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?
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Mar 19, 2021 at 14:57
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    $\begingroup$ @YCor Most of this seems clear but why is it that every nontrivial subgroup of $\mathbb{Z}^X$ has $\mathbb{Z}$ as a quotient? $\endgroup$
    – J.K.T.
    Mar 19, 2021 at 15:51
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    $\begingroup$ @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. $\endgroup$
    – YCor
    Mar 19, 2021 at 16:53

1 Answer 1


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.


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