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Let $k$ be a totally real number field and let $\mathcal{O}_k$ denote its ring of integers. If $H$ is a subgroup of $\text{GL}(n, \mathbb{R})$ let denote with $H(k)$ and $H(\mathcal{O}_k)$ the intersections $H\cap \text{GL}(n, k)$ and $H\cap \text{GL}(n, \mathcal{O}_k)$.

Let $H$ be a semisimple algebraic group defined over $k$. I need to show that $H(k)$ commensurates $H(\mathcal{O}_k)$, i.e. that for every $\gamma\in H(k)$ the subgroup $\gamma H(\mathcal{O}_k)\gamma^{-1}$ is commensurable to $H(\mathcal{O}_k)$.

I do not have any idea of how to proceed. The only observation I have done is that since the commensurator of a subgroup is group, and the constant matrices commensurate $H(\mathcal{O}_k)$, I can suppose that $\gamma$ has entries in $\mathcal{O}_k$, but in this way I probably lose the information of $\gamma$ being an element of $H$.

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    $\begingroup$ Also posted in [math.stackexchange]( math.stackexchange.com/questions/3365855/…) $\endgroup$ – Arturo Magidin Sep 22 '19 at 18:41
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    $\begingroup$ Please do not post the same question to both sites at the same time. Pick one site (the one you think it most appropriate), polst, and wait for a response. Wait longer than 40 minutes on a weekend. If you fail to secure responses in one site and you think you might get them in the other, then indicate in your post that it is a cross-post, to avoid possible duplication of effort. $\endgroup$ – Arturo Magidin Sep 22 '19 at 18:42
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    $\begingroup$ I deleted the one in stack exchange, sorry :) $\endgroup$ – Diego95 Sep 22 '19 at 19:05
  • $\begingroup$ For the question to make sense, you need to fix an embedding of $k$ into $\mathbf{R}$. Also, are you assuming that $H$ is a $k$-subgroup? Otherwise I'm surprised by the generality. $\endgroup$ – YCor Sep 22 '19 at 20:18
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    $\begingroup$ No, I mean that $H$ is defined by algebraic equations with coefficients in $k$. In your generality, $H$ could be a dense subgroup, far from a Zariski-closed one. "$H$ is semisimple": this doesn't make sense for an arbitrary subgroup, but for an algebraic subgroup, or a closed one. $\endgroup$ – YCor Sep 23 '19 at 9:26
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This is well-known, and you don't need semisimplicity for this to hold. You can prove it by considering congruence subgroups of $H(\mathcal O_k)$ as follows. Let $\mathfrak n$ be an ideal of $\mathcal O_k$, assume that $H$ is embedded in $\mathrm{GL}_d$, so we have a well-defined reduction modulo $\mathfrak n$ from $H(\mathcal O_k)$ to $\mathrm{GL}_d(\mathcal O_k/\mathfrak n)$ and define : $$ \Gamma(\mathfrak n) = \{ g \in H(\mathcal O_k), g = \mathrm{Id} \pmod {\mathfrak n}\}. $$ This is a finite-index subgroup of $H(\mathcal O_k)$. Then if $\gamma = (a_{ij}) \in H(k)$, and we write $\gamma^{-1} = a^{ij}$ there exists $\mathfrak n$ such that $a_{ij}\mathfrak n, a^{ij}\mathfrak n \subset \mathcal O_k$ for all $i, j$ and it follows that $\gamma \Gamma(\mathfrak n^2)\gamma^{-1} \subset H(\mathcal O_k)$. In particular $H(\mathcal O_k) \cap \gamma H(\mathcal O_k) \gamma^{-1}$ contains $\Gamma(\mathfrak n^2)$ and so has finite index in $H(\mathcal O_k)$.

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