Let $G$ be a compact topological group and $a \in G$. Suppose $\{a^k\}_{k \in \mathbb{Z}}$ is dense in $G$. Show $G$ is Abelian.
My thoughts for this were that for any $x,y \in G$, $x$ is close to some $a^m$ and $y$ is close to some $a^n$. Since $a^m$ and $a^n$ commute, this should mean $x$ and $y$ commute, but I've had trouble making this work. Note: $G$ might not need to be compact for this to work.
EDIT: new idea: if for any $x,y \in G$ there are sequences $(a^{n_k})_k$ and $(a^{m_k})_k$ converging to $x,y$ respectively, then since multiplication is continuous, we would be done.
EDIT: I think I solved it. The edit above provides a solution assuming the existence of such sequences. We don't know such sequences exist but we do know such a net exists. And we can apply continuity of multiplication for that net, and be done.
A standard claim and proof.
Proposition. Let $G$ be a Hausdorff topological group containing a dense abelian subgroup $H$. Then the group $G$ is abelian too.
Proof. Let $x,y$ be arbitrary elements of the group $G$. Assume that $xy\ne yx$. Pick disjoint neighborhoods $O_{xy}\ni xy$ and $O_{yx}\ni yx$. Since the multiplication on the group $G$ is continuous, there exist open neighborhood $O_x’, O_x’’\ni x$ and $O_y’, O_y’’\ni y$ such that $O_x’O_y’\subset O_{xy}$ and $O_x’’O_y’’\subset O_{yx}$. Since $H$ is dense in $G$, there exist points $a\in O_x’\cap O_x’’\cap H$ and $b\in O_x’\cap O_x’’\cap H$. Then $ab=ba\in O_{xy}\cap O_{yx},$ a contradiction.$\square$
The Haudorfness of the group $G$ is essential, because any (in particular, non-abelian) group endowed with the antidiscrete topology contains a cyclic group $\{e\}$.