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Let $H \leq \mathrm{SL}_3(\mathbb{Z})$ be a finitely generated subgroup which is not Zariski dense, and let $g \in \mathrm{SL}_3(\mathbb{Z})$. Must there be some $a \in \mathrm{SL}_3(\mathbb{Z})$ such that $\langle H,aga^{-1}\rangle$ is not Zariski dense?

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    $\begingroup$ No. If $H$ is the Heisenberg group over $\mathbf{Z}$ (the upper unipotent matrices with integral entries) and $g\in\mathrm{SL}_3(\mathbf{Z})$ is diagonalizable over $\mathbf{R}$ and not $\mathbf{Q}$, it is easy to show that for every $a\in\mathrm{SL}_3(\mathbf{Q})$, the group $\langle H,aga^{-1}\rangle$ is Zariski-dense. Hint: classify $\mathbf{Q}$-defined Zariski closed subgroups containing the upper unipotent subgroup. $\endgroup$
    – YCor
    Jan 17, 2015 at 15:44
  • $\begingroup$ @YCor What if we replace "Zariski dense" by "profinite closure of infinite index"? Is the profinite closure of your example of finite index? $\endgroup$
    – Pablo
    Jan 17, 2015 at 15:49
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    $\begingroup$ In $SL_3(\mathbf{Z})$, Zariski dense is equivalent to having profinite closure of finite index, as I told you in a comment yesterday. $\endgroup$
    – YCor
    Jan 17, 2015 at 16:47
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    $\begingroup$ @Pablo I think so, but it's harder. Isn't it due to Hee Oh? I'm not sure of the reference but I learnt it from Misha Kapovich on MO. $\endgroup$
    – YCor
    Jan 18, 2015 at 8:43
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    $\begingroup$ It is a result of Jacques Tits (an old paper in Comptes Rendus). Hee Oh's result is about any Zariski dense discrete subgroup which intersects the upper triangular unipotents in a finite index subgroup. $\endgroup$ Jan 18, 2015 at 13:35

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