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:

$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$

Let $m, n \in Z$. Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle a^m$	$=$	$\displaystyle a^n$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle a^m \left({a^n}\right)^{-1}$	$=$	$\displaystyle e \land a^n \left({a^m}\right)^{-1} = e$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle a^{m-n}$	$=$	$\displaystyle a^{n-m} = e$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle m - n$	$=$	$\displaystyle 0 \land n - m = 0$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle m$	$=$	$\displaystyle n$	$\displaystyle $	$\displaystyle $	$\displaystyle $

The result follows from transposition.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a^m\)	\(=\)	\(\displaystyle a^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a^m \left({a^n}\right)^{-1}\)	\(=\)	\(\displaystyle e \land a^n \left({a^m}\right)^{-1} = e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a^{m-n}\)	\(=\)	\(\displaystyle a^{n-m} = e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle m - n\)	\(=\)	\(\displaystyle 0 \land n - m = 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle m\)	\(=\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)