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.