All Questions
7 questions
4
votes
1
answer
307
views
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 ...
- 249
5
votes
0
answers
637
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 ...
- 18.7k
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 < ...
- 10.5k
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} ...
- 8,512
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 ...
- 10.5k
7
votes
1
answer
266
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 ...
- 16.9k
8
votes
1
answer
1k
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 ...
- 44.4k