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

Consider the group $\mathrm{Sp}(2g,\mathbb{Z})$ of symplectic square matrices of order $2g$ with integral entries. Suppose furthermore that $g\geq 3$. 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 $G= \mathrm{Sp}(2g,\mathbb{Z})$?

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 $G= \mathrm{Sp}(2g,\mathbb{Z})$?