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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 $xy < xz$ and $yx < zx$ for all $x,y,z \in M$ with $y < z$. In particular, $\mathbb M$ is called an l.o. group if $(M, \cdot)$ is, well, a group, and an l.o. free monoid if $(M, \cdot)$ is the free monoid on an alphabet $X$.

What is known about the following question?

(Q) Let $\mathbb M = (M, \cdot, \le)$ be a linearly ordered free monoid. Does there always exist an embedding of $\mathbb M$ into a linearly ordered group? In more plain words: do there always exist a linearly ordered group $\mathbb G = (G, \cdot, \le)$ and a (monoid) monomorphism $f: (M, \cdot) \to (G, \cdot)$ such that $f(x) < f(y)$ for all $x,y \in M$ with $x < y$?

The answer is affirmative in the case when $\le$ is the lexicographic order induced on $M$ by any well-ordering of the underlying alphabet (this can be proved, e.g., by the "Magnus trick").

But what about the rest? Any reference?

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  • $\begingroup$ Any reference for what you call "Magnus trick"? $\endgroup$ – boumol Apr 9 '14 at 16:01
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    $\begingroup$ As for free groups on finite alphabets, see Section 5 in: D. M. Kim and D. Rolfsen, An Ordering for Groups of Pure Braids and Fibre-type Hyperplane Arrangements, Canad. J. Math. 55 (2002), 822-838 (and the references therein). I don't know if the proof of the general statement: "All free groups are linearly orderable" by the same method (which is as simple or difficult, it is up to you, as the finite case), is explicitly written down somewhere. Does anybody know? $\endgroup$ – Salvo Tringali Apr 9 '14 at 17:03
  • $\begingroup$ For the record: An alternative proof that any free group (and hence any free monoid) is linearly orderable can be found in K. Iwasawa, On linearly ordered groups, J. Math. Soc. Japan 1 (1948), 1-9. Yet, Iwasawa's approach doesn't help much with the OP (as far as I can tell). $\endgroup$ – Salvo Tringali Apr 9 '14 at 17:37
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I suspect this is false, although I don't have a proof. The Thompson group $F$ is generated by $A, B$, which are piecewise-linear homeomorphisms of the interval which change slope at dyadic points, and piecewise have slopes in $2^\mathbb{Z}$. As described in Theorem 4.6 of Cannon-Floyd-Parry, the submonoid generated by $A,B$ is free. The group is linearly ordered (called bi-ordered in the literature), and the spaces of bi-orderings has been classified. I suspect that some of these bi-orderings, when restricted to the submonoid generated by $A, B$, do not extend to the free group generated by $A, B$. I would try one of the 8 isolated bi-orderings of the Thompson group, and see if it can be extended to the free group generated by $A,B$. If it can't, then one can detect this in a ball of finite-radius in the free group.

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  • $\begingroup$ I will try to dig into all of this, but thanks for sharing your insights! A minor note: I wouldn't say that "bi-ordered" is more conventional than "linearly ordered". The latter appears, e.g., in K. Iwasawa's and P. Conrad's papers. The same Conrad coined (?) the term "o-group" in Extensions of Ordered Groups (Proc. AMS, Vol. 6, No. 4 (Aug., 1955), pp. 516-528), which is still in use by a number of authors. Others do simply talk of "ordered groups" (e.g., G. Freiman and coauthors in the additive theory of groups), and the list continues. But yes, some people prefer "bi-ordered". $\endgroup$ – Salvo Tringali Apr 20 '14 at 20:04
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It suffices to show this for l.o. free monoids on finite alphabets, as follows from the Compactness Theorem in logic - which can be found in any text on First Order Logic. [This principle can be applied to a range of similar problems.]

Indeed, let $\mathbb M = (M, \cdot, \le)$ be a linearly ordered free monoid on the alphabet $X$, and consider the first-order language $\mathcal{L} = (\circ, \preceq, \{x_m|m \in M\})$ consisting of a binary function symbol $\circ$ to represent multiplication, a binary relation symbol $\preceq$ representing ordering, plus an individual constant symbol $x_m$ for each element $m$ of the set $M$.

Then let $T$ be the $\mathcal{L}$-theory having the following axioms:

  1. the usual axioms for linearly ordered groups, expressed using $\circ$ and $\preceq$
  2. $x_m \ne x_n$ for all $m, n\in M$ with $m\ne n$ [these axioms ensure that $M$ naturally injects as a subset of any model of $T$ via the interpretation of the $x_m$ constants]
  3. $x_m \preceq x_n$ for all $m, n\in M$ with $m\le n$ [these make this injection an embedding of l.o. sets]
  4. $x_m \circ x_n = x_{m\cdot n}$ for all $m, n\in M$ [which make the embedding a monoid homomorphism].

So any model $\mathbb G = (G, \cdot, \le, \{g_m|m \in M\})$ of $T$ will provide a desired l.o. group, with $m\mapsto g_m$ (being the interpretation of the constant $x_m$ in $\mathbb G$) giving the desired embedding $\mathbb M\to \mathbb G$. And conversely.

By the Compactness Theorem, the theory $T$ admits a model iff every finite subset $\Delta$ of the axioms of $T$ does. Now only finitely many symbols $x_m$ can occur in sentences belonging to such a $\Delta$, and the finitely many elements $m\in M$ so involved can be expressed as words in a finite subalphabet $Y\subseteq X$. This $Y$ generates a l.o. free submonoid $\mathbb S$ of $\mathbb M$ that contains all the $m\in M$ for which $x_m$ figures in statements occurring in $\Delta$. And so any l.o. group that "extends" $\mathbb S$ (in the sense of the OP) will be a model of $\Delta$.

Hence it is enough to consider free l.o. monoids on finite alphabets.

[Incidentally, by the same token, if all finitely generated free groups are linearly orderable, so are all free groups.]

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