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 ? 
 A: Let $A \ast B$ be any non-trivial free product, that is $A,B \neq \{1\}$. Then the normalizer of $A$ (resp. $B$) is $A$ (resp. $B$). However, there is no restriction on the groups $A$ or $B$.
A: 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. 
A: Perhaps a more or less simple (but not as painfully simple as D. Holt's counter example in his comment!) counter example is:
$$G=S_4\;,\;\;H\cong D_4=\text{one of the three Sylow $2$- subgroups of}\;\;G$$
If $\;D_4\lneq H_G(D_4)\;$ , then it must be $\;H_G(D_4)=G\;$ (why?), but this is impossible (why?) , and thus it has to be $\;H=G_H(D_4)\;$, and $\;D_4\;$ isn't abelian.
Remember: a finite group is nilpotent iff every proper subgroup fulfills the normalizer condition, which means it is properly contained in its normalizer. Thus, the idea was to look for a finite non-nilpotent group. $\;S_3\;$ doesn't make the cut since all its proper subgroups are abelian, so let us try with the next one, and etc.
A: 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 $8$.
A: A more structured example that shows a part of a general pattern is to take $H$ to be upper-triangular matrices in $G=GL_2(\mathbb F_q)$.
More generally, $H$ could be upper-triangular matrices in $G=GL_n(\mathbb F_q)$, or even any block-upper-triangular subgroup therein. And the field does not have to be finite.
That is, these $H$'s are the standard parabolic subgroups of those general linear groups.
A: Here is a hint: Consider $H=S_{3}$ and $G=S_{4}$. Then (why?) $N_{G}(H)=H$. The previous hint that I posted was completely wrong. Here is a better hint.
