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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.

ShinyaSakai
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  • Any closure of a subgroup is also a subgroup, for the first part. – awllower Nov 23 '11 at 07:33
  • @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 00:35
  • @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 07:09
  • Thanks for your generous words. – awllower Nov 27 '11 at 02:02
  • @awllower: I am afraid 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. – ShinyaSakai Nov 27 '11 at 19:19

2 Answers2

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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...

Community
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ShinyaSakai
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    It is not true that $[H,H]=\phi(H \times H)$. – Not a grad student Oct 27 '16 at 00:56
  • 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 $[H, H]=\mathrm{Im} ; H^{\times 2n}$. – evgeny Mar 08 '17 at 05:48
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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,

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

Baiqing Zhu
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