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Let $S$ be a monoid. On p. xvii of P.M. Cohn's Free Ideal Rings and Localization in General Rings (CUP, 2006), one reads that

  • an element $u \in S$ is regular if (quote) "[...] it can be cancelled, i.e. $ua = ub$ or $au = bu$ implies $a = b$";
  • $S$ is a conical monoid if (quote) "$ab = 1$ implies $a = 1$ (and so also $b = 1$)".

On p. 53, one then reads that $S$ is an invariant monoid if each of its elements is invariant, an element $c \in S$ being invariant if $c$ is regular and $cS = Sc$ (so in particular, an invariant monoid is cancellative).

Now, with these definitions in mind, Problem 0.9.10 in the same book (p. 58) asks:

Is every invariant conical monoid necessarily commutative?

The problem is numbered "10°" by Cohn, and on p. xiv one reads:

[...] open-ended (or open) problems are marked °, though sometimes this may refer only to the last part; the meaning will usually be clear.

At first, I had mistakenly thought (see the first version of this post) that the problem had a trivial answer, due to the following (flawed) argument:

Start with a semigroup $H$, pick an element $e \notin H$, and let $H^{(e)}$ be the (unique) magma obtained by extending the operation of $H$ to a binary operation on $H^{(e)}$ in such a way that $ex = xe := x$ for every $x \in H^{(e)}$. Clearly $H^{(e)}$ is a monoid (some people would call it an unconditional unitization of $H$); and it is cancellative, commutative, or invariant if and only if so is $H$, resp. On the other hand, $H^{(e)}$ is obviously conical (regardless of whether this is the case with $H$). Therefore, if $H$ is a non-commutative invariant monoid (such as the monoid of non-zero elements of a non-commutative skew-field), then $H^{(e)}$ provides a negative answer to Cohn's question.

The issue here is that $H^{(e)}$ is a cancellative monoid if and only if $H$ is a cancellative semigroup without unity: If $H$ is a monoid with identity $1_H$, then $xe = x1_H = x = 1_H x = ex$ for every $x \in H$ (and, by construction, $e \ne 1_H$). So, my (new) question is:

Q. Is anyone aware of any progress on Cohn's problem since 2006?

Update (12/03/2022). Cohn's problem has been quickly solved by Pace Nielsen below. A second and, in some sense, much easier solution was communicated by George Bergman to Pace Nielsen and is now the bulk of the accepted answer of this thread.

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I mentioned this problem to George Bergman and he offered the following much simpler solution, which he has given permission for me to post. In his own words:

Take any nonabelian group $G$ with a nontrivial homomorphism $f\colon G \to \mathbb{Z},$ and let $S$ be the submonoid of $G$ with underlying set $\{e\} \cup \{x\in G\, |\, f(x) > 0 \}.$

If $G$ is the free group on the two symbols $x,y$, and $f$ is the homomorphism determined by the rule $f(x)=f(y)=1$, then I believe that the monoid $M$ described in my other answer is the "duo-hull" of $x$ and $y$ inside the monoid $S$.

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  • $\begingroup$ Very nice! Just a note for the future me: If $a \in S$ and $x \in f^{-1}(\mathbb Z^+)$, then $f(axa^{-1}) = f(x) > 0$ (since $f$ is a group hom from $G$ to $\mathbb Z$) and hence $axa^{-1} \in f^{-1}(\mathbb Z^+) \subseteq S$, which implies at once that $aS = Sa$. $\endgroup$ Mar 7, 2022 at 10:20
  • $\begingroup$ Another note for the future me: By hp, there are $x,y,z\in G$ s.t. $f(x)\ne 0$ and $yz\ne zy$. We may assume wlog that $f(x)>0$ and $f(y),f(z)\ge 0$, since for all $u,v\in G$ we have $f(u)\le 0$ iff $f(u^{-1})\ge 0$ and $uv\ne vu$ iff $u^{-1}v\ne vu^{-1}$. It follows that $HxH\subseteq S$, where $H$ is the submonoid of $G$ generated by $\{x,y,z\}$. So, if $xy\ne yx$ (resp., $xz\ne zx$), then $x$ and $y$ (resp., $x$ and $z$) are non-commuting elems in $S$; otherwise, $(xy)(xz) =x^2yz\ne x^2zy = (xz) (xy)$, and $xy$ and $xz$ are non-commuting elems in $S$: In both cases, $S$ is non-commutative. $\endgroup$ Mar 7, 2022 at 14:21
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Here is a solution to the problem. (I'm unsure if this problem has been solved elsewhere.)

