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.