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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?

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    $\begingroup$ Did you intend some condition to rule out discrete groups? Otherwise any infinite discrete group is an example. $\endgroup$
    – Jeremy Rickard
    Commented Jul 30, 2015 at 12:32
  • $\begingroup$ Yes. I must add it. $\endgroup$
    – Minimus Heximus
    Commented Jul 30, 2015 at 12:34
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    $\begingroup$ It holds in a more general setting encompassing both the discrete case and the abelian case, namely when G has an open central subgroup, i.e. when "G is central-by-discrete". $\endgroup$
    – YCor
    Commented Jul 30, 2015 at 17:13
  • $\begingroup$ When G is Hausdorff and not central-by-discrete, there are still some cases where it's abelian-by-discrete which might work, possibly not all. $\endgroup$
    – YCor
    Commented Jul 30, 2015 at 17:14

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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 $UH=HU$.

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