Proof of $\mathbb Z/n\mathbb Z\bigotimes_{\mathbb Z}\mathbb Z/m\mathbb Z \cong Hom(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)$ How can we prove that
$\DeclareMathOperator\Hom{Hom}\mathbb Z/n\mathbb Z\bigotimes_{\mathbb Z}\mathbb Z/m\mathbb Z \cong \Hom(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)$ without using the fact that $\mathbb Z/n\mathbb Z\bigotimes_{\mathbb Z}\mathbb Z/m\mathbb Z \cong \mathbb Z/d \mathbb Z$ and $\Hom(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)\cong \mathbb Z/d \mathbb Z$, where $d=$ GCD$(m,n)$? It's not too hard to prove the two latest isomorphisms but it would be interesting to derive one from another using the first isomorphism.
 A: Consider the short exact sequence $0\to\mathbb{Z}\xrightarrow{n} \mathbb{Z}\to \mathbb{Z}/n\to 0$, where the map $\mathbb{Z}\xrightarrow{n}\mathbb{Z}$ is multiplication by $n$. Applying the functors $-\otimes_\mathbb{Z}\mathbb{Z}/m$ and ${\rm Hom}(-,\mathbb{Z}/m)$ to it gives the two exact sequences:
$$
\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m\to \mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m\to 0\\
0\to {\rm Hom}_\mathbb{Z}(\mathbb{Z}/n,\mathbb{Z}/m)\to \mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m
$$
showing that $\mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m$ is the cokernel of $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$ and that ${\rm Hom}_\mathbb{Z}(\mathbb{Z}/n,\mathbb{Z}/m)$ is the kernel of the same map: $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$. The kernel and cokernel are cyclic abelian groups, so it suffices to show that they have the same cardinality. For completeness:
Let $d$ be the size of the cokernel, and $e = |{\rm im}(n)|$, the size of the image of $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$. Then by definition, $m/e = d$. On the other hand, the short exact sequence $0\to \ker(n)\to \mathbb{Z}/m\xrightarrow{n}{\rm im}(n)\to 0$ shows also that $m/|\ker(n)| = e$, whence $d = m/e = |\ker(n)|$.
Of course, with a little more work, you can then prove the greatest common divisor statement for both now.
