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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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. ...
Xodarap's user avatar
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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 $x+z \prec y + z$ ...
Salvo Tringali's user avatar
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3 answers
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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 ...
Harry Baik's user avatar
2 votes
1 answer
429 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}\...
bsog's user avatar
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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} ...
Tom LaGatta's user avatar
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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 $a, b\in H$, ...
Rajnish's user avatar
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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 ...
Rajnish's user avatar
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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 ...
Rajnish's user avatar
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7 votes
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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 ...
Salvo Tringali's user avatar
4 votes
1 answer
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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 < ...
Salvo Tringali's user avatar
2 votes
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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 ...
chros's user avatar
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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 ...
boumol's user avatar
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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 \{(...
Rajnish's user avatar
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Group of divisibility of a commutative domain

We know that the necessary condition for any partially ordered group to be a group of divisibility is that the group must be a directed group. What is the sufficient condtion for partially ordered ...
Rajnish's user avatar
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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 $\...
BN2's user avatar
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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?
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4 votes
0 answers
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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 ...
Tobias Schlemmer's user avatar
7 votes
1 answer
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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 ...
Vladimir Dotsenko's user avatar
8 votes
1 answer
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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 ...
François G. Dorais's user avatar
11 votes
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 $|...
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6 votes
2 answers
358 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 $\...
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1 answer
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Constructing a convex valuation ring/ordered group of rank $n$

I know at least one method of constructing a convex valuation ring of rank $n$ (but it is rather complicated). What are the easiest methods of doing this? Given a natural number $n$ I want to have a ...
Jose Capco's user avatar
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33 votes
7 answers
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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) ...
Andrew Critch's user avatar
2 votes
2 answers
660 views

Automorphisms of the totally ordered group $\mathbb{Z}{^n}$ with lexicographical order

It is easy to see that the totally ordered group $\mathbb{Z}$ (the integers) with the natural order has no non-trivial automorphisms. Is this also true for $\mathbb{Z}^n$ with the lexicographical ...
user717's user avatar
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