Order of Group Element equals Order of Coprime Power

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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 $G$.


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


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


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 $n$.


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

The result follows by definition of coprime integers.


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 $h = g^m$.


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

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: $\left|{g}\right| = \left|{h}\right|$


Sufficient Condition