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 for $G(\Bbb C)$)
  
But I'm struggling to understand some basic stuff,
It seems to me that there can't be finite index subgroup in $G(\Bbb C)$ how can I show that?  

also is there a good source that covers Chevalley groups over $\Bbb Z$?



Finite index subgroup of $\mathrm{GL}_n(\Bbb C)$ and Chevalley groups
(Essentially copied from comments)


It's just Borel-Harish-Chandra + Borel's density theorem (+ definition of Chevalley group). More precisely, $G$ is a $\mathbf{Q}$-defined group without nontrivial rational characters (since it has no nontrivial character at all), so Borel-Harish-Chandra implies $G(\mathbf{Z})$ is a lattice in $G(\mathbf{R})$, and hence so are its finite index subgroups. Then Borel's density theorem says that $G(\mathbf{Z})$ is Zariski-dense in $G(\mathbf{R})$ (and hence in $G(\mathbf{C})$ by Rosenlicht and connectedness of $G$).


As regards the question about finite index subgroups: this argument probably appears several times on this site: any connected real Lie group has no proper finite index subgroup, i.e., each homomorphism to a finite group is trivial: this follows from being generated by 1-parameter subgroups (which satisfy the given property, by divisibility).


