Prove that the normalizer of the subgroup is the subgroup is itself 
Problem: $P$ is a $p$-Sylow subgroup of $G$. $P \subseteq N_G(P) \subseteq H \subseteq G$. $H$ is a subgroup of $G$. Prove that $N_G(H)=H$ , i.e., Normalizer of $H$ in $G$ is $H$ itself.

$H$ and $G$ will have the same Sylow groups for this given prime p.  I observed a few things like the number of p-Sylow groups not contained in $H$ is a multiple of p and so is the number of p-Sylow groups not contained in $N_H(P)$. I couldn't make much progress. Please help.
 A: Let $y\in N_G(H)\implies yHy^{-1}=H\implies yPy^{-1}\subset yHy^{-1}=H\implies P $  and $yPy^{-1}$ are both Sylow-$p$ subgroups of $H$,since any two Sylow $p-$ groups are conjugates hence $ P $  and $yPy^{-1}$  are conjugates in $H$.
Hence $P=x(yPy^{-1})x^{-1}$ for some $x\in H$
$\implies P=(xy)P(xy)^{-1}\implies xy\in N_G(P)\subset H$
Hence $x^{-1}(xy)=y\in H$ as a $H$ is a subgroup [$a,b\in H\implies ab^{-1}\in H$]
A: It's an instance of the so-called Frattini argument:

Let $M$ be a group acting on the set $\Omega$. Suppose the subgroup $H$ acts transitively on $\Omega$. Then for each $\alpha \in \Omega$ we have $M = M_{\alpha} H$.

The proof is immediate: for each $g \in M$ there is $h \in H$ such that $(\alpha^ {g})^{h} = \alpha$, so that $g h \in G_{\alpha}$.
In your case, $\Omega$ is the set of $p$-Sylow subgroups of $M = N_{G}(H)$. Since $P \le H$, and $H \trianglelefteq N_{G}(H)$, by the Sylow theorems $\Omega \subseteq H$ and $H$ acts transitively on $\Omega$.
Now $M_{P} = N_{G}(P)$, so that $M = N_{G}(H) = N_{G}(P) H = H$.
