Order of Power of Group Element

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $g \in G$ be an element of $G$ such that:
 * $\left|{g}\right| = n$

where $\left|{g}\right|$ denotes the order of $g$.

Then:
 * $\forall m \in \Z: \left|{g^m}\right| = \dfrac n {\gcd \left\{{m, n}\right\}}$

where $\gcd \left\{{m, n}\right\}$ denotes the greatest common divisor of $m$ and $n$.

Proof
Let $\gcd \left\{{m, n}\right\}=d$.

From Divide by GCD for Coprime Integers: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$.

Then:

By Element to Power of Multiple of Order is Identity:
 * $\left\vert{g^m}\right\vert \mathrel \backslash n'$.

$\left\vert{g^m}\right\vert = n'' < n'$.

By Bézout's Lemma:


 * $\exists x, y \in \Z: m x + n y = d$

But $d n'' < d n' = n$, contradicting the fact that $n$ is the order of $g$.

Therefore:
 * $\left\vert{g^m}\right\vert = n'$

Recalling the definition of $n'$:


 * $\left\vert{g^m}\right\vert = \dfrac n {\gcd \left\{{m, n}\right\}}$

as required.