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Search options not deleted user 16537
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 …
Salvo Tringali's user avatar
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 …
Salvo Tringali's user avatar
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 $xy < xz$ …
Salvo Tringali's user avatar
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 $ …
Salvo Tringali's user avatar
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 …
Salvo Tringali's user avatar
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 …
Salvo Tringali's user avatar