I think this question should be pretty straightforward but through a combination of being self-taught in group theory and being awful at geometry, something is escaping me. I am trying to understand and prove the following:
Let $G=SL(n,\mathbb{R})$ be an algebraic group and $H$ a subgroup of $G$ isomorphic to the group of integers $\mathbb{Z}^k$ and I know that the Zariski closure of $\mathbb{Z}^k$ is the commutative group $\mathbb{R}^k$. I want to prove that the Zariski closure of $H$ has a commutative Lie algebra of dimension $k$ ( i-e., is the Lie algebra of $\bar{H}$ commutative? ).
The Lie algebra of a commutative group is commutative. See for example answers to this question: link.