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Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$ be a linear group. 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 if $\bar{M}^Z\neq \bar{G}^Z$ then it is a maximal subgroup in the algebraic groups sense? If yes, would it be a maximal subgroup in the abstract group sense?
(I asked this question on MathSE five days ago and got no response, so I re-posted it here)
Thanks in advance for any help.

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    $\begingroup$ How about this: consider any Zariski dense subgroup $H$. For example the non-zero rationals inside $GL_1(\mathbb{C})$. By Zorn's lemma we can pick a maximal (abstract) subgroup $G$ containing $H$. $\endgroup$ Commented Aug 9, 2011 at 11:12
  • $\begingroup$ Yeah, clearly it could happen that $\bar{M}^Z=\bar{G}^Z$, but that's not the interesting case. I edited the question accordingly. $\endgroup$ Commented Aug 9, 2011 at 11:19
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    $\begingroup$ What set do you apply Zorn's lemma to ? Because a union of strict subgroups may not be strict. $\endgroup$ Commented Aug 9, 2011 at 11:21
  • $\begingroup$ @Auguste: I hadn't thought it through - my comment is wrong then. $\endgroup$ Commented Aug 9, 2011 at 12:31

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The answer is no. Assume that $G=\text{SO}(2,\mathbf{R})\ltimes\mathbf{R}^2\subset\text{GL}_3(\mathbf{C})$ and $M=\text{SO}(2,\mathbf{R})$. Then $M$ is maximal in $G$. However the Zariski closures are $\text{SO}(2,\mathbf{C})\subset\text{SO}(2,\mathbf{C})\ltimes\mathbf{C}^2$, so $\text{SO}(2,\mathbf{C})$ is not maximal in the algebraic sense because it stabilizes a line $L$ in $\mathbf{C}^2$ and is thus contained in $\text{SO}(2,\mathbf{C})\ltimes L$.

There are obvious similar examples with $G$ countable. However it seems more subtle to cook up an example with $G$ finitely generated.

(Edit: I understand the confusing assumption "$G\leq\rm{GL}_n(k)$ be a linear group" as "let $G$ be an arbitrary subgroup (in the abstract sense) of the group $\text{GL}_n(k)$.)

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  • $\begingroup$ Ah! Your understanding of the question makes a much better question than mine. I'm sure that you are right, but I'll wait for confirmation from Dennis before deleting mine. $\endgroup$ Commented Aug 9, 2011 at 15:52
  • $\begingroup$ Thanks, that's exactly what I meant. Thanks for your great answer. $\endgroup$ Commented Aug 9, 2011 at 16:52

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