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Suppose we have a countable, torsion-free abelian group $A$ with the property that for each element $a\neq 0$ the set $D_a=\{x\in A|\exists n\in \mathbb{Z}:nx=a\}$ is finite. Is $A$ already a free abelian group?

If one drops the condition "countable" the infinite direct product of countably many copies of $\mathbb{Z}$ is a counterexample.

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I suspect the answer is yes for finitely generated A. Do you know if that is so? Gerhard "Ask Me About. System Design" Paseman, 2012.03.08 – Gerhard Paseman Mar 8 '12 at 16:25
@Gerhard: finitely generated abelian groups are classified in a good undergraduate or basic postgraduate course, and the answer "yes" trivially follows from the classification. – Vladimir Dotsenko Mar 8 '12 at 16:29
@Henrik: to include some context to your question, maybe it is worth mentioning that the property discussed is indeed the key ingredient of the simplest proof that the infinite direct product $\mathbb{Z}^\mathbb{N}$ is not free. (I learned it from…) – Vladimir Dotsenko Mar 8 '12 at 16:31
Thanks, Vladimir. Gerhard "Surpassed Daily Silly Question Quota" Paseman, 2012.03.08 – Gerhard Paseman Mar 8 '12 at 21:08
up vote 13 down vote accepted

No, there are non-free abelian groups of rank 2 (i.e., subgroups of $\mathbb Q^2$) in which every subgroup of rank 1 is free. (I assume you intended $a\neq 0$ in the question; otherwise the only such group would be the zero group.) In fact, such a rank-2 group can be so far from free that its quotient by any pure rank-1 subgroup is divisible. That result is due to L. Fuchs and F. Loonstra in "On the cancellation of modules in direct sums over Dedekind domains" (Indag. Math. 33 (1971) 163-169).

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The journal I cited has two titles, "Indagationes Mathematicae" and "Koninklijke Nederlandse Akademie van Wetenschappen, Proceedings, Series A, Mathematical Sciences." For one of these, the volume number is the 33 in my answer; for the other it seems to be 74. The result I quoted is part of Lemma 2 on page 168. – Andreas Blass Mar 8 '12 at 16:31

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