For a topological group $(G,\mathcal T)$ and a subgroup $H\le G$, we say $\mathcal T$ and $H$ are permutable if for every neighborhood $U$ of $1$, there is a neighborhood $V$ of $1$ with $VH\subseteq HU$.


Do you have an example of an infinite (preferably nonabelian) Hausdorff topological group $(G,\mathcal T)$ such that $\mathcal T$ is not discrete and for every closed subgroup $H$ of $G$, $\mathcal T$ and $H$ are permutable?



A group topology which commutes with closed subgroups
I will give a class of metric examples. 


Let $G\_1$ be an abelian group with a metric $d\_1$.


Let $G\_2$ be a discrete nonabelian group, with metric $d\_2(x,y)=1$ for all $x\not=y$. 


Let $G:= G\_1\times G\_2$, with $d( (x\_1, x\_2), (y\_1, y\_2)) = d\_1(x\_1,y\_1) + d\_2(x\_2, y\_2)$. 


Let $H$ be any subgroup of $G$. Let $U$ be a neighborhood of $1$. Wlog the neighborhood $U$ is a ball of radius $\varepsilon$, and wlog $\varepsilon < 1$. 


Then for all $(x,y)\in U$ we must have $y=1$. So $UH=HU$. 


