Recall that an abelian group $G$ is $Z$-torsionless if for all $a\in G$ ($\neq 0$) there is a homomorphism of $\phi\in Hom(G,Z) = G^\*$ so that $\phi(a)\neq 0$


Suppose $S$ is a subgroup of *torsionless* (edit) abelian $A$ so that $A/S$ is torsionless, and $T$ is a large summand of $S$, say $S=T\oplus \langle s \rangle$. I doubt it's true that $A/T$ needs to be torsionless, but can't see an example. 


An example would have be uncountable, in particular $A/S$ couldn't be countable, otherwise it would be free (as torsionless groups are $\aleph\_1$-free) and so $S$ would be a summand, hence $T$.


Can anyone offer some insight/direction on this?



example re torsionless quotients of abelian groups
I think your question is equivalent to asking whether there is an extension $0\to\mathbb{Z}\to B\to C\to 0$ of abelian groups where $C$ is torsionless but $B$ isn't. 


Given such an extension, take $A$ to be a free abelian group with a surjective map $A\to B$, let $T$ be the kernel of this map, and let $S$ be the kernel of the composition $A\to B\to C$, so that $S\cong T\oplus\mathbb{Z}$. 


To get an example of such an extension, it's sufficient to take the extension corresponding to any non-torsion element of $\operatorname{Ext}^1(C,\mathbb{Z})$ for any torsionless group $C$, as then the last map in the exact sequence
$$0\to\operatorname{Hom}(C,\mathbb{Z})\to\operatorname{Hom}(B,\mathbb{Z})\to\operatorname{Hom}(\mathbb{Z},\mathbb{Z})\to\operatorname{Ext}^1(C,\mathbb{Z})$$
is injective and so $B$ can't be torsionless, as every homomorphism $B\to\mathbb{Z}$ factors through $C$.


I believe that $\operatorname{Ext}^1(C,\mathbb{Z})$ is not a torsion group when $C$ is an infinite direct product of copies of $\mathbb{Z}$, so this will give an example, although maybe not a very explicit one.


