_G(S):S]$ mod $p$ $(\ast)$

(for a proof consider the left multiplication action of $G$ on set of the left cosets of $S$).
Hence if $P \in Syl_p(H)$ and $N_G(P) \subseteq H$ then, since $N_H(P) = N_G(P)$ in this case, we can apply $(\ast)$ to $P$ as subgroup of $G$ and as subgroup of $H$ yielding $[G:P]\equiv [N_G(P):P] \equiv [N_H(P):P] \equiv [H:P]$ mod $p$. Since $P$ is a Sylow $p$-subgroup of $H$, $p$ does not divide $[H:P]$ and we conclude by the previous equivalence that $[G:P]$ has no factor $p$. This means $P$ is a Sylow $p$-subgroup of $G$.

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