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 ? 

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.
