7
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

Malcev's example is orderable. See https://doi.org/10.2307/2036896. So the answer is known and in the negative.

$\endgroup$
0
4
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ @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. $\endgroup$ Dec 14 '12 at 13:55
  • $\begingroup$ Under these general assumptions I'm not sure $\le$ thus defined is the ordinal ordering (so it might be a non-total order). Nevertheless it's correct at least in some cases, e.g., when $O$ is the set of ordinals $<\omega^2$. $\endgroup$
    – YCor
    Dec 7 '20 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.