Quotient of an Abelian group with its torsion subgroup 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!
 A: In fact more is true. Every finitely generated abelian group is a direct sum of its torsion part and the quotient by its torsion part. The structure theorem of finitely generated abelian groups is even stronger than this. 
You can find a proof of the weaker result in Peter May's notes (Theorem 1.11)
http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf
A: Here is the way I hinted at the solution: proving first that a finitely generated torsion-free abelian group is free, then use the fact that free groups are projective.
The argument for the first part is that used by Lang in his Algebra: suppose $A$ is a finitely generated torsion-free abelian group; we may assume it is nonzero. Pick $x_1,\ldots,x_n\in A$ that are maximal with respect to the property that if $\alpha_1x_1+\cdots +\alpha_nx_n=0$, then $\alpha_i=0$ for all $i$. Let $B$ be the subgroup generated by the $x_i$; note that $B$ is free abelian.
Let $y_1,\ldots,y_k$ be a generating set for $A$. For each $i$, 
there exists an integer $d_i\neq 0$ and  integers $\beta_1,\ldots,\beta_n$, not all zero, such that $d_iy +\beta_1x_1 + \cdots +\beta_n x_n=0$; that is, $d_iy_i\in B$. Letting $d=\mathrm{lcm}(d_1,\ldots,d_k)$, we have that $dA$ is contained in $B$. Since $A$ is torsion free, the map $A\to B$ given by $x\mapsto dx$ is an injection, so $A$ is isomorphic to a subgroup of a free abelian group, hence $A$ is free abelian.
Thus, if $A$ is a finitely generated group, and $tA$ is its torsion group, we know that $A/tA$ is finitely generated and torsion-free, hence free. Let $x_1,\ldots,x_k$ be a basis for $A/tA$. For each $i$, let $a_i\in A$ be any element that maps to $x_i$ under the canonical projection $\pi\colon A\to A/tA$. Since $A/tA$ is free, we have a (unique) group homomorphism $\rho\colon A/tA\to A$ that maps $x_i$ to $a_i$. Since $\pi\circ\rho = \mathrm{id}_{A/tA}$, then $\rho$ must be one-to-one, hence $A/tA$ is isomorphic to its image under $\rho$ inside of $A$. Thus, $A$ has a subgroup that is isomorphic to $A/tA$. 
A: Unfortunately the whole group will be generated by the nontorsion elements if the group is infinite. One thing you can do is use the classification theorem for finitely generated abelian groups which says that a finitely generated abelian group is of the form $Z_{m_1} \times ... \times Z_{m_n} \times Z \times ... \times Z$, with finitely many $Z$'s appearing on the right. In this case $tA$ is $Z_{m_1} \times ... \times Z_{m_n}$ and $A$ / $tA$ will be the $Z \times ... \times Z$ which is clearly a subgroup of $A$.
