16.2 (2006)

A group $G$ is subgroup-separable if for any subgroup $H \leqslant G$ and element $x \in G \setminus H$ there is a homomorphism to a finite group $f : G \to F$ such that $f(x) \notin f(H)$. Is it true that a finitely generated solvable group is locally subgroup-separable if and only if it does not contain a solvable Baumslag–Solitar group? Solvable Baumslag–Solitar groups are $BS(1, n) = \langle a, b \mid bab^{-1} = a^n \rangle$ for $n > 1$.
Background: It is known that a finitely generated solvable group is subgroup-separable if and only if it is polycyclic (R. C. Alperin, in: Groups–Korea '98 (Pusan), de Gruyter, Berlin, 2000, 1–5).