Take the 2-minute tour ×

$G$ 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|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

1 Answer 1

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|improve this answer

Your Answer


