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](https://mathoverflow.net/questions/105851/strictly-totally-ordered-semigroups-looking-for-references) for terminology and motivations). Then, my question is:



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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 linearly orderable monoid which does not embed into a linearly orderable group
Malcev's example is orderable. See <https://doi.org/10.2307/2036896>. So the answer is known and in the negative.


