# Morphism from Integers to Group

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## Theorem

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

Let $g \in G$.

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

$\forall n \in \Z: \map \phi n = g^n$.

Then:

If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$.
If $g$ has finite order such that $\order g = m$, then $\phi$ is a group epimorphism from $\struct {\Z, +}$ to $\gen g$ whose kernel is the principal ideal $\paren m$.
Thus $\gen g$ is isomorphic to $\struct {\Z, +}$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.

## Proof

By Epimorphism from Integers to Cyclic Group, $\phi$ is an epimorphism from $\struct {\Z, +}$ onto $\gen g$.

By Kernel of Group Homomorphism is Subgroup, the kernel $K$ of $G$ is a subgroup of $\struct {\Z, +}$.

Therefore by Subgroup of Integers is Ideal and Principal Ideals of Integers, $\exists m \in \N_{>0}: K = \paren m$.

Thus $\gen g \rangle \cong \struct {\Z, +}$.

$\forall m \in \N_{>0}: \order {\Z_m} = m$

So, if $\gen g$ is finite, and if $\gen g \cong \struct {\Z, +}$, then $m = \order g$.

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

If $\gen g$ is infinite, then $m = 0$ and so $\phi$ is a (group) isomorphism from $\struct {\Z, +}$ onto $\gen g$.

$\blacksquare$