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:

\(\ds \paren {g^m}^{n'}\) \(=\) \(\ds \paren {g^{d m'} }^{n'}\) Definition of $m'$
\(\ds \) \(=\) \(\ds \paren {g^{d n'} }^{m'}\)
\(\ds \) \(=\) \(\ds \paren {g^n}^{m'}\) Definition of $n'$
\(\ds \) \(=\) \(\ds e^{m'}\) $n$ is the order of $g$
\(\ds \) \(=\) \(\ds e\)

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 Identity:

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

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:

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


Sources

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