# $G\otimes \mathbb Q=0$ implies torsion group

let $G$ be an abelian group. suppose that $G\otimes \mathbb Q=0$. Does this imply that $G$ is necessarily a torsion group?

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Yes. Note that $G\otimes_\mathbb{Z}\mathbb{Q}\cong (\mathbb{Z}\setminus 0 )^{-1}G$ as $\mathbb{Z}$-modules. Now $\frac{g}{n}=\frac{0}{1}$ by definition means $zg=0$ in $G$ for some $z\in\mathbb{Z}\setminus 0$.
Just want to mention this is true in greater generality (and the proof is essentially the same): if $D$ is an integral domain with field of fractions $F$, and $M$ is any $D$-module, then the kernel of the natural map $M\rightarrow M\otimes F$ is the torsion submodule of $M$. – user641 May 20 '11 at 7:24