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Possible Duplicate:
How to show that for any abelian group $G$, $\\text{Hom}(\\mathbb{Z},G)$ is isomorphic to $G$.

Simple question - is it true that $\mbox{Hom}(\mathbb{Z},G) \simeq G$ (probably by Yoneda Lemma, which I struggle to understand!)

Edit: $G$ is an abelian group

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marked as duplicate by Zev Chonoles, Qiaochu Yuan Apr 28 '11 at 3:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 4 down vote accepted

Yes, it is true and much simpler than Yoneda's Lemma: a homomorphism is uniquely determined by its image on 1 $\in \mathbb{Z}$, since 1 generates $\mathbb{Z}$ as a group. By identifying a $\sigma\in \text{Hom}(\mathbb{Z},G)$ with $\sigma(1)\in G$, you get your isomorphism (bijectivity and the fact that this is a group homomorphism follow straight from group axioms).

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Short and simple, thanks! – Juan S Apr 28 '11 at 2:59

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