When discussing with awllower about this question, I begin to think about another one:

For a topological group $G$ and a subgroup $H$, is it true that $[\overline{H}, \overline{H}] = \overline{[H,H]}$?

where $[H,H]$ denote the derived subgroups of $H$.

I think, if I define the map $\phi: G \times G \rightarrow G, (x,y) \mapsto xyx^{-1}y^{-1}$, then the question will become the equality between $\phi(\overline{H}\times \overline{H})$ and $\overline{\phi(H \times H)}$. Giving $G \times G$ the product topology, we will have $\phi(\overline{H} \times \overline{H}) \supseteq \overline{\phi(H \times H)}$ if $\phi$ is closed; and $\phi(\overline{H} \times \overline{H}) \subseteq \overline{\phi(H \times H)}$ because $\phi$ is continuous. (I hope I am not mistaken in thinking that $\overline{H \times H} = \overline{H} \times \overline{H}$, which is a condition of the above.)

So, a derived question is:

For a topological group $G$, is the map $\phi: G \times G \rightarrow G, (x,y) \mapsto xyx^{-1}y^{-1}$ closed if $G \times G$ is given the product topology?

For the case when $G$ is an algebraic group, the topology will be changed to Zariski topology. Then is $\phi$ a closed map? Does $\phi(\overline{H}\times \overline{H})$ equal $\overline{\phi(H \times H)}$?

In fact I am more concerned with the algebraic group case. But if the first case is dealt, hopefully the algebraic case will become easier.

Thanks to everyone.


2 Answers 2


For algebraic groups, $\overline{[H, H]}=[\overline{H}, \overline{H}]$.

First $H\times H$ is dense in $\overline{H}\times \overline{H}$ for the Zariski topology (this is trivial for the product topology, but needs some proof for Zariski topology). So $[H, H]$ is dense in [$\overline{H}, \overline{H}]$ and $[\overline{H}, \overline{H}]\subseteq \overline{[H, H]}$.

Second $[\overline{H}, \overline{H}]$ is closed. This is a general fact for algebraic groups: the derived group of an algebraic group is an algebraic group.

  • $\begingroup$ Thank you very much. This solves my original question and is very useful for me. Now I only have to prove the density of $H \times H$ in $\overline{H} \times \overline{H}$. I am sorry I can accept only one answer... Thanks again :) $\endgroup$ Nov 28, 2011 at 14:57
  • $\begingroup$ Dear @ShinyaSakai, no problem. Jack's answer is perfect. The next time you post on algebraic groups, you should add the algebraic geometry tag. $\endgroup$
    – user18119
    Nov 28, 2011 at 22:20
  • 1
    $\begingroup$ Why if $H \times H$ is dense in $\widetilde H \times \widetilde H$, the same holds for commutators? I believe that one needs a fact from 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 $[H, H]=\mathrm{Im} \; H^{\times 2n}$. $\endgroup$
    – evgeny
    Mar 8, 2017 at 5:56

First I'll give an answer to what I hope your real question is. Then I'll give a negative answer to your specific question, but hopefully you'll see that it does not matter.

Proposition: If H is a solvable subgroup of the (Hausdorff) topological group G, then the closure of H in G is a solvable subgroup of G.

Special case proof: Suppose H is abelian, so its derived length is k = 1. Let g, h be two (=2k) elements of the closure of H, and let gng and hnh. Since multiplication and inversion are continuous, [ gn, hn ] → [ g, h ], but the former is simply the constant sequence consisting of the identity, so [ g, h ] = 1 as well.

Proof: Since G is Hausdorff, we can consider limits of sequences. Let the derived length of H be k, and let gi for 1 ≤ i ≤ 2k be 2k elements of the closure of H. For each i, let gi,jgi. Since multiplication and inversion are continuous, $$[\ldots[[[g_{1,j},g_{2,j}],[g_{3,j},g_{4,j}]],[[g_{5,j},g_{6,j}],[g_{7,j},g_{8,j}]]]\ldots] \to [\ldots[[[g_1,g_2],[g_3,g_4]],[[g_5,g_6],[g_7,g_8]]]\ldots]$$ but the left hand side is the constant sequence consisting of the identity, and so the right hand side is the identity as well, and the derived length of the closure of H is (no more than, but obviously no less than) k. $\square$

If your spaces are not Hausdorff, then presumably you can do approximately the same thing, as this is only using continuity, not closed-ness of the maps corresponding to the commutators.

In other words, in general one has ${\overline H}^{(k)} \leq \overline{H^{(k)}}$ by topological considerations, but of course $H^{(k)} \leq \overline{H}^{(k)}$ by simple set-theoretic containment. Hence one has in general that:

$$\overline{ \overline{H}^{(k)}} = \overline{H^{(k)}}$$

which should be enough in any sane world (where the closeure of the identity is the identity). I guess in the indiscrete topology on the cyclic group of two elements with G = H, one has

$$ \overline{H}^{(1)} = [G,G] = 1 \quad \text{while} \quad \overline{H^{(1)}} = \overline {1} = G$$

so that your equality need not be true in general.

  • $\begingroup$ Thank you very much. In fact, topology are various, and topological groups are various as a consequence, so I should be more specific. Thank you for the detailed explanation. $\endgroup$ Nov 28, 2011 at 14:43
  • $\begingroup$ It's actually false without Hausdorff (as the closure of the solvable subgroup $\{1\}$ might be arbitrary). $\endgroup$
    – YCor
    Oct 9, 2020 at 22:37

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