## Zmn/Zm isomorphic to Zn

The following is a simple proof of why $\mathbb{Z}_{mn}/\mathbb{Z}_m\cong\mathbb{Z}_n$.

For instance $\mathbb{Z}_6/\mathbb{Z}_2\cong\mathbb{Z}_3$. Note that the tricky part is that $\mathbb{Z}_2$ is not actually the usual {0,1}, but rather {0,3} (considered as part of $\mathbb{Z}_6$). Hence the elements of $\mathbb{Z}_6/\mathbb{Z}_2$ are {0,3}, {1, 4}, {2, 5}, which can be seen to be isomorphic to $\mathbb{Z}_3$.

A sketch of a proof is as follows. Consider $\phi:\mathbb{Z}_{mn}/\mathbb{Z}_m\to \mathbb{Z}_n$, where $\mathbb{Z}_m=\{0,n,2n,\dots,(m-1)n\}$, defined by $\phi (a+\mathbb{Z}_m)=a$.

We may check that it is well-defined since if $a+\mathbb{Z}_m=a'+\mathbb{Z}_m$, then $a\equiv a' \pmod n$, and thus $\phi (a+\mathbb{Z}_m)=\phi (a'+\mathbb{Z}_m)$.

It is a fairly straightforward to check it is a homomorphism, \begin{aligned}\phi (a+\mathbb{Z}_m+a'+\mathbb{Z}_m)&=\phi (a+a'+\mathbb{Z}_m)\\ &=a+a'\\ &=\phi (a+\mathbb{Z}_m)+\phi(a'+\mathbb{Z}_m) \end{aligned}

Injectivity is clear since $\ker \phi=0+\mathbb{Z}_m$, and surjectivity is quite clear too.

Hence, this ends the proof. 🙂