A *torsionless* abelian group $A$ is one for which any element $a\neq 0$ can be sent to a nonzero element of $Z$ by some homomorphism $A\rightarrow Z$ (integers). Equivalently, $A$ can be embedded as a subgroup of a Cartesian power of $Z$, $Z^I$.


A *separable* abelian group $A$ is one for which any element $a$ such that $\langle a \rangle$ is pure can be sent to $1\in Z$ by some homomorphism. Equivalently $A$ can be embedded as a pure subgroup of $Z^I$, equivalently any finitely generated pure subgroup is a direct summand. 


There are torsionless abelian groups which are not separable, for example, take $A = Z^{(\omega)} + 2Z^\omega$. The element $\sum 2e\_n$ generates a pure subgroup but any homomorphism to $Z$ must take it to a mulitple of $2$, since any such homomorphism has finite support in $\omega$, by a theorem of Specker.


I don't know of many other examples however, which leads to my problem:


Suppose $B$ is a countable subgroup of separable group $A$ so that $A/B$ is torsionless. Must $A/B$ be also separable?



Torsionless not separable abelian groups
The answer is no.


Consider your (renamed) example $V = \mathbf{Z}^{(\omega)} + 2\mathbf{Z}^\omega$. It is quotient of the direct sum $$A=\mathbf{Z}^{(\omega)} \oplus 2\mathbf{Z}^\omega\simeq \mathbf{Z}^{(\omega)} \oplus \mathbf{Z}^\omega$$ by a subgroup $B$ isomorphic to the intersection, namely the countable subgroup $2\mathbf{Z}^{(\omega)}$. As you say, $V\simeq A/B$ is torsionless and not separable, while $A$ is separable.


