Almost free group without the Specker group as a subgroup

Question

An Abelian group is almost free whenever every countable subgroup is free Abelian. Famously, the Specker group $\mathbf Z^{\mathbf N}$ is almost free. What are examples of almost free groups that are not free but do not contain the Specker group as a subgroup?

Answer 1

First, just a quick comment about terminology. It's possible that there are varying conventions, but I have seen "almost free" used to mean that all subgroups of an abelian group $G$ that have cardinality less than $|G|$ are free abelian. And I've seen "$\kappa$-free" used to mean that all subgroups of cardinality less than $\kappa$ are free abelian. Using this terminology, the Baer-Specker group is $\aleph_1$-free (every countable subgroup is free abelian), but is only almost free if the continuum hypothesis is true (as it is known to have nonfree subgroups of cardinality $\aleph_1$).

One class of $\aleph_1$-free groups are the subgroups of the Baer-Specker group, which seem to have been studied extensively. Theorem 1 of

Kolman, Oren; Shelah, Saharon, Almost disjoint pure subgroups of the Baer-Specker group, Eklof, Paul C. (ed.) et al., Abelian groups and modules. Proceedings of the international conference in Dublin, Ireland, August 10-14, 1998. Basel: Birkhäuser. Trends in Mathematics. 225-230 (1999). ZBL0943.20054.

states that there are $2^{\aleph_1}$ nonfree subgroups of the Baer-Specker group of cardinality $\aleph_1$ such that no nonfree subgroup of any of them is isomorphic to a subgroup of any of the others. In particular, at most one of them can have a subgroup isomorphic to the Baer-Specker group!

I don't think the proof is very constructive, so I'm not sure whether this counts as "giving examples". But I'm not an expert, and there might well be more elementary facts about subgroups of the Baer-Specker group that would give a more satisfactory answer to your question. Certainly there are people active on MathOverflow who know this subject a lot better than I do, and who might be able to help.