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If $L$ is a finitely generated abelian group, then is $Hom(L,\mathbb{Z})$ a finitely generated abelian group?

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up vote 3 down vote accepted

Yes. In fact we can say even more: $\textrm{Hom}(L, \mathbb{Z})$ will be a finitely-generated free abelian group. This uses the fact that $\mathbb{Z}$ is a principal ideal domain. Indeed, take a finite presentation of $L$, say $$\mathbb{Z}^{\oplus a} \longrightarrow \mathbb{Z}^{\oplus b} \longrightarrow L \longrightarrow 0$$ and then apply the $\textrm{Hom}(-, \mathbb{Z})$ functor to obtain an exact sequence $$0 \longrightarrow \textrm{Hom}(L, \mathbb{Z}) \longrightarrow \textrm{Hom}(\mathbb{Z}^{\oplus b}, \mathbb{Z}) \longrightarrow \textrm{Hom}(\mathbb{Z}^{\oplus a}, \mathbb{Z})$$ but $\textrm{Hom}(\mathbb{Z}^{\oplus b}, \mathbb{Z}) \cong \mathbb{Z}^{\oplus b}$ naturally, and every subgroup of a finitely-generated free abelian group is again a finitely-generated free abelian group, by the classification of finitely-generated modules over a principal ideal domain.

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Let $T$ be a finite group then any homomorphism of it into $\mathbb Z$ is trivial because the elements of $F$ have finite order and those of $\mathbb Z$ don't.

Decompose $L$ into it's torsion and free parts $T \times \mathbb Z^n$ then $\text{Hom}(L,\mathbb Z)$ is isomorphic to $\text{Hom}(Z^n,\mathbb Z)$ because the $T$ part is trivial.

$\text{Hom}(\mathbb Z,\mathbb Z)$ is isomorphic to $\mathbb Z$ because every homomorphism is of the form $x \mapsto zx$ where $z$ ranges over elements of $\mathbb Z$.

The same holds for $\text{Hom}(\mathbb Z^n,\mathbb Z)$ it is simply isomorphic to $\mathbb Z^n$ since any such homomorphism is determined completely by where it's generators are taken, and they can be taken anywhere.

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