# Powers of Infinite Order Element

## 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.

$\blacksquare$