# Cotorsion-freeness in uncountable products of abelian groups

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?
• There exist non split abelian extensions with kernel $\mathbf{Z}$ and quotient $\mathbf{Z}[1/2]$, so $\mathbf{Z}$ is not "cotorsion" in the above sense. Hence no group with $\mathbf{Z}$ as direct factor is cotorsion. In $\mathbf{Z}^X$ for arbitrary $X$, every nontrivial subgroup has $\mathbf{Z}$ as quotient, hence as direct factor, so no nontrivial subgroup is cotorsion. So $\mathbf{Z}^X$ is indeed cotorsion-free.
– YCor
Mar 19, 2021 at 14:57
• @YCor Most of this seems clear but why is it that every nontrivial subgroup of $\mathbb{Z}^X$ has $\mathbb{Z}$ as a quotient? Mar 19, 2021 at 15:51
• @JKT. Let $H$ be a nonzero subgroup. Take a nonzero element $f\in H$. There exists $x\in X$ such that $f_x\neq 0$. Then consider the homomorphism $H\to\mathbf{Z}$, $g\mapsto g_x$: it is nonzero.
– YCor
Mar 19, 2021 at 16:53