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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.

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There is a prominent example in the non-commutative setting.

Take some set O of ordinals which is closed under addition and contains 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:

$0+\omega = \omega$, $1+\omega = \omega$.

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.

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@Frank Stephan. Sorry for the delay in replying. This doesn't count as an example. The OP refers to ordered sgrps, termed linearly ordered sgrps, where strict inequalities are compatible with the composition law. – Salvo Tringali Dec 14 '12 at 13:55

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