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

Question

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.

Answer 1

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 $$\text{Hom}(A,\mathbb Z)\cong \text{Hom}(\mathbb Z^n\times T,\mathbb Z)=\text{Hom}(\mathbb Z,\mathbb Z)^n\times \text{Hom}(T,\mathbb Z)$$

$\text{Hom}(T,\mathbb Z)=0$ because $T$ is torsion and $\text{Hom}(\mathbb Z,\mathbb Z)=\mathbb Z$. This means that $\text{Hom}(A,\mathbb Z)=\mathbb Z^n$.

Answer 2

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.