TITLE: When does the Commensurator of a subgroup of a group $G$ not equal $G$?
QUESTION [5 upvotes]: Let $H\leq G$ be two groups. I'm interested in the Commensurator $$\mathrm{comm}_G(H)=\{g\in G : gHg^{-1} \cap H \text{ has finite index in both}\}.$$
Obviously, $\mathrm{comm}_G(H)\leq G$. I read on wiki, that the equality holds for any compact open group $G$. But does anyone have an easy example, where $1\ne\mathrm{comm}_G(H)\neq G$? Is this possible, since we have normal subgroup of finite index in both $H,gHg^{-1}$, if one of them already has finite index in $G$, namely $\bigcap\limits_{g\in G} gHg^{-1}$. This implies that $H,gHg^{-1}$ are both of finite index in $G$. So how do we get $1\neq\mathrm{comm}_G(H)\neq G$ for a groups $H\leq G$?
Thanks for help.

REPLY [4 votes]: For many examples, take $H$ to be an infinite subgroup of $G$ which is malnormal, meaning that if $g \in G \setminus H$ then $H \cap gHg^{-1}$ is trivial. Or one can take $H$ to be infinite and virtually malnormal, meaning that the intersection is finite.
For examples of malnormal subgroups: If $G$ is a free group and $H$ is a proper, nontrivial free factor, then by Grushko's theorem, $H$ is malnormal. 
For more examples of virtually malnormal subgroups: If $G$ is a word hyperbolic group and $H$ is a quasiconvex group then $H$ is malnormal. Specifically: take $G$ to be the fundamental group of a closed surface of genus $\ge 2$, and take $H$ to be any finitely generated, infinite, infinite index subgroup (such an $H$ must be a free group).