# 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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questions

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### 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**

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**4**answers

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### 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

**1**answer

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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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**3**answers

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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 (...

**14**

votes

**1**answer

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### 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**

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**1**answer

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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 ...

**11**

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**1**answer

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### 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**

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**1**answer

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### 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**

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**1**answer

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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 ...

**8**

votes

**2**answers

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**

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**1**answer

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

**2**answers

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

**2**answers

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**

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**1**answer

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**

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**2**answers

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**

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**1**answer

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

**1**answer

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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-...

**5**

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**0**answers

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**

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**0**answers

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 ...

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### 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**

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**1**answer

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### 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**

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**3**answers

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**

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### 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**

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**1**answer

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-...

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**1**answer

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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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### 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**

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**1**answer

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 < ...

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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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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 ...

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**0**answers

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 ...

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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 ...

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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**

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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**

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**1**answer

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 ...

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**1**answer

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### 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**

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**1**answer

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 ...

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**1**answer

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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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**1**answer

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**

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**1**answer

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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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### 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 ...

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### 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 ...

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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 ...

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### 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).
...

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**2**answers

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 ...

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**1**answer

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### 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 $...

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**1**answer

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 ...

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**1**answer

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### 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 \{(...

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**1**answer

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 $\...

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**2**answers

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**

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**1**answer

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}\...