On solvable group and normalizer

Question

" Let $G$ be finite group, $H$ is a subgroup of $G$, and $P$ is a Sylow-p Subgroup of $G$. If $N_G(P) \leq H$, show that $N_G(H)=H.$ "

This problem appears in Martin Isaacs book under the chapter about solvable group, and I have no idea how to solve it using the concept of solvability. Anyone can give a hint?

Thanks in advance.

Answer 1

The inclusion $H \subseteq N_G(H)$ is trivial.

Now, you have that:

1. $P$ is a Sylow p-subgroup of $H$ (since $P \subseteq N_G(P) \subseteq H$)
2. $H$ is normal in $N_G(H)$ by definition of the normalizer

Hence by Frattini's argument $$N_G(H)=N_G(P)H \subseteq H$$