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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.

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    This is not true as stated: I think you want $N_G(P)$ in your containment? – 2017-01-08
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    Oh! sorry that was my mistake. I'll edit it. – 2017-01-08
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    OK, now try using the fact that any element that normalizes $H$, must conjugate $P$ to a Sylow subgroup of $H$... – 2017-01-08

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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 $M = N_{G}(H) = N_{G}(P) H = H$.

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    This is nice but, perhaps, a little too abstract for someone asking a question this elementary. Besides, your link to "Frattini Argument" only gives the usual, standard formulation, not the one with actions of groups on sets. – 2017-01-08
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    @DonAntonio, you are right on both count, will probably delete. – 2017-01-08
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    Please don't. I think this can be valuable for other people...say, for me. Reading this I remembered how nice stuff is actions of groups on sets. +1 – 2017-01-08
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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 [$a,b\in H\implies ab^{-1}\in H$]