Powers of Infinite Order Element
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Theorem
Let $G$ be a group whose identity is $e$.
Let $a \in G$ have infinite order in $G$.
Then:
- $\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
Proof
Let $m, n \in \Z$.
Then:
\(\ds a^m\) | \(=\) | \(\ds a^n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^m \paren {a^n}^{-1}\) | \(=\) | \(\ds e\) | |||||||||||
\(\, \ds \land \, \) | \(\ds a^n \paren {a^m}^{-1}\) | \(=\) | \(\ds e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{m - n}\) | \(=\) | \(\ds a^{n - m} = e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{\size {m - n} }\) | \(=\) | \(\ds e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {m - n}\) | \(=\) | \(\ds 0\) | Definition of Infinite Order Element: there exists no $x \in \Z_{>0}$ such that $a^x = e$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds n\) |
The result follows from transposition.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 38.2$ Period of an element