Group homomorphism/Z to Z mod d/Directly/Example

Let ${}d\in \mathbb {N} _{+}$. We consider the set

${}\mathbb {Z} /(d)=\{0,1,\ldots ,d-1\}\,,$

together with the addition described in exercise, which makes it a group. The mapping

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(d)$

that sends an integer number ${}n$ to its remainder after division by ${}d$ is a group homomorphism. For, if ${}m=ad+r$ and ${}n=bd+s$ are given with ${}0\leq r,s<d$, then

${}m+n=(a+b)d+r+s\,.$

Here, it may happen that ${}r+s\geq d$. In this case,

${}\varphi (m+n)=r+s-d\,,$

and this coincides with the addition of ${}r$ and ${}s$ in ${}\mathbb {Z} /(d)$. This mapping is surjective, but not injective.