All Questions
Tagged with semigroups-and-monoids ordered-groups
14 questions
5
votes
0
answers
107
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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
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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
1
vote
0
answers
43
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Interleaving in Viennot's Heaps models?
I am looking for past results on interleaving of heaps (in the sense of Viennot). For a very simplified example, suppose I have two pieces, (b1 a1 b1), and (b2 c2 b2), where the letter represents a ...
- 11
3
votes
1
answer
157
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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
3
votes
0
answers
134
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Partial orders on $\mathbb{N}^m$ compatible with addition
I'm looking for a classification (or just non-trivial examples) of partial orders on monoid $\mathbb{N}^{m}$ that are compatible with addition. That is, partial orders $\leq$ satisfying two additional ...
- 645
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
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
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 ...
- 63.9k
4
votes
1
answer
307
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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 ...
- 63.9k
8
votes
1
answer
1k
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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 ...
- 63.9k
4
votes
1
answer
518
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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 < ...
- 63.9k
4
votes
2
answers
393
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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 < ...
- 63.9k
7
votes
1
answer
266
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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 ...
- 63.9k
8
votes
2
answers
826
views
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 ...
- 63.9k