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A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, we say that $S$ is a left (resp., right) cancellative semigroup if, for each $a \in S$, left (resp., right) multiplication by $a$ is an injective function on $S$. Then, we call $S$ cancellative if it is both left and right cancellative.

I recently asked (here) if a cancellative semigroup embeds into a group. Pace Nielsen promptly answered the question in the affirmative (here). In a comment to Pace's answer (here), I then raised the following question:

Q. If a duo semigroup is left/right cancellative, is it cancellative? Any reference?

There is a certain abundance of non-commutative, cancellative duo semigroups "in nature" (see here for a short list). Yet, one-sided cancellative, duo semigroups appear to be "rare" (if any exist). Among other things, note that every finite left/right cancellative duo semigroup is a group.

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Here is a right cancellative duo monoid that is not cancellative.

Let $A$ be a free abelian group of countable rank with basis $e_0,e_1,\ldots$. Let $\Phi\colon A\to A$ be the endomorphism given by $\Phi(e_0)=0$ and $\Phi(e_i) = e_{i-1}$ for $i>0$. Note that $\Phi$ is surjective.

Let $M=\langle t\rangle$ be a free monoid generated by $t$. Form the semidirect product $N=A\rtimes M$ where $t$ acts via $\Phi$. I claim that $M$ is right cancellative and duo but not cancellative. Let's write $f$ for the identity of $M$.

We need to check that each element is right cancellable. If $(b,t^r)(a,t^n)=(c,t^s)(a,t^n)$, then $r+n=s+n$, and so $r=s$. Also, $b+\Phi^r(a)=c+\Phi^r(a)$, and so $b=c$. Therefore $(b,t^r)=(c,t^s)$.

Note that $(0,t)(0,f) = (0,t) = (0,t)(e_0,f)$, and so $N$ is not left cancellative.

To see that $M$ is duo, compute $$(a,t^n)(b,t^r) = (a+\Phi^n(b),t^{n+r})=(a+\Phi^n(b)-\Phi^r(a),t^r)(a,t^n)$$ and so $(a,t^n)N\subseteq N(a,t^n)$.

Next consider $(c,t^s)(a,t^n) = (c+\Phi^s(a),t^{s+n})$. Since $\Phi$ is surjective, we can find $d\in A$ with $\Phi^n(d) = c+\Phi^s(a)-a$. Then $(a,t^n)(d,t^s) = (a+\Phi^n(d),t^{n+s}) = (c+\Phi^s(a),t^{n+s})$. Therefore, $N(a,t^n)\subseteq (a,t^n)N$. It follows that $N$ is duo.

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  • $\begingroup$ (A note for my future self.) In this answer, $A$ is, up to iso, the direct sum of countably infinite copies of the additive grp of ints, and $M$ is the additive monoid of non-negative ints. Next, we consider the (left) (unitary) action of $M$ on $A$ that maps a pair $(n,x)\in M\times A$ to $\Phi^n(x)$, where $\Phi^n$ is the $n$-th iterate of the surjective endo of $A$ sending an element $x=(a_0,a_1,\ldots) \in A$ (uniquely represented by the integer coordinates $a_i$ relative to the canonical basis of $A$ as a free left $\mathbb Z$-module) to the element $(a_1,a_2,\ldots)\in A$. [...] $\endgroup$
    – Salvo Tringali
    Commented Oct 15 at 2:57
  • $\begingroup$ [...] Lastly, we look at the semidirect product $N := A \rtimes M$ of $A$ by $M$ relative to the given action, which is the Cartesian product $A \times M$ endowed with the binary multiplication defined by $$(x, m) (y, n) := (x + \Phi^m(y), m+n), \qquad\text{ for all }x, y \in A\text{ and }m, n \in \mathbb N.$$ The bottom line is that $N$ is a right cancellative, duo monoid that is not left cancellative. $\endgroup$
    – Salvo Tringali
    Commented Oct 15 at 3:10
  • $\begingroup$ In fact, all what you need in this proof is seemingly that $\Phi$ is a surjective, non-injective endomorphism of a group $G$ and $N$ is the semidirect product of $G$ by the additive monoid $\mathbb N$ of non-negative integers wrt the left unitary action of $\mathbb N$ on $G$ given by $(m,a) \mapsto \Phi^m(a)$. AFAICS, it's not even necessary to assume that $G$ is abelian. $\endgroup$
    – Salvo Tringali
    Commented Oct 15 at 6:49
  • $\begingroup$ Yes that is true. I think you might even be able to get away with M being any cancellative and duo monoid with action by surjective endomorphisms on a group not all of which are injective $\endgroup$
    – Benjamin Steinberg
    Commented Oct 15 at 9:46
  • $\begingroup$ Sorry I originally used a instead of t and then realized it could be confusing because the group was A. I missed though chasing a to t in that one line. I fixed it $\endgroup$
    – Benjamin Steinberg
    Commented Oct 15 at 12:52

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