# 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.

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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.)

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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 $\langle a \rangle \cap \langle b \rangle = \{1\}$.

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Nice. Thank you. –  Peter May 25 '12 at 14:52
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)$. – user641 May 25 '12 at 19:16 For many examples, take$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).

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