All Questions
Tagged with order-theory ordered-groups
30 questions
-1
votes
0
answers
73
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Why is there in theory no morphism/isogenies when enlarging a prime field sharing a common suborder/subgroup? [closed]
Simple question : I have a prime field having modulus $p$ where $p−1$ contains $O$ as prime factor, and I have a larger prime field $q$ also having $O$ as its suborder/subgroup. Why are there no ...
- 12.9k
5
votes
0
answers
107
views
Structure of well-ordered commutative monoids
Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where
$\forall a\in M,\ 0\leq a$
$\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$
The first condition means $M$ will be ...
- 18.7k
2
votes
1
answer
176
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 ...
- 3,738
16
votes
4
answers
1k
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 (...
- 4,413
33
votes
7
answers
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) ...
- 3,738
10
votes
1
answer
821
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 ...
3
votes
1
answer
157
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 ...
- 18.7k
7
votes
2
answers
544
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 ...
- 63.9k
11
votes
1
answer
529
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 $|...
- 63.9k
4
votes
0
answers
280
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 ...
- 63.9k
4
votes
4
answers
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} ...
- 63.9k
3
votes
0
answers
589
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 ...
- 63.9k
6
votes
2
answers
362
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 $\...
- 63.9k
3
votes
1
answer
231
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 ...
- 63.9k
4
votes
1
answer
204
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-...
- 63.9k
2
votes
2
answers
474
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 ...
- 63.9k
2
votes
1
answer
236
views
Example involving partially ordered Abelian groups
Definition 1:
Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\...
- 63.9k
5
votes
1
answer
302
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-...
- 63.9k
2
votes
0
answers
417
views
An equivariant Hahn embedding theorem?
The Hahn Embedding Theorem asserts that for any (linearly) ordered abelian group $\Lambda$, there exists a linearly ordered indexing set $\Omega$ such that $\Lambda$ admits an order-preserving group ...
- 63.9k
1
vote
1
answer
184
views
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 ...
- 63.9k
3
votes
2
answers
291
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.
...
- 63.9k
3
votes
1
answer
315
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 ...
- 63.9k
2
votes
1
answer
314
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 $\...
- 63.9k
4
votes
1
answer
518
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 < ...
- 63.9k
4
votes
2
answers
393
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 < ...
- 63.9k
0
votes
1
answer
370
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 ...
- 63.9k
15
votes
1
answer
1k
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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 ...
- 63.9k
0
votes
1
answer
268
views
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 &...
CommunityBot
- 1
3
votes
2
answers
695
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 ...
- 63.9k
1
vote
1
answer
541
views
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 ...
- 63.9k