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A group generated by all the root subgroups above ideal of $\mathbb Z[\alpha]$ is of finite index in $G(\mathbb Z[\alpha])$
Let $G$ be a simply connected complex Chevalley group and let $T$ be some maximal torus in $G$ with $\dim T\geq 2$.
From "A note on generators for arithmetic subgroups of algebraic groups" by ...
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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 ...
- 3,399
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Finite index subgroup of $\mathrm{GL}_n(\Bbb C)$ and Chevalley groups
I'm trying to show that if $G$ is a Chevalley group, then every finite index subgroup of $G(\Bbb Z)$ is Zariski dense in $G(\Bbb C)$. ($G(\Bbb Z)$ is the Chevalley group over $\Bbb Z$ and similarly ...
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