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.
Best Answer
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.