Just to confirm that this is correct and is a result from first ring isomorphism theorem right?
The map I used is $f: n\Bbb Z \to \Bbb Z/(m/n\Bbb Z)$ by $f(kn)=k\mod m/n$.
This is apparently surjective, and $\ker{f}=\{kn,k\equiv 0 \pmod{m/n}\}=m\Bbb Z$.
But this seems to be wrong: consider $2\Bbb Z/10\Bbb Z=\{2,4,6,8,0\}$. The unity is $6$. But by the induced isomorphism above, $6=2\cdot3$ and $3 \mod 5(=10/2)=3$ instead of $1$ as required. What is wrong here?
Suppose $2Z/4Z\cong Z/2Z$ where $2Z/4Z:=\{0+4Z,2+4Z\}$. The only nonzero element satisfies $(2+4Z)^2=0+4Z.$ But the only nonzero element of $Z/2Z$ is $1$ and does not have this property. So the claim is false and the proof fail simply because it does not preserve multiplication.