Suppose $G$ is an abelian group totally bi-ordered by $\leq$ and by $\leq'$. Does it follow that $\leq'$ is either equal to or the converse of $\leq$?
2026-02-22 23:09:15.1771801755
Uniqueness of total orders on abelian groups
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No. For example, consider $\mathbb{Z} \oplus \mathbb{Z}$.
One ordering is given by $(a,b)> (c,d)$ iff $a>c$ or $a = c,b>d$.
Another ordering is given by $(a,b)> (c,d)$ iff $b>d$ or $b = d, a>c$.
Morally, if $\phi$ is an automorphism of an ordered abelian group $(G,\leq)$, then $\leq'$, defined by $a \leq' b$ iff $\phi(a) \leq \phi(b)$ is an ordered abelian group. This can yield plenty of ordered abelian group structures on a general $G$.