It was used in the middle of a theorem's proof and I am not sure how to prove this fact.
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You need some other separation axiom like Hausdorff, otherwise there are simple counterexamples. If $G$ is Hausdorff, then $D = \{(x,x):x \in G\}$ is closed in $G \times G$. Hence $C_y = \{x \in G: xy = yx\}$ is closed, because it is the preimage of $D$ under the map $x \mapsto (xy,yx)$. Hence $C_{a^n}$ is closed, and contains the dense subset $\{a^n:n\in \mathbb Z\}$. Hence $C_{a^n} = G$. Hence for any $x \in G$, we have that $x \in C_{a^n}$, which implies that $a^n \in C_x$. Hence $C_x$ is closed and contains the dense subset $\{a^n:n\in \mathbb Z\}$. Hence $C_x = G$.
Stephen Montgomery-Smith
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Assume $G$ Hausdorff.
Given $z=xy\in G$ and an open set $A\ni z$ there exist open sets $U\ni x$ and $V\ni y$ in $G$ such that $UV\subset A$. Indeed the inverse image of $A$ under the multiplication map is open in $G\times G$ and thus contains some basic open set $U\times V$.
Suppose $xy\neq yx$ and let $A\in xy$ and $B\ni yx$ open sets separating the two points. Up to taking intersections, we may assume that $UV\subset A$ and $VU\subset B$.
But by density of $\langle a\rangle$ in $G$, we can find $a^m\in U$ and $a^n\in V$. Thus $a^{m+n}\in UV\cap VU$ which is a contradiction.
Andrea Mori
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For Metric Spaces: Because this set is dense, any $g,h \in G$ can be written as $\lim_{k \to \infty} a^{n_k}$ for sequence $\{n_k^{(g)}\}_{k \in \Bbb N},\{n_k^{(h)}\}_{k \in \Bbb N} \subset \Bbb Z$. Because $G$ is a topological group, we have $$ gh = \lim_{k \to \infty}a^{n_k^{(g)}}a^{n_k^{(h)}} $$
In terms of nets: Note that the sequences $\{a^{n_k^{(g)}}\}$ and $\{a^{n_k^{(h)}}\}$ are convergent nets in $G$, so that $\{a^{n_k^{(g)}} \times a^{n_k^{(h)}}\}$ converges to $(g,h)$ in $G \times G$.
Because the group operation is continuous, it maps the convergent net in $G \times G$ to a convergent net in $G$, and the limit of this net is $gh$.
If $G$ is Hausdorff, then the limit of this net is unique, and we may indeed say that $$ gh = \lim_{k \to \infty}a^{n_k^{(g)}}a^{n_k^{(h)}} $$ As before.
Ben Grossmann
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That works if the topology comes from a metric. Does it hold for general topological groups? Maybe you have to use limits of nets instead of limits of sequences. – Stephen Montgomery-Smith Aug 24 '15 at 18:26
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@StephenMontgomery-Smith I may have secretly forgotten that not all spaces are metric spaces. After looking at your answer, I am doubtful that such an approach will work in general. – Ben Grossmann Aug 24 '15 at 18:28
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1I think it would work with nets instead of sequences. You need the additional axiom of Hausdorff to be sure that limits of nets are unique, if they exist. – Stephen Montgomery-Smith Aug 24 '15 at 18:29
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I was thinking about your previous answer, but your edition made it correct. Although it is not necessary a metric space, thank you for your answer. – B. Rivas Aug 24 '15 at 18:30
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@StephenMontgomery-Smith oh, with the Hausdorff assumption, sure. I'll try to think of something (not used to working with nets, though). – Ben Grossmann Aug 24 '15 at 18:31
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@B.Rivas you're welcome – Ben Grossmann Aug 24 '15 at 18:31
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Yeah, nets can be tricky. I don't know too much about them either. – Stephen Montgomery-Smith Aug 24 '15 at 18:43
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See my latest edit. – Ben Grossmann Aug 24 '15 at 18:45
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If a subgroup H is dense in a group G, then the derived subgroup of H is dense in the derived subgroup of G.
In particular, if H is abelian, then the derived subgroup of G is trivial, so G is also abelian.
Mariano Suárez-Álvarez
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