Tensor product of a finitely generated abelian group and the field of rational numbers

Let $G$ be a a finitely generated abelian group. Then $G\otimes_\mathbb{Z} \mathbb{Q} = 0$ if and only if $G$ is a finite group. The "if" part is easy. The "only if" part can be proved using the fundamental theorem of finitely generated abelian groups. Can we prove it without using the theorem?

Let $F= \{f_1, \ldots, f_n\} \subseteq G$ a generating set. As $f_i \otimes 1 = 0$ for each $i$, there is a $m_i \in \mathbb Z$ such that $m_if_i = 0$. Hence $$G \subseteq \left\{ \sum_{i=1}^n k_i f_i \biggm| k_i\in \mathbb Z, \left|k_i\right| \le \left|m_i\right| \right\}$$ and $G$ is finite.
Addendum: Consider the short exact sequence $0 \to K \to \mathbb Z \to G \to 0$ where $1 \mapsto f_i$ is the map from $\mathbb Z$ to $G$. Tensoring with $\mathbb Q$ gives us the short exact $K \otimes \def\Q{\mathbb Q}\Q \to \mathbb Q \to 0 \to 0$. Hence $K \otimes \mathbb Q \to \mathbb Q$ is onto, but $K$ is a subgroup of $\mathbb Z$, so $K \ne 0$, say $K = n \mathbb Z$ with $n \ne 0$, giving $nf_i = 0$.
• Could you explain why there exists such $m_i$? – Makoto Kato Dec 10 '13 at 9:39
• I'm afraid that the sequence $0 \to K \to \mathbb Z \to G \to 0$ is not exact in general. Regards, – Makoto Kato Dec 10 '13 at 21:13
Let $A$ be an abelian group, then $A\otimes \mathbb{Z}_m\cong A/mA$ as abelian groups. If $A$ is $\mathbb{Q}$, then clearly $\mathbb{Q}/m\mathbb{Q}$ is trivial.
As you request, the other direction is proved without using the fundamental theorem. Let $G$ be a finitely generated abelian group, then at least one of the generator has infinite order. Therefore $\mathbb{Z}\hookrightarrow G$, and $\mathbb{Q}\cong \mathbb{Q}\otimes\mathbb{Z}\hookrightarrow \mathbb{Q}\otimes G\ne0$.