Step 1: First construct a duo monoid in an almost free manner, starting from two generators.

To do this, let $V_0=\{x,y\}$ be a set containing two variables, and let $R_0=\emptyset$ be the collection of relations that we assume they must satisfy.

Now, assuming that $V_n$ and $R_n$ have been defined for some integer $n\geq 0$, we recursively define $$ V_{n+1} = V_n\cup \{u_{n+1,s,t},v_{n+1,s,t}\, :\, s,t\in V_n\}. $$ So, for each pair of variables $s,t\in V_n$ we have created two new variables. Moreover, we recursively define $$ R_{n+1}=R_n\cup \{u_{n+1,s,t}s=st,\ tv_{n+1,s,t}=st \, :\, s,t\in V_n\}. $$ This guarantees that variables can push past each other (to either the left or right).

Now, take $V_{\omega}=\bigcup_{n\geq 0}V_n$ and $R_{\omega}=\bigcup_{n\geq 0}R_n$. The monoid $M_{\omega}$ generated by $V_{\omega}$ subject to the relations in $R_{\omega}$ is duo.

Step 2: We next force cancellativity.

The monoid $M_{\omega}$ is not cancellative. (Indeed, we have $u_{1,x,y}x=u_{2,x,y}x$ but $u_{1,x,y}\neq u_{2,x,y}$.) So if $ab=ac$ in $M_{\omega}$, we add $b=c$ to the set of relations, and similarly if $ba=ca$, we add $b=c$ to the set of relations.

Let $M_{\omega+1}$ be the monoid generated by $V_{\omega}$ subject to the relations in $R_{\omega+1}$. It might be the case that $M_{\omega+1}$ is guaranteed to be cancellative; however, I doubt it. In any case, we can repeat the previous process, adding new relations to $R_{\omega+1}$ if necessary. After repeating this process countably many times, we have a set of relation $R_{\omega\cdot 2}$, and the monoid $M=M_{\omega\cdot 2}$ generated by $V_{\omega}$ and subject to these relations is both cancellative and duo.

It is important to note that after forcing cancellativity, this construction is now truly a free construction. There is no freedom in the introduction of the new variables forcing the duo conditions; for if $us=st$ and $u's=st$, then $u=u'$ (and similarly on the other side)

Step 3: The conical condition comes for free!

Treating each variable as having degree/grade $1$, we can check that in each of the previous two steps, all of the relations are homogeneous in the grading. Thus, in particular $M$ is conical; the only one-sided unit is $1$, by a simple degree argument

Step 4: Check that $x$ and $y$ don't commute in $M$.

We do this by noting that the ring of real quaternions, $Q$, is a domain. Thus $Q-\{0\}$ is a duo, cancellative monoid (but clearly not conical). From freeness, we can map $M$ to $Q-\{0\}$ by sending $x$ and $y$ to nonzero, noncommuting elements. Since their images don't commute in $Q$, the relations in $R_{\omega\cdot 2}$ cannot force commutativity of $x$ and $y$.

