Linearly ordered group

A group with a two-sided invariant total order. Thus it is a lattice-ordered group whose join and meet choose the larger and smaller element. An order on only one side is not sufficient here. Every linearly ordered group is torsion-free: a positive element has strictly increasing positive powers, and a negative element has a positive inverse.

An ordered group is Archimedean when $a\ge 1$ and $a^n\le b$ for all positive integers force $a=1$. In an additive Abelian group this excludes a positive element whose every integer multiple stays below one fixed bound. This generalizes the Archimedean property of the real numbers.

An Archimedean ordered group with a two-sided invariant total order is commutative. If there is a least positive element, its cofinal powers exhaust the group. Otherwise each $c>1$ has $d>1$ with $d^2<c$. To see this, choose $1<s<r<c$ and use $r{s}^{-1}$ if its square is below $r$; if not, the inequality $(r{s}^{-1}{)}^2\ge r$ gives $s^2\le r<c$. If positive $g,h$ satisfied $gh<hg$, put $c=(gh{)}^{-1}hg$ and choose such a $d$. Cofinality gives $d^m\le g<d^{m+1}$ and $d^n\le h<d^{n+1}$. Two-sided order invariance then gives both $hg<d^{m+n+2}$ and $hg=ghc>d^{m+n+2}$, a contradiction. Replacing negative elements by their inverses reduces arbitrary commutativity to the positive case.

For ordered Abelian groups, compare tuples at their first differing coordinate. This gives a two-sided invariant total order on their direct product. The additive group ${\mathbb{Z}}^2$ with its first coordinate dominant is not an Archimedean ordered group, since $n(0,1)<(1,0)$ for every positive integer $n$.

For a torsion-free abelian group of finite rational rank $r$, each convex subgroup of an ordered group is pure: if $ng\in C$ for a positive integer $n$, then $g$ lies between zero and $ng$ or their negatives and hence belongs to $C$. A proper inclusion of convex subgroups therefore increases rational rank. They form a chain, so there are at most $r+1$, including zero and the whole group. A lexicographically ordered copy of ${\mathbb{Z}}^r$ attains the bound.

A group admits a two-sided invariant total order exactly when every finite list of nonidentity elements can be signed so that the conjugation-invariant semigroup generated by the signed elements excludes the identity. A compactness theorem argument supplies a compatible global sign assignment. Omitting the conjugates changes the criterion to a different orderability problem.