Let $G$ be a group (for now discrete). A subgroup $H$ of $G$ is called a commensurated subgroup of $G$, if $H\cap xHx^{-1}$ is a finite index subgroup of $H$ for all $x\in G$. These subgroups are also called Hecke subgroups or almost normal subgroups. My question is:

Is there any non-elementary, closed, discrete and commensurated subgroup $H$ of a non-discrete locally compact group $G$?

Elementary examples are finite subgroups and normal subgroups.

Edit: Regarding Jean Raimbault's comment, we consider all closed discrete nearly normal subgroups of $G$ as elementary examples too. (nearly normal means it is commensurable to some normal subgroup).

I just noticed that another class of very elementary examples can be considered as follows: Consider $G=\Gamma\times \Delta$, where $\Gamma$ is discrete and $\Delta$ is an arbitrary non-discrete locally compact group and $H$ is a commensurated subgroup of $\Gamma$.

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    $\begingroup$ We exclude normal subgroups $\endgroup$ – user23860 Sep 30 '13 at 9:26
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    $\begingroup$ Do you want $H\cap xHx^{-1}$ to have finite index in $H$, rather than $G$? $\endgroup$ – Amritanshu Prasad Sep 30 '13 at 10:54
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    $\begingroup$ You should maybe add finite--index subgroups of normal subgroups to your list of "elementary examples" $\endgroup$ – Jean Raimbault Sep 30 '13 at 11:35
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    $\begingroup$ non-elementary should be that the subgroup is not commensurate to a normal subgroup. Interesting examples are for instance if you take $\Gamma$ with a dense homomorphims into $SL_n(\mathbf{Q}_p)$ and consider the inverse image of $SL_n(\mathbf{Z}_p)$. Example: $SL_n(\mathbf{Z})\subset SL_n(\mathbf{Z}[1/p])$. $\endgroup$ – YCor Sep 30 '13 at 14:22
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    $\begingroup$ @Vahid: see arxiv.org/abs/0911.1966 $\endgroup$ – YCor Sep 30 '13 at 19:08

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