User:Dfeuer/Archimedean Totally Ordered Group is Abelian/Lemma 1

Theorem

Let $G$ be an Archimedean totally ordered group with a smallest strictly positive element, $\epsilon$.

Then $G$ is the infinite cyclic group generated by $\epsilon$, with the natural ordering.

Let $e$ be the identity element of $G$.

Let $x\in G$.

Since $G$ is totally ordered, either $x=e$, $x > e$, or $x < e$.

$e$ is in the group generated by $\epsilon$ by ....

Suppose that $x>e$.

Then $x\ge \epsilon$ by the premise.

Since $G$ is Archimedean, there is a smallest $n \in \N_{>0}$ such that $\epsilon^n \ge x$.

Suppose for the sake of contradiction that $\epsilon^n > x$.

Then $\epsilon^n x^{-1} > e$.

Thus by our choice of $\epsilon$, $\epsilon^n x^{-1} \ge \epsilon$, so $\epsilon^{n-1} \ge x$, a contradiction.

Thus we conclude that $x = \epsilon^n$

Suppose instead that $x < e$.

Then $x^{-1} > e$, so for some $n \in\N_{>0}$, $x^{-1}=\epsilon^n$.

Thus $x = \epsilon^{-n}$.

Finally, since $\epsilon > e$, $x\epsilon > x$ for each $x\in G$, so $G$ has the natural ordering.

$\blacksquare$