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## Question related to the abelianization of simplectic groups

Let $H \subset \mathrm{Sp}(\mathbb{Z})$ be a subgroup of the simplectic group of (square of even dimension) matrices with integer entries, and let $H^{(\ell)}$ denote its pro-$\ell$ completion for some prime $\ell$. Then does $H$ being Zariski dense in $\mathrm{Sp}(\mathbb{Z})$ imply that the abelianization $(H^{(\ell)})^{ab}$ is finite?

Or does the converse hold (i.e. $(H^{\ell})^{ab}$ is finite for some $\ell$ implies $H < \mathrm{Sp}(\mathbb{Z})$ is Zariski dense)?

I have no idea how to go about solving this, and half expect that there are easy counterexamples that I'm not thinking of.

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The converse seems to be disproved by considering e.g. any finite subgroup of Sp$_n(\mathbb{Z})$. – Tom Church Nov 10 2011 at 21:29
The Tits alternative gives lots of free subgroups which are Zariski dense, and their abelianizations are long way from being finite (see Theorem 3 of Tits' paper). For the converse, Tom Church gave examples of finite groups, but there are infinite examples as well, for example, $\mathrm{Sp}_{2n}(\mathbf{Z})$ naturally contains a copy of $\mathrm{SL}_n(\mathbf{Z})$, and the latter has finite abelianization for all $n \ge 3$. – Lavender Honey Nov 10 2011 at 23:44
@Jeff: Could you please give some motivation for the question? – Alain Valette Nov 11 2011 at 11:09