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    $\begingroup$ I'll need a fresh mind to check some details, but I'm convinced that $M$ is a duo monoid: Since $V_\omega$ is a generating set of $M$, it suffices to check that, for all $s,t\in V_\omega$, there exist $p,q\in M$ s.t. $st=ps$ and $ts=sq$. To this end, fix $s,t\in V_\omega$; let $i,j\in\mathbb N$ be the smallest non-negative integers s.t. $s\in V_i$ and $t\in V_j$; and note that $st=u_{n+1,s,t}s$ and $ts=sv_{n+1,t,s}$, where $n:=\max(i,j)$. $\endgroup$ Mar 5, 2022 at 20:52
  • $\begingroup$ In the recursive def of $R_{n+1}$, do you really want to enlarge $R_n$ by adding the rels $u_{n+1,s,t} s = t u_{n+1,s,t} = st$ for all $s, t \in V_n$? Or do you rather want to add the same rels for all $s,t\in V_n$ with $s\ne t$? $\endgroup$ Mar 5, 2022 at 21:05
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    $\begingroup$ (Minor point: That second $u$ variable you wrote should be a $v$ variable instead.) At first, I tried to be really restrictive about which pairs of variables to allow when enlarging the relation set. But then I realized none of that mattered, for two reasons. Firstly, it was extremely difficult to control exactly when I should or should not add new variables and relations. Secondly, the cancellative property collapses all of those elements together anyway. So there was no need to make that (or any other) restriction. If $s=t$, then cancellativity makes $u_{n+1,s,t}=s=t=v_{n+1,s,t}$. $\endgroup$ Mar 6, 2022 at 3:58
  • $\begingroup$ (Yes, sorry for the typo: I meant $u_{n+1,s,t} s = t v_{n+1,s,t}=st$.) Fine, I see your point. I haven't yet got to convince myself that $M$ is cancellative, but I'll give it another try later. $\endgroup$ Mar 6, 2022 at 8:53
  • $\begingroup$ Perhaps this will help. In $M$, suppose we have an equality $ab=ac$ for some words $a,b,c$ in the variables $V_{\omega}$. This equality holds by using only finitely many relations in $R_{\omega\cdot 2}$, and so all of those relations belong in $R_{\omega+n}$ for some natural number $n$. Thus, the equality $b=c$ belongs to $R_{\omega+n+1}$. $\endgroup$ Mar 6, 2022 at 16:45
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Let $\mathcal G = (G, \preceq)$ be a [pre]ordered group, by which we mean that (i) $G$ is a (multiplicatively written) group; (ii) $\preceq$ is a partial [pre]order on (the underlying set of) $G$ such that, if $x \preceq y$ and $u \preceq v$, then $xy \preceq uv$; (iii) the last inequality is strict whenever $x \prec y$ or $u \prec v$.

It is readily checked that the positive cone $\mathcal G^+ := \{x \in G \colon 1_G \prec x\}$ of $\mathcal G$ is a non-unital subsemigroup of $G$ that is closed under conjugation in the sense that $a^{-1}\mathcal G^+ a \subseteq \mathcal G^+$ for every $a \in G$ (if $x \in \mathcal G^+$ and $a \in G$, then $1_G = a^{-1} 1_G a \prec a^{-1} x a$); cf. the first half of Lemma 1 in

  • K. Iwasawa, On linearly ordered groups, J. Math. Soc. Japan 1 (1948), No. 1, 1-9.

It follows that $\mathcal G^+$ is a non-unital, duo subsemigroup of $G$ (a semigroup $S$ is duo if $aS = Sa$ for each $a \in S$) and hence the unitization $\{1_G\} \cup \mathcal G^+$ of $\mathcal G^+$ is a conical, invariant submonoid of $G$ in the sense of the OP (every subsemigroup of a group is obviously cancellative).

Suppose now that there exist non-commuting elements $x, y \in G$ such that $1_G \prec x \preceq y$, and note that this condition is certainly satisfied if $\mathcal G$ is a non-commutative, totally ordered group, i.e., $G$ is a non-commutative group and $\preceq$ is a total order (many examples are listed in another MO thread). It is then clear that $\{1_H\} \cup \mathcal G^+$ is a non-commutative submonoid, and so we are done.


This fixes the faulty argument originally mentioned in the OP and recovers Bergman's solution (as reported by Pace Nielsen in the accepted answer of the thread): If there is a group homomorphism $f$ from $G$ to the additive group of the integers, then taking $\preceq$ to be the pullback of the usual order $\leq$ on $\mathbb Z$ along $f$ (so that $a \preceq b$, for some $a, b \in G$, iff $f(a) \le f(b)$) makes $(G, \preceq)$ into a totally preordered group.

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