# Order of Power of Group Element

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $g \in G$ be an element of $G$ such that:

$\order g = n$

where $\order g$ denotes the order of $g$.

Then:

$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n} }$

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

## Proof

Let $\gcd \set {m, n} = d$.

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

Then:

 $\displaystyle \paren {g^m}^{n'}$ $=$ $\displaystyle \paren {g^{d m'} }^{n'}$ Definition of $m'$ $\displaystyle$ $=$ $\displaystyle \paren {g^{d n'} }^{m'}$ $\displaystyle$ $=$ $\displaystyle \paren {g^n}^{m'}$ Definition of $n'$ $\displaystyle$ $=$ $\displaystyle e^{m'}$ $n$ is the order of $g$ $\displaystyle$ $=$ $\displaystyle e$
$\order {g^m} \divides n'$.

Aiming for a contradiction, suppose $\order {g^m} = n'' < n'$.

$\exists x, y \in \Z: m x + n y = d$
 $\displaystyle g^{d n''}$ $=$ $\displaystyle g^{\paren {m x + n y} n''}$ $\displaystyle$ $=$ $\displaystyle \paren {g^{m n''} }^x \paren {g^n}^{y n''}$ $\displaystyle$ $=$ $\displaystyle e^x \paren {g^n}^{y n''}$ $n''$ is the order of $g^m$ $\displaystyle$ $=$ $\displaystyle e^x e^{y n''}$ $n$ is the order of $g$ $\displaystyle$ $=$ $\displaystyle e$

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

Therefore:

$\order {g^m} = n'$

Recalling the definition of $n'$:

$\order {g^m} = \dfrac n {\gcd \set {m, n} }$

as required.

$\blacksquare$

## Examples

### Order of Powers of $x$ when $\order x= 20$

Let $G$ be a group.

Let $x \in G$ be such that:

$\order x = 20$

where $\order x$ denotes the order of $x$ in $G$.

Then:

 $\text {(1)}: \quad$ $\displaystyle \order {x^4}$ $=$ $\displaystyle 5$ $\text {(2)}: \quad$ $\displaystyle \order {x^{10} }$ $=$ $\displaystyle 2$ $\text {(3)}: \quad$ $\displaystyle \order {x^{12} }$ $=$ $\displaystyle 5$