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:

\(\displaystyle a^m\) \(=\) \(\displaystyle a^n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^m \paren {a^n}^{-1}\) \(=\) \(\displaystyle e\)
\(\, \displaystyle \land \, \) \(\displaystyle a^n \paren {a^m}^{-1}\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{m - n}\) \(=\) \(\displaystyle a^{n - m} = e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{\size {m - n} }\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {m - n}\) \(=\) \(\displaystyle 0\) Definition of Infinite Order Element: there exists no $x \in \Z_{>0}$ such that $a^x = e$
\(\displaystyle \leadsto \ \ \) \(\displaystyle m\) \(=\) \(\displaystyle n\)

The result follows from transposition.

$\blacksquare$


Sources