Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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

1 Answer 1

up vote 12 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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.