Order of Group Element not less than Order of Power

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Theorem

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

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

Let $\left\lvert{g}\right\rvert$ denote the order of $g$ in $G$.


Then:

$\forall m \in \Z: \left\lvert{g^m}\right\rvert \le \left\lvert{g}\right\rvert$

where $g^m$ denotes the $m$th power of $g$ in $G$.


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 $n$.

The result follows from Greatest Common Divisor is at least 1.

$\blacksquare$


Proof 2

Let $g^n = e$.

Let $h = g^m$.

Then:

$h^n = g^{m n} = \paren {g^n}^m = e^m = e$

Hence by definition of order of group element:

$order h \le n$

Hence the result.

$\blacksquare$