Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

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 $Hom(A,\mathbb Z)=\mathbb Z^n$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.