TITLE: Subgroups of Sp(2g,Z) that map onto all Sp(2g,Z/m)
QUESTION [8 upvotes]: 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 $G= \mathrm{Sp}(2g,\mathbb{Z})$?

REPLY [12 votes]: 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.