If the normalizer of a subgroup in a group is equal to the subgroup then is the subgroup abelian? If $H$ is a proper subgroup of $G$ such that $H=N_G(H)$ ( the normalizer of $H$ in $G$ ) , then is it true that $H$ is abelian ? 

Since no one posted a simple counterexample yet here it goes:
Consider $G=S_4$ and $H=S_3$ considered as a subgroup of $G$ (i.e. as the stabilizer of one of the $4$ points $G$ naturally acts on). Then there are $4$ conjugates of $H$ (the stabilizers of the $4$ points are all conjugate) but also $4=[G:H]$ hence $N_G(H)=H$ and $H$ is obviously not abelian. 
