Morphism from Integers to Group

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

Let $$g \in G$$.

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

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

Then:


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

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

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

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

By Canonical Epimorphism from Integers by Principal Ideal:

$$\forall m \in \mathbb{N}^*: \left|{\mathbb{Z}_m}\right| = m$$

So, if $$\left \langle{g}\right \rangle$$ is finite, and if $$\left \langle{g}\right \rangle \cong \left({\mathbb{Z}_m, +}\right)$$, 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 $$\left({m}\right)$$

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