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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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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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 ...
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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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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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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}\...
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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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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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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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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 $a, b\in H$, ...
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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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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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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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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 < ...
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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 ...
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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 ...
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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 ...
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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 ...
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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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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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Ordered group acting freely on partially ordered set
Let $(G, <)$ be a totally ordered group, and let $<$ be left-invariant. Let $G$ act (freely?) on a partially ordered set $(S, <)$, such that this group action preserves the ordering:
$$ s_1 &...
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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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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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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 ...
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Realizing certain affine functions on Choquet simplices on dimension groups
This is a question that is a bit outside my usual mathematical comfort zone, but I feel like an expert might know the answer.
Recall that a dimension group is an ordered abelian group $G$ with ...
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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 ...
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