If a subgroup of an algebraic group is solvable, is its closure necessarily solvable? $G$ is an algebraic group, and $H$ is a subgroup which is solvable. $\overline{H}$ is its closure in $G$.
Then $\overline{H}$ is also a subgroup of $G$. Is it also solvable?
For any algebraic group $G$, denote $[G,G]$ the derived subgroup of $G$. Then is it true that $\overline{[H,H]} = [\overline{H}, \overline{H}]$? If this is true, I think the solvability of $\overline{H}$ might be proved by dimension comparision.
Many thanks~ Special thanks to @awllower for the enlightening comments.
 A: Many thanks to @QiL for the answer to another question.
In fact the answer to this question is affirmative.
For any algebraic group $G$, and its solvable subgroup $H$:


*

*$H \times H$ is dense in $\overline{H} \times \overline{H}$;

*The map $\phi: G \times G \rightarrow G, (x,y) \mapsto xyx^{-1}y^{-1}$ is continuous. Thus $[H,H] = \phi(H \times H)$ is dense in $[\overline{H}, \overline{H}]=\phi(\overline{H} \times \overline{H})$;

*$[\overline{H}, \overline{H}]$ is a closed subgroup of $\overline{H}$, thus is closed in $G$. $\overline{[H,H]} \subseteq [\overline{H}, \overline{H}]$. From (2), $\overline{[H,H]} \supseteq [\overline{H}, \overline{H}]$, so the two are equal.

*As $H$ is solvable subgroup, it has a derived series: $H_0 \supsetneq H_1 \supsetneq \cdots \supsetneq H_n =1$, where $H_0 = H$, and $H_{i+1}$ is the derived subgroup of $H_i$ for $i=0, \cdots, n-1$. From (3), it can be seen that $\overline{H_0} \supsetneq \overline{H_1} \supsetneq \cdots \supsetneq \overline{H_n} =1$ is a derived series of $\overline{H}$. So $\overline{H}$ is solvable.
All the above are the same for Hausdorff topological groups. Although algebraic groups are in general not Hausdorff... I don't know if this can be more general. At least, the 4th step requires that the set $\{ 1 \}$ being closed.
I hope I am not mistaken...
A: Following the proof of @ShinyaSakai, try to prove this result:
For $H_{1}\subset H_{2}$ two subgroups (not necessarily closed) of $G$, we have $[\overline{H_{1}},\overline{H_{2}}]=\overline{[H_{1},H_{2}]}$.
The reason for the assumption $H_{1}\subset H_{2}$ is that the commutator of two closed subgroups is not closed generally. 
There is a tricky point in the proof of this result. As is pointed out by @evgeny under the answer of @ShinyaSakai, generally speaking, $[H_{1},H_{2}]\neq \phi(H_{1}×H_{2}) $. A correct proof is the following,

*

*$H_{1}×H_{2}$ is dense in $\overline{H_{1}}×\overline{H_{2}}$.

*$\phi(H_{1}×H_{2})$ is dense in $\phi(\overline{H_{1}}×\overline{H_{2}})$.

*$[\overline{H_{1}},\overline{H_{2}}]$ is a closed subgroup of $G$, generated by $\phi(\overline{H_{1}}×\overline{H_{2}})$, hence containing $\overline{[H_{1},H_{2}]}$.

*$\overline{[H_{1},H_{2}]}$ is the closure of the geoup generated by $\phi(H_{1}×H_{2})$, hence containing $\phi(\overline{H_{1}}×\overline{H_{2}})$. By 3, it contains $[\overline{H_{1}},\overline{H_{2}}]$.

You can use this result to prove that the closure of a solvable (nilpotent) algebraic group is still solvable (nilpotent).
