Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$G$ 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.

share|cite|improve this question
Any closure of a subgroup is also a subgroup, for the first part. – awllower Nov 23 '11 at 7:33
@awllower: Thank you very much for your comment~ It is true that the closure of a subgroup is still a subgroup. I think for any subgroup $H$ of $G$, and the closure $\overline{H}$ of $H$, if it is true that the closure of the derived group of $H$ is just the derived group of $\overline{H}$, then I can prove from the solvability of $H$ that $\overline{H}$ is solvable by dimension comparision. But I don't know the correctness of $\overline{(H,H)}= (\overline{H}, \overline{H})$... – ShinyaSakai Nov 24 '11 at 11:19
Indeed I was wondering if there is some way we can relate the derived subgroups and the closures of them; so far no much progress is in hand. Sorry I cannot provide an answer. Does it make much difference to work with algebraic groups, from working with just topological groups? Maybe you can change the tag? Thanks for listening. – awllower Nov 25 '11 at 0:35
@awllower: Thank you very much for the enlightment. Algebraic group is a special type of topological groups, so similar results on topological groups in general may shed light on this problem. I will edit the tag :) – ShinyaSakai Nov 25 '11 at 7:09
Thanks for your generous words. – awllower Nov 27 '11 at 2:02

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...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.