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Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$. Assume that $M<G$ is a maximal subgroup (in the abstract group sense). Denote by $\bar{G}^Z$ the Zariski closure of $G$ in $\rm{GL}_n(k)$. Is it true that $\bar{M}^Z<\bar{G}^Z$ is a maximal subgroup in the algebraic groups sense? If yes, would it be a maximal subgroup in the abstract group sense?
Thanks in advance for any help.

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Do you mean by linear group, an (affine) algebraic group ? – user18119 Oct 29 '12 at 21:56
That was redundant, since it just meant that $G\leq{\rm{GL}}_n(k)$. BTW, I found a nice counter-example since then. I just didn't delete the question since I hoped that someone will answer it eventually (didn't want to answer my own question) – Dennis Gulko Oct 29 '12 at 23:46
Thanks! I asked the question because a linear subgroup is always closed for the Zariski topology. It would be nice if you could give your counterexample. – user18119 Oct 30 '12 at 8:42
Take $G={\rm{SO}}_2(\mathbb{R})\ltimes\mathbb{R}^2$. Then $H={\rm{SO}}_2(\mathbb{R})$ is a maximal subgroup of $G$. Then $\bar{G}^Z={\rm{SO}}_2(\mathbb{C})\ltimes\mathbb{C}^2$ and $\bar{H}^Z={\rm{SO}}_2(\mathbb{C})$, which is no longer maximal, since it stabilizes a line $L=\langle(1,i)\rangle$, and hence is a subgroup of ${\rm{SO}}_2(\mathbb{C})\ltimes L$ – Dennis Gulko Oct 30 '12 at 9:04

The following is a different type of a counterexample. Not sure that it qualifies or matches with what you were looking for. Let $$ G=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{Z}\right\}. $$ We view this as a subgroup of $GL_2(\Bbb{C})$. It is isomorphic to the additive group of integers, so the subgroup $M=M_p$ consisting of the elements of $G$ such that $m$ is divisible by a fixed prime $p$ is a maximal subgroup.

But both $G$ and $M$ have as their Zariski closure the group $$ \overline{G}=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{C}\right\}=\overline{M}. $$ So $\overline{M}$ is not a maximal subgroup of $\overline{G}$ in any sense.

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An infinite cyclic group of diagonal matrices will work, too. – Jyrki Lahtonen Sep 23 '13 at 19:09

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