Let $G$ be a finite group and $P<G$ be a Sylow p-subgroup. Let $N_{G}(P)$ be the normalizer of $P$ in $G$. Let $H<G$ be a subgroup containing $N_{G}(P)$. Prove that $N_{G}(H)=H$.
I've been playing around with this for awhile. I've been considering group actions and the fact that $P \leq N_{G}(P) \leq H \leq N_{G}(H)$ but am getting nowhere.
Observe that we need only show that $N_G(H) \leq H$. Let $g \in N_G(H)$. Then $$ gPg^{-1} \subseteq gHg^{-1} = H. $$ Note that $gPg^{-1}$ is a subgroup of $H$. It is, in fact, a Sylow $p$-subgroup of $H$! Therefore by the Sylow Conjugacy Theorem, $gPg^{-1} = hPh^{-1}$, for some $h \in H$. Then $$ (g^{-1}h)P(g^{-1}h)^{-1} = P, $$ so $$g^{-1}h \in N_G(P) \leq H.$$ But that implies that $$g^{-1}h h^{-1} = g^{-1} \in H,$$ so $g \in H$, too.