Let $m$ and $n$ be positive integers and $\otimes=\otimes_\mathbb{Z}$. Denote the GCD of $m$ and $n$ by $(m, n)$.
I proved $\mathbb{Z}_m \otimes \mathbb{Z}_n \cong \mathbb{Z}_{(m,n)}$ by considering a $\mathbb{Z}$-bilinear map $\iota: \mathbb{Z}_m \times \mathbb{Z}_n \rightarrow \mathbb{Z}_{(m,n)}$ and check the universal property of tensor product. Here $\iota$ is given by $\iota(\bar x, \bar y)=\overline{xy}$.
And I also know that for an abelian group $A$, we have $A \otimes \mathbb{Z}_{n} \cong A/nA$. Combining these two facts, we can conclude that $\mathbb{Z}_m /n\mathbb{Z}_m \cong \mathbb{Z}_{(m,n)}$
It seems that this result can be proved more easily. So I tried to find a group epimorphism $\phi \colon \mathbb{Z}_m \rightarrow \mathbb{Z}_{(m,n)}$ with kernel $n\mathbb{Z}_m$, for which we can apply the first isomorphism theorem.
At first I consider a map of the form $\overline x \mapsto \overline{kx}$ but I've stocked. Am I missing something obvious?
After a moment of thought, I realized that $\overline x \mapsto \overline{x}$ is the one I want to find.
I miss the fact $n\mathbb{Z}_m = d\mathbb{Z}_m$ , where $d = (m,n)$. This can be showed as follows:
$n\mathbb{Z}_m \subset d\mathbb{Z}_m$ : obvious
$d\mathbb{Z}_m \subset n\mathbb{Z}_m$ : This is because there exist $r, s \in \mathbb{Z}$ such that $rm+sn=d$.