45
votes
Accepted
The sum of two well-ordered subsets is well-ordered
Ramsey theory! Suppose $A + B$ is not well-ordered. Then there is a strictly decreasing sequence $a_1 + b_1 > a_2 + b_2 > \cdots$. Observe that for any $i < j$, either $a_i > a_j$ or $b_i &...
- 42.8k
31
votes
Accepted
Characterizing $\mathbf{R}$ as an ordered group
The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following ...
- 12.5k
16
votes
The sum of two well-ordered subsets is well-ordered
A more general result:
Let $G$ be an abelian totally ordered group. Let $E \subseteq G^+$. Write $E^*$ for the semigroup generated by $E$. If $E$ is well-ordered, then $E^*$ is well-ordered.
The ...
- 41.1k
15
votes
The sum of two well-ordered subsets is well-ordered
There's also a more elementary proof than the one given by Nik Weaver (although I also enjoy the use of the countably infinite Ramsey's theorem!). First prove that an ordered set is well-ordered if ...
- 2,519
13
votes
Accepted
Are Artin-Tits groups ordered groups?
Your question is answered in Mulholland and Rolfsen's article Local indicability and commutator subgroups of Artin groups. On the one hand, the Artin-Tits group $A(I_2(n))$ is not bi-orderable because ...
- 8,401
11
votes
Accepted
Groups with three conjugacy classes that define an ordering
There are currently no known examples of bi-orderable groups where all positive elements are conjugate. The question of their existence appears as Problem 3.31 of the 2009 problem list Unsolved ...
- 1,089
8
votes
Accepted
Is Thompson's group definably orderable?
Yes, Thompson's group $F$ is definably bi-orderable.
Let $a$ be some element of $F$ with the support of $a$ equal to $(0,1/2)$. Let $b$ be some element of $F$ with the support of $b$ equal to $(1/2,1)$...
- 191
6
votes
Accepted
Are the Baumslag–Solitar groups locally indicable groups?
Even more is true: All torsion-free 1-relator groups are locally indicable:
Brodskij, S. D., Equations over groups, and groups with one defining relation, Sib. Math. J. 25, 235-251 (1984); translation ...
- 12.2k
6
votes
Characterization of Archimedean linearly ordered monoids
By a classic result of Alimov (1950), a fully ordered semigroup $S$ is order isomorphic to a subsemigroup of the additive group of real numbers if and only if it is cancellative and contains no ...
- 6,453
5
votes
Is it true that the structure of a commutative ordered semiring is unique on a commutative ordered monoid?
The answer is no.
Consider the set $\mathbb{N}^{<\omega}$, consisting of $\omega$-tuples with only finitely many nonzero entries. This set is totally ordered under the relation $\prec$, where $(...
- 18.7k
5
votes
Accepted
Generating totally ordered free commutative monoids
For your first question, the answer is positive with lexicographic ordering (as said by Chris).
Your monoid is $\mathbb{N}^{(A)}$ (i.e. the set of mappings $\alpha: A\to \mathbb{N}$ with finite ...
- 3,738
4
votes
Groups with three conjugacy classes that define an ordering
I have proposed a positive solution to this problem in a preprint entitled Hyperexponentially closed fields, to be found here, more precisely in Sections 10.1 and 10.2.
The solution is based on work ...
- 2,519
4
votes
Accepted
Ordered group acting freely on partially ordered set
Yes, such a partial order can be extended to a total order: we can amend the proof of the Szpilrajn extension theorem which essentially establishes the result in the case $G=1$.
Starting with the ...
- 1,089
3
votes
Orderable subgroup of the braid groups over the 2-sphere
A google search for "braid groups orderable" gives a wealth of information. In particular, Juan Gonzalez-Meneses (arXiv preprint here) proved that pure braid groups on closed orientable surfaces are ...
- 25k
3
votes
How can you order a free group?
Let us first clarify the relationship between scattered, discrete, and dense orders. The last two notions are standard in the theory of ordered groups.
An order (left or two-sided) is discrete if ...
2
votes
Topology of the Malcev-Neumann group ring
It is unclear to me whether you are working with commutative or non-commutative objects, but I will take $G$ to be Abelian and linearly ordered, and $R$ to be possibly non-commutative.
The topology ...
- 2,519
2
votes
Least positive elements of discrete preorders on surface groups
Theorem. Let $G$ be the surface group of an oriented surface. Then an element $a$ is a least positive element for some discrete preordering on $G$ if and only if $a$ is not a proper power.
Proof. Let $...
- 357
2
votes
For which groups is (non-)left orderability decidable?
This is a very nice question, which I don't know how to answer. But I hope the following (shameless self-promotion) will be of interest.
You seem to be looking for some sort of local condition that ...
- 25k
2
votes
Accepted
Name of the class of linearly ordered groups with no minimal positive element
I assume that the order is left-invariant. All orders here are total.
First, by homogeneity, the order is either dense or discrete (in the sense that the order topology is discrete). Beware that ...
- 63.9k
2
votes
Accepted
Suppose $G$ is Conradian right-ordered group. If $x < z$ and $y < z$, can we say $xy < z^2$?
(Since I usually use left-orderings, the following answer uses left-orderings)
No, when $<$ is a left-ordering which is not a bi-ordering then you can always find elements $x,y,z$ with
$x <z $ ...
- 548
1
vote
Accepted
Real exponentiation in the quotients of rings of continuous functions by prime ideals
Fix prime ideal $I$, $r\in\mathbb{R}_{>0}$ and $a = f+I = g+I$ with $f, g\geq 0$. I'll prove that $f^r+I = g^r+I$.
First, note if $r\in\mathbb{R}_{>0}$ and $a = 0$, then $f^r = f^{\frac{1}{n}}f^{...
- 1,201
1
vote
Accepted
Realizing certain affine functions on Choquet simplices on dimension groups
Yes. Form Aff $(T)$; since $T$ is metrizable, Aff $(T)$ contains a countable dense set. Adjoin $f$ and the constant function $1$ to the countable dense set (creating a larger countable dense set), and ...
- 4,679
1
vote
Can Suslin (or Aronszajn) lines ever be orderings of abelian groups?
Gruenhage, Heath and Poerio showed that a Suslin Line cannot even carry the structure of a cancellative topological semigroup. The same is true for those Aronsajin Lines that are Lindelöf. See:
...
- 4,413
1
vote
Every abelian torsion-free group is strictly totally orderable (via the compactness theorem)
A group $G$ is not bi-orderable if and only if for some finite subset $J$ of $G\smallsetminus\{1\}$, for every $e\in \{-1,1\}^J$ there exists $n\ge 1$, group elements $c_1,\dots,c_n$, and a function $...
- 63.9k
Only top scored, non community-wiki answers of a minimum length are eligible