Order of Power of Group Element

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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 $g$
\(\displaystyle \) \(=\) \(\displaystyle e\)

By Element to Power of Multiple of Order is Identity:

$\order {g^m} \divides n'$.


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

By Bézout's Lemma:

$\exists x, y \in \Z: m x + n y = d$
\(\displaystyle g^{d n''}\) \(=\) \(\displaystyle g^{\paren {m x + n y} n''}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {g^{m n''} }^x \paren {g^n}^{y n''}\)
\(\displaystyle \) \(=\) \(\displaystyle e^x \paren {g^n}^{y n''}\) $n''$ 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:

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


Sources