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 ?
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 $B$.
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.