Is the following question obvious ?

Let $G$ be an abelian group, such that for any finite abelian group $A$, we have $G\otimes_{\mathbf{Z}}A=0$, does it mean that $G$ is a $\mathbf{Q}$-vector space ?


closed as off-topic by Eevee Trainer, Cesareo, YiFan, Song, Antinous Mar 12 at 13:36

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You have $G\otimes(\Bbb Z/n\Bbb Z)=0$ for all $n\in \Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then $G\otimes(\Bbb Z/n\Bbb Z)=0$ and so $G\otimes A=0$ for all finitely generated Abelian groups.

But not all divisible Abelian groups are $\Bbb Q$-modules: they may have torsion. As an example, let $G=\Bbb Q/\Bbb Z$.


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