My question is 

Let $A$ be a finitely generated abelian group. The structure theorem says that $A$ is isomorphic to $F \times T$, where $F$ is isomorphic $\mathbb Z^m$, some $m \geq 0$, and $T $ is isomorphic to $\mathbb Z^{n\_1} \times\dots\times \mathbb Z^{n\_k}$ , $n\_i \geq 2$. Show that $\operatorname{Hom}(A,\mathbb Z)$ is a free abelian group. 

I'm having a problem of understanding exactly what a free abelian group is and how I'm suppose to show that it is a free abelian group.