I'm learning about topological groups from Arhangelskii and Tkachenko "Topological groups and related structures" and this is one of the exercises there.
A paratopological group is a group with topology such that $(x, y)\mapsto xy$ is continuous.
Let $G$ be a $T_1$ paratopological group and $H\subseteq G$ a commutative dense subgroup. It's to show that $G$ is commutative.
Some of my observations:
- $G$ need not be Hausdorff (so various approaches like $H\times H\ni (x, y)\mapsto xy$ extends to unique continuous function don't work, any argument working with limits of nets shouldn't work either).
- $G_0 = \{(x, x^{-1}) : x\in G\}\subseteq G\times G$ with obvious group operations is a Hausdorff topological group and $h:G_0\to G$ given by $h(x, x^{-1}) = x$ is a continuous group isomorphism.
- The set $\{(x_0, y_0)\in G_0\times G_0 : [x_0, y_0] = e\}$ is closed in $G_0\times G_0$.
- Let $H_0 = h^{-1}[H]$. If we could show that $H_0$ is dense in $G_0$, then we would obtain that $G$ is commutative since $H_0\times H_0\subseteq \{(x_0, y_0) : [x_0, y_0] = e\}$.
- If $G$ is the Sorgenfrey line with addition, then $H = \mathbb{Q}$ is a dense abelian subgroup, but $G_0$ is discrete so that $H_0$ isn't dense in $G_0$.
Any hints, suggestions, counter-examples appreciated.
Edit: I've communicated with Alex Ravsky that this result is false. I want people who get stuck on this exercise in the future to know it's wrong, especially since they can be new to topological groups like me, so here is the counter-example (by Alex Ravsky):
Equip $G = \mathbb{R}\times \mathbb{R}_+$ with operation $(x, y)\cdot (t, z) = (x+yt, yz)$. Then $G$ is a non-abelian group. Say $U\subseteq G$ is open iff $U = \emptyset$ or $U = \mathbb{R}\times (r, \infty)$ for some $r > 0$. This is a topology on $U$, it's $T_1$ and makes multiplication in $G$ continuous. So $G$ is a $T_1$ paratopological group.
Then $H = \{0\}\times \mathbb{R}_+ \subseteq G$ is a dense abelian subgroup of $G$, but $G$ is not abelian.