# Zariski closure of an algebraic subgroup of finite index [closed]

Let $$G$$ be an algebraic group and $$H \subset G$$ a subgroup of $$G$$. Let $$H_0$$ be a subgroup of $$H$$ of finite index. Then I guess the Zariski closure of $$H_0$$ is exactly the Zariski closure of $$H$$ in $$G$$, since intuitively it should be.

But I have no clue how to prove it, anything will be helpful.

Note that this is equivalent to asking whether any finite index subgroup of $$H$$ is dense. In particular there's no real reason to consider the situation within an ambient algebraic group $$G$$.
If you assume irreducibility, or even connectedness, of $$H$$, the claim is true as the only finite index closed subgroup of $$H$$ is $$H$$ itself (This follows as $$H$$ is the union of the cosets of any such subgroup), yet the closure of any finite index subgroup will be a finite index subgroup as well.
Without assuming connectedness of $$H$$ the claim is not true. Consider for example $$H_0\subsetneq H\subsetneq G$$ finite groups.
• If $H$ is not irreducible, it is still true that some proper subgroup $H_0 \subset H$ have the same Zariski closure? Commented Mar 3 at 6:01
• @finiteness No, take $H$ to have two elements. Commented Mar 3 at 6:03
• In the first paragraph, are you reducing to $G = H$ or $G = \overline{H}$? What about the case where $H$ is not closed but $\overline{H}$ is irreducible? Commented Mar 4 at 13:27