User:Dfeuer/Archimedean Totally Ordered Group is Abelian

Theorem

Let $G$ be an Archimedean totally ordered group.

Then $G$ is abelian.

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 $G$ be a non-trivial totally ordered group with no smallest strictly positive element.

Let $g$ be a strictly positive element of $G$.

Let $n \in \N_{>0}$.

Then there is a strictly positive $h \in G$ such that $h^n \le g$

If $G$ has a smallest strictly positive element then by User:Dfeuer/Archimedean Totally Ordered Group is Abelian/Lemma 1, $G$ is a cyclic group and thus abelian.

Suppose that $G$ has no smallest strictly positive element.

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

Suppose for the sake of contradiction that $a,b \in G$ and $ab \ne ba$.

Then clearly, neither $a$ nor $b$ is the identity, and $a \ne b$.

Then multiplying on the left and on the right by $a^{-1}$, $ba^{-1} \ne a^{-1}b$.

We can do similarly for $b$. Thus we can assume WLOG that $a > e$ and $b > e$.

By swapping $a$ with $b$ if necessary, we can ensure that $ab < ba$.

Let $c = b^{-1}a^{-1}ba$, so $abc = ba$.

Since $G$ is Archimedean, there are $n,m\in\N_{>0}$ such that

$(1)$	$:$				$\ds d^n < a \le d^{n+1} $
$(2)$	$:$				$\ds d^m \le b < d^{m+1} $

Then multiplying the first inequality by the second yields

$(3)\quad d^{n+m} < ab$

while multiplying the second by the first yields

$(4)\quad abc = ba < d^{m+n+2} < d^{m+n}c$

Multiplying $(4)$ on the right by $c^{-1}$ yields $ab < d^{m+n}$.

But we have already shown that $d^{m+n} < ab$, so we have reached a contradiction.

Thus we conclude that $ab = ba$.

As this holds for all $a, b \in G$, $G$ is abelian.

$\blacksquare$

\((1)\)	$:$				\(\ds d^n < a \le d^{n+1} \)
\((2)\)	$:$				\(\ds d^m \le b < d^{m+1} \)