An abelian group $A$ is *cotorsion* provided that whenever $A \leq G$ with $G$ abelian and $G/A$ is
torsion-free, we have $G \cong A \oplus B$ for some $B \leq G$. An abelian group $A$ is
*cotorsion-free* if it contains no non-trivial cotorsion subgroup.


It seems that $\mathbb{Z}^{\omega}$ is cotorsion-free.


1. What about an uncountable direct product like $\mathbb{Z}^{\mathfrak{c}}$? Is this group cotorsion-free?
2. More generally, if $G\_i$, $i \in I$ are slender abelian groups, must $\prod\_{i\in I}G\_i$ be cotorsion-free?



Cotorsion-freeness in uncountable products of abelian groups
In fact, more is true.


Every direct product of cotorsion-free abelian groups is cotorsion-free. This is clear because the cotorsion-free abelian groups are precisely those that have no nonzero homomorphisms from any cotorsion group.


This answers the question, since slender abelian groups are cotorsion-free.


