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).
• The set $\{(x, y, x^{-1}, y^{-1}) : xy = yx, x, y\in G\} = \{(x, y, z, t) : xz = yt = xyzt = e\}$ is closed in $G\times G\times G\times G$.
• If $G$ is not a topological group, then the map $(x, y, x^{-1}, y^{-1})\mapsto (x, y)$ from above set to $\{(x, y) : xy = yx\}$ is not a homeomorphism, so that we can't conclude that $\{(x, y) : xy = yx\}$ is closed from this.
• If we could show that $\{(x, y, x^{-1}, y^{-1}) : xy = yx, x, y\in H\}$ is dense in $G\times G\times G\times G$, then the result would follow.

Any hints, suggestions, counter-examples appreciated.