Assume $\Gamma$ is a discrete subgroup of some $GL\_n$, and let $G$ be its Zariski closure. 
Let $H$ an algebraic cocompact normal subgroup of $G$. Do we have that $H\cap \Gamma$ is of finite index in $\Gamma$.
I'm not sure this is true (actually i think it should be false), i'm far from being an expert in this field, and i don't have any couterexample yet.


Many thanks



on the zariski closure and cocompactness of discrete groups
This is not true. There are irreducible (arithmetic) lattices $\Gamma=G({\mathbb Z})$ in $$SL \_2({\mathbb R})\times SU(2)=G({\mathbb R})$$ such that $\Gamma$ projects densely into $SU(2)$ and its projection to $SL\_2({\mathbb R})$ is a cocompact lattice. $G$ is an algebraic $\mathbb Q$-subgroup of $GL\_n$ and the Zariski closure of $\Gamma$ is not $H= SL\_2({\mathbb R})$ but it is $G({\mathbb R})$. $H$ is an algebraic co-compact normal subgroup of $G({\mathbb R})$ but its intersection with $\Gamma$ is only the trivial element. 


