Let $A$ be a finitely generated abelian group. Show that $\operatorname{Hom}(A,Z)$ is a free abelian group.

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.

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So your book/class doesn't define "free abelian group?" –  Thomas Andrews Dec 11 '13 at 23:51
A free abelian group is a group that is isomorphic to $\mathbb{Z}^n$ for some $n \in \mathbb{N}$. Remember also that $Hom(\cdot,\mathbb{Z})$ preserves finite products(in this case is $\mathbb{Z}$ but any other abelian group also works). –  sjvega Dec 11 '13 at 23:52
@ThomasAndrews No, not that I could find. I know that free abelian group and free Z-module mean the same thing. –  Lynn Dec 11 '13 at 23:55

From an epimorphism $\mathbb{Z}^n\to A$ we get the induced monomorphism $$\operatorname{Hom}(A,\mathbb{Z})\to\operatorname{Hom}(\mathbb{Z}^n,\mathbb{Z})$$ and $\operatorname{Hom}(\mathbb{Z}^n,\mathbb{Z})\cong\mathbb{Z}^n$. Every subgroup of a free abelian group is free.
Abelian group is free if its of the form $\bigoplus_I \mathbb Z$. If $I$ is finite then $\bigoplus_I \mathbb Z\cong \mathbb Z^{|I|}$. If $A$ is finitely generated abelian group then by structural theorem $A\cong \mathbb Z^n\times T$ where $T$ is torsion group. We have $$Hom(A,\mathbb Z)\cong Hom(\mathbb Z^n\times T,\mathbb Z)=Hom(\mathbb Z,\mathbb Z)^n\times Hom(T,\mathbb Z)$$
$Hom(T,\mathbb Z)=0$ because $T$ is torsion and $Hom(\mathbb Z,\mathbb Z)=\mathbb Z$. This means that $Hom(A,\mathbb Z)=\mathbb Z^n$.