Prove that $(\mathbb{Z}/n\mathbb{Z})/m(\mathbb{Z}/n\mathbb{Z})=(\mathbb{Z}/n\mathbb{Z})/(\gcd(m,n)\mathbb{Z}/n\mathbb{Z})$

Question

I assume it's very easy to explain.

I want to prove that: $(\mathbb{Z}/n\mathbb{Z})/m(\mathbb{Z}/n\mathbb{Z})=(\mathbb{Z}/n\mathbb{Z})/(\gcd(m,n)\mathbb{Z}/n\mathbb{Z})$.

Here $m,n$ are integers $>0$ and $\gcd(m,n)$ is the greatest common divisor of $m$ and $n$.

I am referring to the second answer here: Calculating Ext and Tor for $\mathbb Z/m\mathbb Z$ and $\mathbb Z/n\mathbb Z$. The only step I can't see it the one above and I can't comment in the answer there directly.

Answer 1

We can show that $m(\Bbb Z/n\Bbb Z) = \text{gcd}(m, n)\Bbb Z/n\Bbb Z$.

"$\subseteq$" is obvious.

Suppose $[x] \in \text{gcd}(m, n)\Bbb Z/n\Bbb Z$. Let $g = \text{gcd}(m, n)$. So, $g | x$, suppose $x = gk$. By Bezout's identity, there exist integers $a$ and $b$ such that $g = am + bn$. So, $x = k(am + bn)$. Reducing modulo $n$, we get, $[x] = [kam] \in m(\Bbb Z/\Bbb n\Bbb Z)$. Thus, $m(\Bbb Z/n\Bbb Z) = \text{gcd}(m, n)\Bbb Z/n\Bbb Z$.