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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.

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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.

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  • $\begingroup$ If $H$ is not irreducible, it is still true that some proper subgroup $H_0 \subset H$ have the same Zariski closure? $\endgroup$
    – finiteness
    Commented Mar 3 at 6:01
  • $\begingroup$ @finiteness No, take $H$ to have two elements. $\endgroup$
    – Etropy
    Commented Mar 3 at 6:03
  • $\begingroup$ 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? $\endgroup$
    – ronno
    Commented Mar 4 at 13:27

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