Group Isomorphism/Examples/Congruence Modulo Initial Segment of Natural Numbers

Example of Group Isomorphism
Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:
 * $\N_{<m} = \set {0, 1, \ldots, m - 1}$

Let $\RR_m$ denote the equivalence relation:
 * $\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$

For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$:
 * $\eqclass a m := \set {a + z m: z \in \Z}$

Let $\Z_m$ be the set defined as:
 * $\Z_m := \set {\eqclass a m: a \in \N_{<m} }$

Let $+_\PP$ denote the operation induced on $\powerset \Z$ by integer addition.

Let $\phi_m: \struct {\N_m, +_m} \to \struct {\Z_m, +_\PP}$ be the mapping defined by:
 * $\forall a \in \N_m: \map {\phi_m} a = \eqclass a m$

where $\struct {\N_m, +_m}$ denotes the additive group of integers modulo $m$.

Then $\phi_m$ is a (group) isomorphism.

Proof
We have that:

demonstrating that $\phi_m$ is a homomorphism.

Then we have that:

Hence $\phi_m$ is injective.

We also have that $\phi_m$ is trivially surjective.

Hence $\phi_m$ is a bijection.

Hence the result.