# When does the Commensurator of a subgroup of a group $G$ not equal $G$?

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.

Firstly, $H \leq \operatorname{comm}_G(H)$ always, so if $H$ is a non-trivial subgroup, its commensurator is non-trivial. A natural question to ask, then, is how much bigger than $H$ its commensurator in $G$ is.

The basic situations of $H \leq G$ where $\operatorname{comm}_G(H)$ is much bigger than $G$, but smaller than $G$, is when $G$ is a Lie group and $H$ is an arithmetic lattice in $G$. E.g. if $H = \mathrm{SL}_n(\mathbb Z) \subset \mathrm{SL}_n(\mathbb R) = G$, then $\operatorname{comm}_G(H)$ is equal to $\mathrm{SL}_n(\mathbb Q)$, so it is much bigger than $H$, but much smaller than $G$.

Of course, as you more-or-less have observed, if $H$ has finite index in $G$, then $\operatorname{comm}_G(H) = G$. (One way to phrase this is to say that if $H$ has finite index in $G$, then $H$ and $G$ are commensurable subgroups of $G$, and so their commensurators in $G$ coincide.)

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).

I'm no expert on the topic of Lie groups, so no example springs to mind in this context. For abstract groups, however, we can construct an example as follows: Define $G$ as the finitely presented group $$G := \langle a,b,c ~|~ cac^{-1} = b \rangle$$ and let $H := \langle a \rangle \le G$. Then $c \not\in \text{comm}_G(H)$, since $cHc^{-1} = \langle b \rangle$ and $\langle a \rangle \cap \langle b \rangle = \{1\}$.

• But then $\text{comm}_G(H)=1$, right? Since $|G:H|=\infty$, right? So I'm searching for a subgroup where $1<\text{comm}_G(H)$<G$. – Peter May 25 '12 at 14:55 • We have$a\in\text{comm}_G(H)\$. – user641 May 25 '12 at 19:16