# Questions tagged [ordered-groups]

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

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### Compatibility with multiplication of a cyclic order on a ring

I am copying my question from here: https://math.stackexchange.com/q/3233462/427611. Is it correct that $\mathbb Z/3\mathbb Z$ and $\mathbb Z/4\mathbb Z$ are the only rings with three or more ...
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### Is Thompson's group definably orderable?

Is Thompson's group $F$ definably left-orderable? definably bi-orderable? Orderability definitions: Recall that a group $G$ is left-orderable (resp. bi-orderable) if it admits a left-invariant (resp. ...
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### How can you order a free group?

A left order on a (discrete) group $G$ is a total order on $G$ satisfying $\forall g,h,k \in G: g < h \implies kg < kh$. A right order is defined symmetrically, and a biorder is an order that is ...
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### Characterization of Archimedean linearly ordered monoids

In this question, it is shown that all Archimedean ordered groups are isomorphic to an ordered subgroup of $\mathbb R$. Additionally, it is shown that if such a group is complete, then it is ...
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### Braided lobsters

If $(X,m)$ is a median algebra, then for each $x\in X$, define an operation $\wedge_{x}$ by letting $y\wedge_{x}z=m(x,y,z)$. Then $(X,\wedge_{x})$ is a meet-semilattice with least element $x$. Define ...
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### Pure (ordered) subgroups

Let $H,G$ be abelian groups with $H \leq G$. We say that $H$ is a pure subgroup of $G$ if for every $n \in \mathbb N$ and $h \in H$ the following holds: If $h$ is $n$-divisible in $G$, then $h$ is $n$-...
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### Generating totally ordered free commutative monoids

Let’s say I have a set $A$. I build the free commutative monoid $M$ generated by $A$. When can a well-order on $A$ be extended to $M$, in a way that is compatible with its monoid structure? I am ...
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### Relations between $\Omega$-groups, locally indicable groups, and right-orderable groups

We know that the class of right-orderable groups $\mathit{RO}$, is contained in the class of $\Omega$-groups (read it from "A note on group rings of certain torsion-free groups" by Burns-Hale). A ...
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### Is there a name for this kind of structure? (Not quite a lattice-ordered group)

I'm looking at a certain class of groups $G$ that come with a partial order $\le$ on the elements. So far it looks like $(G,\le)$ has the following properties: The partial order is invariant under ...
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### For which groups is (non-)left orderability decidable?

Mainly, my question is in the title, but let me be more precise here. Let $G$ be a finitely presented group with solvable word problem. If G is not left-orderable, is there an finite-time algorithm ...
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### Uncountable divisible groups and the existence of order-preserving isomorphisms of their subsets

Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$. Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an ...
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### Extending homomorphisms between ordered abelian groups

Let $\Omega$ be a linearly (i.e. fully) ordered set, and let $\Lambda_{\Omega}$ be the ordered abelian group consisting of those $(\lambda_\omega)_{\omega\in\Omega}\in\mathbb{R}^{\Omega}$ with well-...
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### Unique product groups (and semigroups)

A group $G$ is called a u.p.-group (short for unique product group) if for all nonempty finite subsets $A,B\subseteq G$, there exists an element $g\in A \cdot B$ which can be uniquely written as a ...
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### Existence of an orbit of exponential growth for group acting on the real line

Let G be a non-abelian finitely generated subgroup of increasing homeomorphisms of the real line having a fixed point free element $h$ ($hx>x$ for all $x$ in the line). Is there a real number $a$ ...
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### Compatible total orderings of the group $\mathbb{Z}^\mathbb{N}$

Given the additive group of the module $\mathbb{Z}^\mathbb{N}$ and a total ordering of the group that is compatible with addition and where $\chi_{\{n\}} > 0$ for all $n \in \mathbb{N}$, can we say ...
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### Name of the class of linearly ordered groups with no minimal positive element

Is there a special name for a linearly ordered group $G$ such that for every positive element $g\in G$ there exists an element $h\in G$ such that $e<h<g$?
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### Spliting of short exact exact sequences of partially ordered groups

Consider a short exact sequence of partially ordered groups $$0 \longrightarrow H \stackrel{\alpha}{\longrightarrow} G \stackrel{\beta} {\longrightarrow} G/H \longrightarrow 0 ,$$ where $H$ is a ...
363 views

### Unique product group which is not right orderable

(1) I am looking for an example of a u.p (unique product) group which is not right orderable (RO). Almost any group I pick up (obviously torsion-free, as u.p. group cannot have nontrivial torsion ...
250 views

### Normal subgroup of a totally ordered group

A totally ordered group is a group equipped with a compatible total order, that is, $x\leq y$ and $z\leq t$ imply $x+z\leq y+t$ for all $x,y,z,t$ in the group. Is it true that every totally ordered ...
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### Is there a left-orderable profinite group?

Is there a nontrivial profinite group $G$ with a binary transitive relation $<$ such that $x<y$ implies $x\neq y$, and for any different $x,y \in G$ either $x < y$ or $y < x$ and such ...
291 views

### Quasi-isometry and left invariant orderability for groups

Is the property of left invariant orderability for finitely generated groups preserved by quasi-isometrics? More precisely, if $G$ is a left orderable (finitely generated) group and $H$ is a torsion-...
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### Can Suslin (or Aronszajn) lines ever be orderings of abelian groups?

I am interested in realizing linear orders as orderings of abelian groups. In particular, can Suslin lines (and other classes of line) be realised in this way? Let $\mathcal{C}$ be a class of (...
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### 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-...
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### 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 ...
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### 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 $x \prec y$....
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### 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 ...
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### 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. ...
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### 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 $x+z \prec y + z$ ...
648 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 ...
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., $m,n\in\mathbb{Z}\... 4answers 718 views ### Why do we choose the standard total order on the integers? I understand why the set of natural numbers$\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} ... 1answer 224 views ### Subgroup of lattice-ordered group 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$a, b\in H$, ... 2answers 170 views ### Direct limit of lattice-ordered groups In general, any abelian group can be expressed as a direct limit of its f.g. subgroups. For the case of$\ell$-group (lattice-ordered group) is that true or not? As an abelian group we do not have ... 1answer 341 views ### Lattice-ordered group of rational rank 1 Does there exist a lattice-ordered, not totally ordered, group of rational rank$1$? Rational rank 1 means isomorphic to a nonzero subgroup of$\mathbb{Q}$. There exist totally ordered groups of ... 2answers 251 views ### A linearly orderable monoid which does not embed into a linearly orderable group It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable ... 1answer 360 views ### Strictly totally ordered semigroups - Looking for references Let$\mathfrak A = (A, \cdot)$be a semigroup (written multiplicatively). We say that$\mathfrak A$is linearly orderable if there exists a total order$\le$on$A$such that$ac < bc$and$ca < ...
I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e. for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I ...