most recent 30 from http://mathoverflow.net 2013-05-24T21:35:47Z http://mathoverflow.net/feeds/question/72467 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72467/does-the-zariski-closure-of-a-maximal-subgroup-remain-maximal Dennis Gulko 2011-08-09T10:15:37Z 2011-08-09T15:49:42Z

<p>Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$ be a linear group. Assume that $M&lt; G$ 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?<br> (I asked this question on MathSE five days ago and got no response, so I re-posted it here)<br> Thanks in advance for any help. </p>

http://mathoverflow.net/questions/72467/does-the-zariski-closure-of-a-maximal-subgroup-remain-maximal/72479#72479 Yves Cornulier 2011-08-09T15:29:33Z 2011-08-09T15:29:33Z

<p>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 $\text{SO}(2,\mathbf{C})\ltimes L$. </p> <p>There are obvious similar examples with $G$ countable. However it seems more subtle to cook up an example with $G$ finitely generated.</p> <p>(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 $\text{GL}_n(k)$.)</p>