Order of Group Element not less than Order of Power/Proof 2
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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
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.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(iv)}$