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. 



Does the Zariski closure of a maximal subgroup remain maximal?
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 $\text{GL}\_n(k)$.)


