# Order of Group Element equals Order of Coprime Power

## Contents

## Theorem

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

Let $g \in G$ be an element of $g$.

Let $\order g$ denote the order of $g$ in $G$.

Then:

- $\forall m \in \Z: \order {g^m} = \order g \iff m \perp \order g$

where:

- $g^m$ denotes the $m$th power of $g$ in $G$
- $\perp$ denotes coprimality.

### Corollary

Let $H \le G$ be a subgroup of $G$.

Let $\order g = n$.

Let $g^m \in H$.

Let $m$ and $n$ be coprime.

Then $g \in H$.

## Proof 1

Let $\left\lvert{g}\right\rvert = n$.

Then from Order of Power of Group Element:

- $\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{{m, n}\right\}}$

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

Thus:

- $\left\lvert{g^m}\right\rvert = \left\lvert{g}\right\rvert \iff \gcd \left\{{m, n}\right\} = 1$

The result follows by definition of coprime integers.

$\blacksquare$

## Proof 2

Let $\left|{g}\right| = n$.

### Necessary Condition

Let $m \perp n$.

Then by Bézout's Lemma:

- $\exists x, y \in \Z: x m + y n = 1$

Let $h = g^m$.

Then:

- $h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$

and so $g$ is also a power of $h$.

Hence from Order of Group Element not less than Order of Power:

- $\left|{g}\right| \le \left|{h}\right| \le \left|{g}\right|$

and it follows that: $\left|{g}\right| = \left|{h}\right|$

$\Box$