Is the dual $\mathbb{Z}$-module of $\mathbb{Z}/n\mathbb{Z}$, that is ${\rm Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})$ isomorphic to $\mathbb{Z}/n\mathbb{Z}$.
Looking at it briefly I think it should be since there are only n possibilities of for maps?


You can't map a torsion module into a torsion-free module, so it's isomorphic to $0$. For any map $\phi:\mathbb Z/ n\mathbb Z\to\mathbb Z$, $$ \phi(1)=m\implies \phi(0)=n\cdot \phi(1)=nm=0\quad\forall n\in\mathbb Z\implies m=0. $$

For the set $\operatorname{Hom}_\mathbb Z(\mathbb Z,\mathbb Z/n\mathbb Z)$, maps are completely determined by where $1$ is sent. You can send $1$ to any element since $\mathbb Z$ is the free abelian group on one element.

That is, for any set morphism $\{1\}\to A$, where $A$ is an abelian group, there is a unique extension to an abelian group homomorphism (i.e. a $\mathbb Z$-module homomorphism) $\phi:\mathbb Z\to A$. This tells you about $\operatorname{Hom}_\mathbb Z(\mathbb Z,A)$ in general.

  • $\begingroup$ What about the converse then, that is $Hom_Z(Z,Z/nZ)$? $\endgroup$ – AIM_BLB Feb 14 '14 at 0:40
  • $\begingroup$ Check the addition. $\endgroup$ – Ian Coley Feb 14 '14 at 0:53

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