# Questions tagged [ordered-groups]

Groups (possibly semigroups) endowed with possibly left/right/bi-invariant partial/total orderings. Study of such orders on groups.

68 questions
Filter by
Sorted by
Tagged with
78 views

### Partial orders on $\mathbb{N}^m$ compatible with addition

I'm looking for a classification (or just non-trivial examples) of partial orders on monoid $\mathbb{N}^{m}$ that are compatible with addition. That is, partial orders $\leq$ satisfying two additional ...
240 views

292 views

401 views

### An equivariant Hahn embedding theorem?

The Hahn Embedding Theorem asserts that for any (linearly) ordered abelian group $\Lambda$, there exists a linearly ordered indexing set $\Omega$ such that $\Lambda$ admits an order-preserving group ...
342 views

### Do all right orderable groups have the Haagerup property?

Do all right orderable groups have the Haagerup property? Recall that a group is right orderable if there exists a total order $\leq$ on it such that $a\leq b\Rightarrow ac\leq bc$. This property is ...
298 views

### Amenable groups acting on the real line, that are not subexponentially-amenable

In the literature, there are several examples of solvable groups acting faithfully by order-preserving homeomorphisms of the real line. There are also examples of groups of intermediate growth with ...
202 views

Definition 1: Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\... 1answer 175 views ### Totally right preorderable groups Are there any known non-trivial sufficient conditions, or full characterizations, of a totally right-preorderable group? More precisely: totally right-preorderable: has a non-trivial total right-... 1answer 370 views ### Bi-orderability of Baumslag-Solitar group$\langle a,b \mid a^{-1} b^m a = b^n\rangle$and of$\langle a,b \mid a^{-1} b a^m = b^n\rangle$We say that a group$(A, \cdot)$is bi-orderable if there exists a total order$\preceq$on$A$such that$xz \prec yz$and$zx \prec zy$for all$x,y,z \in A$with$x \prec y$. Let$m,n$be non-zero ... 1answer 367 views ### Is$x + y \ne y+nx$for$x \ne 0$and$n \ge 2$(in an ordered group)? Let$(A, +, \preceq)$be an ordered group, namely$(A, +)$is a group and$\preceq$is a total order on$A$such that$x + z \prec y + z$and$z + x \prec z + y$for all$x,y,z \in A$with$x \prec y$.... 1answer 190 views ### Extensions of partial orders to linear orders on (nonabelian) groups If$G$is a group with a (left) linear order, does every (left) partial order on$G$extend to a (left) linear order? The answer is affirmative on abelian groups, where being torsion-free is ... 2answers 270 views ### Conditions for a group to be lattice-ordered Given a set$S$with a group operation$\cdot$and a lattice ordering$\leq$, I wish to know when we can say that$\cdot$preserves$\leq$, i.e.$(x\vee y)z=xz\vee yz$and similarly for meets. ... 2answers 863 views ### Every abelian torsion-free group is strictly totally orderable (via the compactness theorem) Let$\mathbb G = (G, +)$be a group. We say that$\mathbb G$is strictly totally orderable (others would say bi-orderable) if there exists a total order$\preceq$on$G$such that$x+z \prec y + z$... 3answers 705 views ### Group action on the real line I was wondering about the following question: if you have a faithful action of a group$G$on the real line$\mathbb{R}$by orientation-preserving homeomorphisms, it is easy to construct a new action ... 1answer 364 views ### Ordered groups - examples Let$G=\operatorname{BS}(m,n)$denote the Baumslag–Solitar group defined by the presentation$\langle a,b: b^m a=a b^n\rangle$. We assume that$G$is non-abelian, i.e.,$m,n\in\mathbb{Z}\...
I understand why the set of natural numbers $\mathbb N = \{ 0, 1, 2, \cdots \}$ is equipped with a total order. Indeed, every monoid has a pre-order, where n' \succeq n \quad \mathrm{if~and~only~if} ...
Let $H$ be a subgroup of a lattice-ordered group $G$. Suppose that $H$ with the induced order is a lattice (but a priori not a sublattice), so that $H$ is a lattice-ordered group too. For $a, b\in H$, ...