Order of Group Element equals Order of Coprime Power

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 $\order g = n$.

Then from Order of Power of Group Element:

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

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

Thus:

$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$

The result follows by definition of coprime integers.

$\blacksquare$

Proof 2

Let $\order g = n$.

Necessary Condition

Let $m \perp n$.

Then by Bézout's Identity:

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

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:

$\order g \le \order h \le \order g$

and it follows that:

$\order g = \order h$

$\Box$

Sufficient Condition

We prove the contrapositive.

Suppose $m \not \perp n$.

Then:

$\exists d \in \Z_{>1}: \exists a, b \in \Z: m = a d \land n = b d$

Let $h = g^m$.

Then:

$h^b = \paren {g^m}^{n/d} = g^{a n} = \paren {g^n}^a = e^a = e$

Hence:

$\order h \le b < n$

and it follows that:

$\order g \ne \order h$

$\blacksquare$