Let $A$ be a finitely generated Abelian group. Let $tA$ denote the torsion subgroup. Prove that $A$ has a subgroup isomorphic to $A/tA$.
I know that $A/tA$ is torsion free, so my thinking so far has been to take the non-torsion elements of $A$ and consider the subgroup generated by them. Then show that subgroup and $A/tA$ are isomorphic. (I'm doing all this in additive notation, just so you know.)
Let $A'=\{a\in A\,|\, na\neq 0 \text{ for all } n\in \mathbb{Z}^+\}$ and consider the subgroup $B=\langle A'\rangle$. Since $A$ is finitely generated, so is $B$. Let the generators be $\{a_1, \ldots , a_n\}$.
Restrict the natural projection $\pi: A\to A/tA$ to a map $\varphi: B\to A/tA$. Since $\pi$ is a well-defined homomorphism, so is $\phi$.
Now I'd like to show that $\varphi$ is bijective, but I'm a little stuck. First of all, I haven't convinced myself it's even true. Secondly, for injectivity, does it suffice to only consider two generators $a_i$ and $a_j$ and suppose their images are distinct and go from there? Or do I have to take two arbitrary elements of $B$?
Any help would be great, thanks!