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$G$ 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.

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    @awllower: I am a$f$raid algebraic groups are not topological groups because the topology given on product groups are different, the former Zariski topology and the latter product topology. I posted another question and hopefully it will get some attention. http://math.stackexchange.com/questions/86128/for-a-topological-group-g-and-a-subgroup-h-is-it-true-that-overlineh Thanks for the enlightening discussion.2011-11-27

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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 $H$:

  1. $H \times H$ is dense in $\overline{H} \times \overline{H}$;

  2. 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 $[\overline{H}, \overline{H}]=\phi(\overline{H} \times \overline{H})$;

  3. $[\overline{H}, \overline{H}]$ 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.

  4. 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...

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    I believe it is written somewhere in Springer's Linear algebraic groups that for an affine algebraic group the number of necessary commutators to form an element of a commutant is bounded independently of an element, say by $n$, so $[H, H]=\mathrm{Im} \; H^{\times 2n}$.2017-03-08