Morphism from Integers to Group

Theorem
Let $G$ be a group whose identity is $e$.

Let $g \in G$.

Let $\phi: \Z \to G$ be the mapping defined as:


 * $\forall n \in \Z: \phi \left({n}\right) = g^n$.

Then:


 * If $g$ has infinite order, then $\phi$ is an isomorphism from $\left({\Z, +}\right)$ to $\left \langle{g}\right \rangle$.
 * If $g$ has finite order such that $\left|{g}\right| = m$, then $\phi$ is an epimorphism from $\left({\Z, +}\right)$ to $\left \langle{g}\right \rangle$ whose kernel is the principal ideal $\left({m}\right)$.
 * Thus $\left \langle{g}\right \rangle$ is isomorphic to $\left({\Z, +}\right)$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.

Proof
By Epimorphism from Integers to a Cyclic Group, $\phi$ is an epimorphism from $\left({\Z, +}\right)$ onto $\left \langle{g}\right \rangle$.

The kernel $K$ of $G$ is a subgroup of $\left({\Z, +}\right)$.

Therefore by Subgroup of Integers is Ideal and Principal Ideals of Integers, $\exists m \in \N^*: K = \left({m}\right)$.

Thus $\left \langle{g}\right \rangle \cong \left({\Z_m, +}\right)$.

By Canonical Epimorphism from Integers by Principal Ideal:


 * $\forall m \in \N^*: \left|{\Z_m}\right| = m$

So, if $\left \langle{g}\right \rangle$ is finite, and if $\left \langle{g}\right \rangle \cong \left({\Z_m, +}\right)$, then $m = \left|{g}\right|$.

Furthermore, $m$ is the smallest (strictly) positive integer such that $g^m = e$, since $m$ is the smallest (strictly) positive integer in $\left({m}\right)$ from Principal Ideals of Integers.

If $\left \langle{g}\right \rangle$ is infinite, then $m = 0$ and so $\phi$ is an isomorphism from $\left({\Z, +}\right)$ onto $\left \langle{g}\right \rangle$.