Let $G$ be an abelian element. If $G$ is a f.g. $\mathbb{Z}$-module, then $Hom(G,\mathbb{Z})$ is either a free $\mathbb{Z}$-module or $0$. But what if $G$ is not f.g.? Can it have torsion elements?
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What do you think? What did you try? How do you prove that $\text{Hom}(G, \mathbb{Z})$ is free for finitely generated $G$? – Oct 15 '18 at 14:57
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The group $\operatorname{Hom}(G,\mathbb{Z})$ embeds in $\mathbb{Z}^G$, direct product of copies of $\mathbb{Z}$.
egreg
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The addition on $Hom (A, B)$ is pointwise. Therefore if $f\in Hom (A, B)$ is of finite order then so is $f(x)$ for all $x\in A$.
In the particular case of $B=\mathbb{Z}$ (or any other torsion free group) this implies that $f(x)=0$ for all $x\in A$, i.e. $f$ is the neutral element.
freakish
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It cannot have torsion elements, which is what the other answers address.
But it may not be free. For instance, if $G:=\bigoplus_{\mathbb{N}}\mathbb{Z}$, then $\mathrm{Hom}(G;\mathbb{Z}) \simeq \prod_{\mathbb{N}}\mathbb{Z}$, which is not free.
Aloizio Macedo
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