Separable topology on a group 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?
 A: 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.
