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.

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
