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In the paper "Continuous isomorphisms onto separable groups", Applied General Topology, (13) 2012, 135--150, L. Morales Lopez proved Theorem: Let $G$ be an Abelian group with $|G| \leq 2^{2^{\aleph_0}}$. Then $G$ admits a separable, precompact, Hausdorff group topology. It is not true for general non-abelian groups by Shelah's results.

Is it true that any solvable group admits a separable Hausdorff group topology? Are there any published results about sufficient conditions under which a group admits a separable Hausdorff group topology?

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  • $\begingroup$ I guess the condition that the group has cardinal $\le 2^c$ is missing in the sentence "It is not true..." (otherwise this is trivially true) and in the question "It is true..." (otherwise there are trivial counterexamples). $\endgroup$
    – YCor
    Commented Nov 22, 2018 at 15:37
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    $\begingroup$ By the way it is easy to see that every abelian group of cardinal $\le 2^c$ embeds as a subgroup of $(\mathbf{Q}\times\mathbf{Q}/\mathbf{Z})^c$, which is is a separable Hausdorff topological group. This is, still, not a trivial proof of the result you quote, because a subgroup of a separable group, with the induced topology, is in general not separable. $\endgroup$
    – YCor
    Commented Nov 22, 2018 at 19:12
  • $\begingroup$ Sure. Even a closed subgroup of a separable group, with the induced topology, is in general not separable. $\endgroup$
    – Leiderman
    Commented Nov 22, 2018 at 21:30

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The question is only reasonable if one assumes the groups to be Hausdorff, and also to restrict to cardinal $\le 2^c$, since any Hausdorff separable space has cardinal $\le 2^c$.

In sharp contrast to the abelian case, we have:

For every uncountable cardinal $\alpha$, there exists a 2-step nilpotent group of cardinal $\alpha$ that has no Hausdorff separable group topology. It can be chosen torsion-free, or locally finite of exponent $4$ or odd prime $p$.

Indeed, fix a countable (possibly finite) non-abelian 2-step nilpotent group $F$ (e.g., of exponent 4 or odd $p$, or torsion-free). Let $G$ be the group $F^{(\alpha)}$: the subgroup of $F^\alpha$ of finitely supported families. It has cardinal $\alpha$. It has the property that every countable subset has a non-abelian centralizer.

On the other hand, on a Hausdorff topological group, if $D$ is a dense subset, then the centralizer of $D$ is equal to the center, and thus is abelian; in particular if a group $H$ admits a Hausdorff separable group topology, then it has a countable subset with abelian centralizer. So $G$ has no such topology.

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