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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32
votes
6answers
4k views

What's a non-abelian totally ordered group?

Because I have heard the phrase "totally ordered abelian group", I imagine there should be non-abelian ones. By this I mean a group with a total ordering (not to be confused with a well-ordering) ...
26
votes
4answers
2k views

The sum of two well-ordered subsets is well-ordered

Apologies if the answer is trivial, this is far from my domain. In order to define the field of Hahn series, one needs the following fact: if $A,B$ are two well-ordered subsets of $\mathbb{R}$ (or any ...
19
votes
1answer
555 views

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. ...
15
votes
3answers
947 views

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 (...
14
votes
1answer
789 views

Characterizing $\mathbf{R}$ as an ordered group

A standard characterization of $\mathbf{R}$ uses the order and the field structure: any linearly ordered field that is archimedean and complete is isomorphic to $(\mathbf{R}, +, \times, <)$ as an ...
12
votes
1answer
377 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 ...
11
votes
1answer
516 views

Partial word orders on groups

This is a followup question related to this question. Recall that a left-invariant partial order on a finitely generated group $G$ is called a partial word order if for every $a\le b\le c$ we have $|...
9
votes
1answer
334 views

Left orderable linear groups

Are all torsion-free finitely generated linear groups over $\mathbb{C}$ left orderable? In particular, are torsion-free congruence subgroups of $SL_n(\mathbb{Z})$ left orderable?
8
votes
1answer
288 views

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 ...
8
votes
2answers
764 views

Which semigroups can be linearly ordered?

As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the ...
8
votes
1answer
978 views

Lattice-ordered commutative monoids

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two ...
7
votes
2answers
287 views

Residual $p$-finiteness of principal congruence subgroups

Let $\Gamma(N)$ be the principal congruence subgroup of level $N$ in $\mathrm{SL}_n(\mathbf{Z})$, where $n\geq 3$. Then $\Gamma(N)$ is residually $p$-finite for all primes $p$ dividing $N$. Can $\...
7
votes
2answers
411 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 ...
7
votes
1answer
246 views

Positive cone of a subgroup of $\mathbb{Z}^n$

This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a ...
6
votes
2answers
349 views

orders and length functions on finitely generated groups

Let $G$ be a finitely generated group with the natural word length function ($|x|$ is the length of the shortest word in generators of $G$ representing $x$). We call a partial left invariant order $\...
6
votes
1answer
262 views

Groups with three conjugacy classes that define an ordering

Consider the following property for a group $(\mathcal{G},\cdot,1)$: There are exactly three conjugacy classes $\{1\}$, $\mathcal{C}_1$, $\mathcal{C}_2$ in $\mathcal{G}$, and we have $\mathcal{C}_1 \...
5
votes
1answer
253 views

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-...
5
votes
0answers
246 views

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 ...
5
votes
0answers
498 views

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 ...
4
votes
4answers
1k 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} ...
4
votes
1answer
347 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 ...
4
votes
3answers
751 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 ...
4
votes
2answers
426 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 ...
4
votes
1answer
180 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-...
4
votes
1answer
126 views

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 ...
4
votes
2answers
308 views

Embedding a linearly ordered free monoid into a linearly ordered group

A linearly ordered (shortly, l.o.) monoid is a triple $\mathbb M = (M, \cdot, \le)$ for which $(M, \cdot)$ is a (multiplicatively written) monoid and $\le$ is a total order on $M$ such that $xy < ...
4
votes
1answer
387 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 < ...
4
votes
0answers
80 views

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 ...
4
votes
0answers
43 views

Closed and bounded intervals of definably complete ordered groups

True or false? All closed and bounded intervals of definably complete ordered groups are definably compact. Let $G$ be an ordered abelian group. Then, a definable subset $D ⊆ G$ is said to be ...
4
votes
0answers
303 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 ...
4
votes
0answers
273 views

What is known about orbifolding ordered groups and sets? Who has been involved? Links to Lee metrics?

In mathematical music theory several ordered groups are considered. Some examples contain the frequency space or Tonnetzes. Other groups (commutative and non-commutative ones) are discussed by Dawid ...
3
votes
2answers
330 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-...
3
votes
2answers
935 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 $x+z \prec y + z$ ...
3
votes
1answer
268 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 ...
3
votes
1answer
115 views

Is it true that the structure of a commutative ordered semiring is unique on a commutative ordered monoid?

Is it true that the structure of a commutative ordered semiring with identity is unique on a commutative ordered monoid (i.e., the structure of the monoid and the order are consistent)? I am not ...
3
votes
1answer
113 views

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 ...
3
votes
1answer
146 views

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 ...
3
votes
1answer
268 views

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 ...
3
votes
1answer
195 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 ...
3
votes
0answers
93 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 ...
3
votes
0answers
68 views

Extending a representation of a free group to an extension of a mapping torus

Given a free group on $n$ generators, $F_n$, $\phi$ an automorphism of $F_n$, and a non-trivial representation $\rho: F_n \rightarrow \operatorname{Homeo}_+(\mathbb{R})$, are necessary and sufficient ...
3
votes
0answers
574 views

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 ...
3
votes
0answers
77 views

A Krull-Schmidt theorem for partially ordered groups

If $G$ is a po-group (ie. partially ordered group), we say that $G$ is po-indecomposable if it's not the direct product of two non trivial subgroups (such subgroups are necessary convex and normal). ...
2
votes
2answers
434 views

Non-archimedean group over the reals

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 ...
2
votes
1answer
184 views

Is there a non-right-orderable torsion-free quotient group of the braid group on 3 strands?

The braid group on 3 strands has the presentation $\langle x,y \;|\; xyx=yxy\rangle$. A group $G$ is called right-orderable if there is a total order $<$ on the set $G$ such that if $a<b$ then $...
2
votes
1answer
124 views

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 ...
2
votes
1answer
312 views

Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?

I was looking some lattice-ordered group structure. I have kind of difficulty to figure out about the group $\mathbb{Z}^{2}$ with positive cone is $\mathbb{N}_{>0} \times \mathbb{N}_{>0} \cup \{(...
2
votes
1answer
295 views

When do infinitesimals split in dimension groups?

Let $G$ be a dimension group (i.e. a directed, unperforated abelian group satisfying the Riesz interpolation property) with order unit $u\in G^{+}$. There is a canonical positive group homomorphism $\...
2
votes
2answers
271 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. ...
2
votes
1answer
372 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}\...