Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$. Assume that $M
Thanks in advance for any help.
Does the Zariski closure of a maximal subgroup remain maximal?
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group-theory
algebraic-geometry
algebraic-groups
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0Do you mean by linear group, an (affine) algebraic group ? – 2012-10-29
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0That 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) – 2012-10-29
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0Thanks! 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. – 2012-10-30
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3Take $G={\rm{SO}}_2(\mathbb{R})\ltimes\mathbb{R}^2$. 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 ${\rm{SO}}_2(\mathbb{C})\ltimes L$ – 2012-10-30