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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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
1 Answers
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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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0An infinite cyclic group of diagonal matrices will work, too. – 2013-09-23