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I stumbled into the following problem. I apologize for being a bit naive.

For $g\geq 3$, consider the group $\mathrm{Sp}(2g,\mathbb{Z})$ of symplectic square matrices of order $2g$ with integral entries. For every integer $m\geq 2$ let $p_m:\mathrm{Sp}(2g,\mathbb{Z})\rightarrow\mathrm{Sp}(2g,\mathbb{Z}/m)$ be the natural projection.

Let $G$ be a finitely-generated subgroup of $\mathrm{Sp}(2g,\mathbb{Z})$ such that $p_m(G)=\mathrm{Sp}(2g,\mathbb{Z}/m)$ for all $m\geq 2$ (really "all $m$" and not just "all but finitely many $m$").

(1) Can I say that $G$ is the whole $\mathrm{Sp}(2g,\mathbb{Z})$?

If the answer to (1) is no, then:

(2) what are typical counterexamples?

(3) is there some further (non-tautological) hypothesis that would ensure that $G= \mathrm{Sp}(2g,\mathbb{Z})$?

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  • $\begingroup$ I might also be interested in the case $g=2$, but it's not my main concern. $\endgroup$
    – Gabriele Mondello
    Commented Sep 30, 2020 at 10:59
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    $\begingroup$ By Weisfeiler every Zariski-dense subgroup has surjective congruence projections for all but finitely many $m$. I'd be therefore inclined to believe that there exist many subgroups for which surjectivity holds for all $m\ge 1$. However I don't know examples. (Note: f.g. doesn't matter: if a countable subgroup has surjective projections, it has some finitely generated subgroup with surjective projection as well.) $\endgroup$
    – YCor
    Commented Sep 30, 2020 at 11:17
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    $\begingroup$ Yes, if $G$ is Zariski-dense, then $p_m$ is surjective for all primes $m$ except possibly finitely many. Also, Margulis-Soifer showed that there are maximal subgroups $H$ of infinite index in $\mathrm{Sp}(2g,\mathbb{Z})$, whose projection to $\mathrm{Sp}(2g,\mathbb{Z}/m)$ is thus surjective for every $m$. However, I don't see why such $H$ should be finitely generated. $\endgroup$
    – Gabriele Mondello
    Commented Sep 30, 2020 at 11:25
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    $\begingroup$ So, this completes the picture. No, $H$ is not finitely generated, but being Zariski-dense, it has a f.g. subgroup $K$ that is Zariski dense. So $K$ has all projections surjective but finitely many. So taking the subgroup of $H$ generated by $K$ and finitely elements, you get what you want. $\endgroup$
    – YCor
    Commented Sep 30, 2020 at 11:35
  • $\begingroup$ I don't immediately see how it leads to a conclusion. The problem is that it surjects onto all but finitely many primes, but I have to take care of powers of primes too. So this potentially requires infinitely many generators to be added. $\endgroup$
    – Gabriele Mondello
    Commented Sep 30, 2020 at 11:54

1 Answer 1

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The answer to your question as to whether $G=Sp_{2g}({\mathbb Z})$ is NO. $Sp_{2g}({\mathbb Z})$ contains a pro-finitely dense FREE subgroup (hence has infinite index) which is also finitely generated. In fact the number of generators may be taken to be four: https://arxiv.org/pdf/1205.1140.

The same conclusion holds for any arithmetic subgroup of $G({\mathbb Q})$ for a ${\mathbb Q}$-simple semi simple algebraic group defined over ${\mathbb Q}$ which satisfies the congruence subgroup property.

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