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 semigroup (is cancellative and torsion-free, and) embeds into a linearly orderable monoid (see here for terminology and motivations). Then, my question is:


Is it known whether or not a linearly orderable monoid embeds into a linearly orderable group?


The answer is surely yes in the commutative setting (by the construction of the Grothendieck group). But I've no clue about the non-commutative case. Thank you in advance for any hint.
 A: Malcev's example is orderable. See https://doi.org/10.2307/2036896.  So the answer is known and in the negative.
A: There is a prominent example in the non-commutative setting.
[Edit (YCor): here linearly ordered monoid is interpreted as a monoid endowed with a total ordering satisfying: $a\le b,c\le d$ imply $ac\le bd$. As mentioned in the comments, this does not imply that left/right multiplication preserve the strict ordering.]
Take some set $O$ of ordinals which is closed under addition and
contains $0$, $1$ and $\omega$, where addition is defined the way Cantor did it.
$(O,+)$ has a first-order definable order:
$x \leq y$ if and only if there is a $z$ such that $y = x+z$.
It is easy to see that $x \leq y$ implies $x+z \leq y+z$ and $z+x \leq z+y$.
However, there is no group $(G,+)$ having $(O,+)$ as a subgroup.
The main reason is that adding $\omega$ cannot be inverted:
$$0+\omega = \omega, \quad 1+\omega = \omega.$$
If $(G,+)$ is a group extending $(O,+)$, then there would be an inverse $g$ to $\omega$ in $G$ and $0 = \omega+g = (1+\omega)+g = 1+(\omega+g) = 1+0 = 1$, a contradiction.
