Let $G$ be a topological group and $A,B$ be two subgroups of $G$ such that $AB$ is a closed subset. I wonder if for any element $c\in G$, the subset $AcB$ is always closed?
In the original question I was thinking about, there is either a $T_1$ condition on $G$ or better, $G$ is locally compact Hausdorff. So please feel free to add such mild conditions if the above assertion fails.
The set $AcB$ can be non-closed. For instance, let $G$ be a Cartesian product of topological spaces $\Bbb R$ and $\Bbb Z$ endowed with a specific operation of a semidirect product $(x,n)(y,m)=(x+n\sqrt{2}y, m+n)$. It is easy to check that $G$ is a topological group. Now if we put $A=B=\Bbb Z\times \{0\}$ and $c=(0,1)$, we obtain that $c^{-1}AcB=(\Bbb Z+\sqrt{2}\Bbb Z)\times \{0\}$ is a non-closed subgroup of $G$.