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. 🙂

Do check out some Recommended Books on Undergraduate Mathematics, and also download the free SG50 Scientific Pioneers Ebook, if you haven’t already.

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