Abelian group Rank ιs equal to the dimension of the tensor product

Question

If $G$ is a finitely generated abelian group then why it's rank is ιs equal to the dimension of $G\otimes_\mathbb{Z} \mathbb{Q}$ as a vector space over $\mathbb{Q}$?

Thank you in advance.

Answer 1

Hint: For a finite product $$ C_1\times C_2\times\cdots\times C_n $$ of cyclic groups (each of them either finite or infinite), tensoring with $\Bbb Q$ will kill the finite factors and make the infinite factors into copies of $\Bbb Q$.

Answer 2

That is because any finitely generated abelian group of rank $r$ is the direct sum of its torsion subgroup $T$ and a torsionfree finitely generated subgroup. The latter is a free group $L$ of rank $r$, and $$T\otimes_{\mathbf Z}\mathbf Q=\{\mkern 1mu0\mkern1mu\},\quad\text{whereas}\quad L\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf Q^r.$$