The following proof should show that $\phi$ is a homomorphism, by making use of a modular arithmetic property:
\begin{equation} (A+B) \mod {C} = (A \text{ mod C} + B \text{ mod C}) \mod {C} \end{equation}
I would like to get some feedback if the below proof makes sense and if I used the modular arithmetic property in a correct way.
Proposition:
For any integer $n \ge 2$, the mapping $\phi$ is a homomorphism.
\begin{align*} \phi: (\mathbb{Z}, +) &\longrightarrow (\mathbb{Z}_n, +_n) \\ k &\longrightarrow k_{\text{mod}\ n} \end{align*}
(Note that the notation $k_{\text{mod}\ n}$ is the remainder of $k$ divided by $n$.)
Proof:
- Let $n$ be an integer with $n \ge 2$, and let $\phi$ be the mapping defined above. Let $r, s \in \mathbb{Z}$. Now:
\begin{align} \phi(r) +_n \phi(s) &= r_{\text{mod}\ n} +_n s_{\text{mod}\ n} \nonumber \\ &= (r \text{ mod n} + s \text{ mod n}) &\mod {n} \tag{1} \\ &= (r + s) &\mod {n} \tag{2} \end{align}
$(1)$ is by definition of $+_n$ and $k_{\text{mod}\ n}$, then $(2)$ holds because of the addition property of modular arithmetic.
- Also, we find $\phi(r + s) = (r + s) \mod n$.
- Thus $\phi(r) +_n \phi(s) = \phi(r + s)$ is true, which shows that $\phi$ is a homomorphism.