All Questions

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 ...

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} ...

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 < ...