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Let $G \leq SL_2(\mathbb{R})$ be a lattice, let $H \leq G$ be a finitely generated subgroup of infinite index, and let $n \in \mathbb{N}$. Must there be some $H \leq U \leq G$ such that $n \leq [G : U] < \infty$ ?

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    $\begingroup$ What you ask in the question is weaker than LERF, namely whether the profinite closure of every f.g. subgroup of infinite index has infinite index. The answer is yes, surface groups are LERF by a theorem of Peter Scott. See e.g. arxiv.org/abs/1204.5135 $\endgroup$
    – YCor
    Jan 16, 2015 at 16:16
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    $\begingroup$ @YCor: It seems to me that you just provided a perfectly fine answer. Why did you choose to put it as a comment instead of an answer? $\endgroup$ Jan 16, 2015 at 16:31
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    $\begingroup$ @Pablo are you worried about torsion, or about $SL$ vs $PSL?$ $\endgroup$
    – Igor Rivin
    Jan 16, 2015 at 16:52
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    $\begingroup$ Yes it dramatically changes. By Weisfeiler, any Zariski-dense subgroup of $SL_3(\mathbf{Z})$ is has finite index closure for the pro-congruence topology, which coincides with the profinite topology by Bass-Milnor-Serre. By Margulis-Soifer, there are Zariski-dense subgroups of infinite index. $\endgroup$
    – YCor
    Jan 16, 2015 at 19:28
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    $\begingroup$ @Pablo it follows from the 1-generator version of the Tits alternative, due to Burnside. But you don't need this, since Margulis-Soifer directly construct a finitely generated free Zariski dense subgroup. $\endgroup$
    – YCor
    Jan 16, 2015 at 20:02

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