abstract algebra question with p-subgroup

Question

Let $G$ be a finite group, $H<G$ a subgroup. Suppose that $H$ is a $p$-group, $p$ a prime, and $p\mid[G:H]$. Show that $p\mid[N_G(H):H]$ and $H<N_G(H)$. In particular, if $G$ is a $p$-group and $H<G$, then $H<N_G(H)$.

I think I will have to use the $3$rd Sylow Theorem, but do not know how to do it specifically. Can someone give an idea?

Answer 1

Consider $[G:N_G(H)][N_G(H):H]=[G:H]$. Then either $p$ divides $[G:N_G(H)]$ or $p$ divides $[N_G(H):H]$. First suppose the former case is true and derive a contradiction.

Edit: fixed mistake. sorry about that.

Answer 2

Hint: Set $\Omega=\{xHx^{-1}|x\in G\}$. You can see easily that $G$ acts on $\Omega$. You should try to follow the following steps:

1.Since $|\Omega|=[G:N_G(H)]$, we have two cases $|\Omega|>1$ and $|\Omega|=1$.

2.If $|\Omega|>1$, then put $\Delta=\Omega-\{H\}$ and so $H$ acts on $\Delta$ by conjugation.

3. There is $x\in G$ such that for every $h\in H$, $xhx^{-1}\in N_G(H)$ and so $h\in N_G(x^{-1}Hx)$.

4. Switch $x^{-1}Hx$ by $H$ in $4$.