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It is to be shown that Hom$_\mathbb Z(\frac{\mathbb Z}{n\mathbb Z},\frac{\mathbb Z}{m\mathbb Z})\simeq \frac{\mathbb Z}{(n,m)\mathbb Z} $, i.e., isomorphic as modules over $\mathbb Z$.

I am very new to module theory. I tried to prove it the following way:

Let $f:\text{Hom}_\mathbb Z(\frac{\mathbb Z}{n\mathbb Z},\frac{\mathbb Z}{m\mathbb Z})\to \frac{\mathbb Z}{(n,m)\mathbb Z}$ be defined by $f(\phi)=\phi(\bar 1)$ for every $\phi$ in the domain.

$\phi$ is a group homomorphism as well and therefore $|\phi(\bar 1)|| (n,m)$, which implies that $\phi(\bar 1)\in \frac{\mathbb Z}{(n,m)\mathbb Z}.$ This $f$ is clearly well defined.
For any $\phi, \psi$ in domain of $f$ and any $z\in \mathbb Z$,

$f(\psi+\phi)=(\psi+\phi)(\bar 1)=\psi(\bar 1)+\phi(\bar 1)=f(\psi)+f(\phi)$ and $f(z\phi)=z\phi(\bar 1)=zf(\phi)$.

It follows that $f$ is a homomorphism. $f$ is surjective by its construction. Since mapping $\bar 1\in \mathbb Z/n\mathbb Z$ determines $\phi$, it follows that $f$ is 1-1. Therefore, $f$ is a module isomorphism. This proves the result.

Is my proof correct? Thanks.

Koro
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    Yes the proof looks correct. If you want an alternative method without constructing functions you can use the well-known fact that $\text{hom}(M,N) = M^\otimes N$. Now the dual $M^$ is isomorphic to itself (in your example) and so the hom-module is isomorphic to the tensor product between Z/n and Z/m which is Z/(n,m). – Nicolas Bourbaki Oct 18 '22 at 07:33
  • @NicolasBourbaki: Thanks a lot for reviewing the proof :-). – Koro Oct 18 '22 at 07:38
  • $\phi(\bar 1)\in \frac{\mathbb Z}{m\mathbb Z},$ not $ \frac{\mathbb Z}{(n,m)\mathbb Z}.$ Your definition of $f$ and argument for its injectivity are unclear. – Anne Bauval Oct 18 '22 at 07:44
  • @AnneBauval: $\phi(\bar 1)$ is in $Z/mZ$ so $|\phi(\bar 1)|| m$. Since, $\phi$ is a module homomorphism, $\phi (\bar n)=0=n\phi(\bar 1)\implies |\phi(\bar 1)|| n$. Therefore, the order of $\phi (\bar 1)$ divides $(n,m)$. It follows that $(n,m)\phi(\bar 1)=0$. Then, I thought that since $\bar 2\in Z_2$ and $\bar 2\in Z_4$ as well, one can say that $\phi(\bar 1)\in Z/(n,m)Z$, which I don't know how to justify properly. (Contd.) – Koro Oct 18 '22 at 07:57
  • Assuming this issue to be resolved, $f(\phi)=f(\psi)\implies \phi(\bar 1)=\psi(\bar 1)\implies \bar z\phi(\bar 1)=\bar z\psi (\bar 1)\implies \phi (\bar z)=\psi(\bar z)$ for all $\bar z\in Z/n Z\implies \phi=\psi$. So $f$ is one-one. – Koro Oct 18 '22 at 07:57

1 Answers1

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Your proof does not make sense, as pointed in the comments.

However, you nearly proved that for any $\phi\in{\rm Hom}(\mathbb Z/n\mathbb Z,\mathbb Z/m\mathbb Z),$

$\phi(\bar1)$ belongs to the subgroup $G$ of $\mathbb Z/m\mathbb Z$ generated by $\frac m{(n,m)}\bmod m.$

Let $g:\mathbb Z/(n,m)\mathbb Z\to G$ be the isomorphism which sends $\bar1$ to that generator.

The map $$f:{\rm Hom}(\mathbb Z/n\mathbb Z,\mathbb Z/m\mathbb Z)\to\mathbb Z/(n,m)\mathbb Z,\;\phi\mapsto g^{-1}(\phi(\bar1))$$ is easily seen to be an isomorphism.

Anne Bauval
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