Q: How to show a Hom group is isomorphic to some modulo group? Let $\mathbb{Z}_{n}[m]=\{a\in{\mathbb{Z}_{n}|ma=0\}}$, where $m$ is a positive integer. Then how do we establish the following?
$$
\mathbb{Z}_{n}[m]\cong{\mathbb{Z}_{\gcd(m,n)}}
$$

A: Let $\mathbb{Z}_{n}[m]$ be as you defined. Consider $da,d=gcd(m,n)$. Since we know $ma=0,na=0$, then we know $da=0$ by Euclidean algorithm. Therefore there is a map $\mathbb{Z}_{n}[m]\rightarrow \mathbb{Z}_{d}$. 
We claim this map is surjective. Let $h\in \mathbb{Z}_{d}$. I claim that $h\frac{n}{d}\in \mathbb{Z}_{n}[m]$ as well. This is not well defined as $h\in \mathbb{Z}_{d}$, what I really meant is $h\frac{n}{d}$'s value as a number in $\mathbb{Z}$. But this is automatic since $n|mh\frac{n}{d}$ is the same as $1|h\frac{m}{d}$, which we know it does by $d|m$. 
