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 ? 

In general, if $H$ is a subgroup of $G$ then $H \unlhd G$ iff $G=N_G(H)$. $H$ is abelian iff $H \subseteq C_G(H)$. Here $C_G(H)=\{g \in G : gh=hg \text { for all } h \in H\}$, the centralizer of $H$ in $G$. Note that $C_G(H) \subseteq N_G(H)$. So $H=C_G(H)$ would indeed imply $H$ being abelian.
I will now give you a general counterexample to your original OP. I will use a well-known fact about Sylow $p$-subgroups: if $P \in Syl_p(G)$ and $H=N_G(P)$ then $N_G(H)=H$. So pick a group $G$ with a non-abelian Sylow $p$-subgroup $P$, and let $H=N_G(P)$. Then $H$ cannot be abelian, since $P \subseteq H$ and $P$ was chosen to be non-abelian. A concrete example? See for example Timbuc's example. Or take one of the Sylow $2$-subgroups of $A_6$, being isomorphic to $D_4$, the dihedral group of order $8$.
