Index of p-subgroup divisible by p implies normalizer index divisible by p 
Show that if $H$ is a $p$-subgroup of $G$ with $p$ dividing $[G:H]$, then $[N_G(H):H]$ is divisible by $p$.

By considering the action of $H$ on $G/H$. I've considered the $\varphi:H\to S_{G/H}$ and tried a lot of things, I just fail to see how the relevance of the normalizer will solve this problem.
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Consider the action of $H$ by multiplication on the right on the set of cosets
$$
G/H = \Set{ H x : x \in G}.
$$
So $h \in H$ sends the coset $H x$ to the coset $H x h$.
Since $H$ is a (finite) $p$-group, of order $p^{n}$, say, the orbits will be of size a divisor of $p^{n}$. Let $s_{i}$ be the number of orbits of size $p^{i}$, then
$$
\Size{G : H} = \Size{G/H} = \sum_{i=0}^{n} s_{i} p^{i}. 
$$
Since $\Size{G : H}$ is divisible by $p$, we obtain that $s_{0}$ is divisible by $p$.
Now you know that there is at least one orbit of size $1$, namely the orbit of the coset $H = H 1$. So $s_{0} > 0$, and since it is divisible by $p$, $s_{0} \ge p$.
Now for all $H x$ such that $H x$ has length $1$ we have $H x H = H x$.
Now show that $x \in N_{G}(H)$. Conversely, show that if $x \in N_{G}(H)$ we have that $H x$ has an orbit of length $1$.
A: It's sufficient to assume Z(G) is a subgroup of H and note that Z(G) is non-trivial 
Now u can use induction on o(G) and get a simple.proof for finite groups by considering  G/Z(G) and H/Z(G)
Also u can consider X is the set of all conjugates of H. Then G acts in X by conjugation and hence Card(X)= [G:N(H)] now if u assume N(H)=H then H also acts on X and H is an orbit of size 1 there are at least p-1 orbits of size 1 . Thus there exists some conjugate say gHg^-1 not equal to H which is invariant under conjugacy by each element of H thus u get g belongs to N(H) and hence gHg^-1 = H a contradiction
