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If $\Lambda \cap U$ is Zariski-dense in $U$, then $\Lambda$ contains $U(k\mathbb Z)$ for some $k ≥ 1$?
If $G$ is a $\mathbb Q$-defined subgroup of $\operatorname{GL}_n(\mathbb C)$, $\Lambda$ is a subgroup of $G(\mathbb Z)$, and $U$ is a unipotent subgroup of $G(\mathbb C)$ such that $\Lambda \cap U$ is ...
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