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7
votes
2
answers
538
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 sem …
4
votes
2
answers
383
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 $xy < xz$ …
4
votes
1
answer
505
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 $ca < cb$ f …
3
votes
1
answer
171
views
Well-foundedness of divisibility vs well-foundedness of right- and left-divisibility
Say that a preorder (i.e., a reflexive and transitive binary relation) $\preceq$ on a set $X$ is
artinian if there is no sequence $(x_n)_{n \ge 1}$ of elements of $X$ with $x_{n+1} \prec x_n$ for eac …
2
votes
0
answers
116
views
Isomorphic subcategories of directed graphs and presets
For the purposes of this post, a digraph (directed graph) has neither loops nor multiple parallel edges, and a preset is an ordered pair consisting of a set $S$ and a preorder (viz., a reflexive and t …
2
votes
What's a non-abelian totally ordered group?
Not really an example (see the edit below), but somehow related to Greg Kuperberg's one: Let $\mathbb A = (A, +, \cdot, \preceq)$ be a strictly totally orderable semiring (*) and for a fixed integer $ …