A topological group is called minimal if it admits no strictly weaker Hausdorff group topology. By Prodanov-Stoyanov Theorem, a complete Abelian topological group is minimal if and only if it is compact. Since each non-compact complete topological group contains a non-compact closed separable subgroup, we obtain the following
Fact: A complete Abelian topological group $X$ is minimal if and only if each closed separable subgroup of $X$ is minimal.
Therefore, the minimality of complete Abelian toplogical groups is recognizable at the level of closed separable subgroups. Is the same true for non-commutaive groups?
Problem. Is a complete topological group $X$ minimal if each closed separable (or $\omega$-narrow) subgroup of $X$ is minimal?
A topological group $X$ is called $\omega$-narrow if for any neighborhood $U\subset X$ of the unit there exists a countable subset $A\subset X$ such that $X=AU=UA$.
