Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

Question

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the taking of homomorphic images within the category of magmas, every semigroup is (isomorphic to) a subdirect product of subdirectly irreducible semigroups.

What about cancellative semigroups? More precisely:

Q. Is every cancellative semigroup a subdirect product of subdirectly irreducible, cancellative semigroups?

If a semigroup $T$ is the homomorphic image of a cancellative semigroup $S$, then $T$ need not be cancellative (otherwise we would be done). However, I am hoping that there is a generalization of Birkhoff's theorem that applies not only to (finitary) algebras in the sense of Birkhoff's paper, but also to algebras (such as cancellative semigroups) that satisfy certain equational implications.

UPDATE 1. Below, Keith Kearnes proves that every cancellative commutative semigroup is a subdirect product of (subdirectly irreducible) groups. This provides a strong affirmative answer to my question in the commutative setting, and it is perhaps worth remarking that nothing similar can generally be true in the non-commutative setting: otherwise, every cancellative semigroup would embed into a group, which is not the case, as first proved by Mal'cev in [Math. Ann. 113 (1937), No. 1, 686-691]. It thus seems natural to ask what happens in the specific situation when $S$ embeds into a group, and I did it in a separate thread (here).

Answer 1

I am hoping that there is a generalization of Birkhoff's theorem that applies not only to algebras (in the sense of Birkhoff's paper) but also to algebras that satisfy certain equational identities (such as cancellative semigroups).

There is a version of Birkhoff's Theorem that applies to quasivarieties, such as the quasivariety of cancellative semigroups, but it doesn't give you what you are asking for here. Namely, if $\mathcal{Q}$ is a quasivariety and $\mathbf{A}\in \mathcal{Q}$, then call $\theta$ a $\mathcal{Q}$-congruence on $\mathbf{A}$ if $\mathbf{A}/\theta\in \mathcal{Q}$. The $\mathcal{Q}$-congruences on $\mathbf{A}$ form an algebraic lattice, and that is enough for the '$\mathcal{Q}$-version' of Birkhoff's Theorem to hold: every algebra in $\mathcal{Q}$ is a subdirect product of '$\mathcal{Q}$-subdirectly irreducibles'. An algebra $\mathbf{S}\in \mathcal{Q}$ is $\mathcal{Q}$-subdirectly irreducible if it has a least nonzero $\mathcal{Q}$-congruence. In particular, every cancellative semigroup is a subdirect product of cancellative semigroups that are subdirectly irreducible relative to the class of cancellative semigroups.

But a $\mathcal{Q}$-subdirectly irreducible might not be subdirectly irreducible in the absolute sense, so this doesn't answer the question that you asked. [For example, if $\mathcal{Q}$ is the quasivariety of torsion-free abelian groups, then every member of $\mathcal{Q}$ is a subdirect product of $\mathcal{Q}$-subdirectly irreducibles. But no torsion-free abelian group is subdirectly irreducible in the absolute sense, so the members of $\mathcal{Q}$ are not subdirect products of (absolutely) subdirectly irreducible torsion-free abelian groups.]

Let me at least give an affirmative answer to the commutative version of your question.

Claim. If $S$ is a commutative cancellative semigroup, then $S$ is a subdirect product of subdirectly irreducible cancellative semigroups. (In fact, $S$ is a subdirect product of subdirectly irreducible abelian groups considered as semigroups.)

Reasoning. Embed $S$ in its universal group $U$, which is an abelian group. For each $a\neq b$ in $S$ ($\subseteq U$), choose a group congruence $\theta$ on $U$ that is maximal for $(a,b)\notin \theta$. The group $U/\theta$ is a subdirectly irreducible abelian group, hence it is isomorphic to a subgroup of some Prüfer group $\mathbb Z_{p^{\infty}}$. The composite map $f_{a,b}\colon S\to U\to U/\theta\to \mathbb Z_{p^{\infty}}$ separates $a$ and $b$ and has image $f_{a,b}(S)$ that is a subsemigroup of $\mathbb Z_{p^{\infty}}$. Every subsemigroup of $\mathbb Z_{p^{\infty}}$ is a subdirectly irreducible group that is also subdirectly irreducible as a semigroup. Thus, $\prod f_{a,b}\colon S\to \prod f_{a,b}(S)$ represents $S$ as a subdirect product of subdirectly irreducible abelian groups. \\\

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Edit. Let me respond to a request in the comments by giving a reference for the relative version of Birkhoff's Theorem. See Theorem 1.1 of

Finite basis theorems for relatively congruence-distributive quasivarieties
Don Pigozzi
Trans. Amer. Math. Soc. 310 (1988), 499-533.

Answer 2

Sorry for answering my own question, but YCor's construction in a related thread (here) gave me a lightbulb moment. Hopefully, it's not a broken lightbulb.

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The answer to the question asked in the OP appears to be no.

Let $S$ be the positive cone of a non-abelian, totally ordered group $(G, \leq)$ all of whose proper quotients are abelian (the existence of such an object is proved by YCor as part of his construction). In particular, this means that $\leq$ is a total order on $G$ with the further property that $a \leq b$ implies $xay \leq xby$ for all $x, y \in G$, and $S$ is the subsemigroup $\{a \in G: 1_G \lneq a\}$ of $G$ (everything here is written multiplicatively).

Our goal is to prove the following:

Theorem. The semigroup $S$ is not a subdirect product of subdirectly irreducible, cancellative semigroups.

The proof is going to demand some work. I will organize it into a series of lemmas.

Lemma 1. The positive cone $H$ of any totally ordered group is a (cancellative) duo semigroup, that is, $aH = Ha$ for all $a \in H$.

Proof. See, for instance, here.

Lemma 2. Every homomorphic image of a duo semigroup is duo.

Proof. Let $f \colon H \to K$ be a surjective semigroup homomorphism, and assume $H$ is duo. If $u, v \in K$, then $u = f(x)$ and $v = f(y)$ for some $x, y \in H$; and since $H$ is duo, there are $a, b \in H$ such that $xy = ax = yb$. It follows that $uv = f(xy) = f(ax) = f(a) u$ and, in a similar way, $uv = v f(b)$. Thus, $K$ is duo. []

For the next results, recall that a subset $I$ of a semigroup $H$ is a left (resp., right) ideal if $HI \subseteq I$ (resp., $IH\subseteq I$); some people would also require that $I$ is non-empty, but we don't. In addition, $I$ is a (two-sided) ideal if it is both a left and a right ideal.

Lemma 3. If a cancellative duo semigroup $H$ has a non-empty minimal ideal, then it is a group.

Proof. Let $M$ be a non-empty minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Exercise 22.4A in Lam's A First Course in Noncommutative Rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a non-empty minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid and $e$ is its identity. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group. []

Lemma 4. Every non-empty, subdirectly irreducible semigroup $H$ has a non-empty minimal ideal.

Proof. Let $\mathfrak I(H)$ be the family of all non-empty ideals of $H$. Clearly, $M := \bigcap \mathfrak I(H)$ is itself an ideal of $H$; and by construction, it is contained in any non-empty ideal. We are left to show that $M \ne \emptyset$.

To this end, denote by $\theta_I$ the Rees congruence on $H$ induced by an ideal $I$ of the semigroup itself (see, for instance, here), and note that $I$ is empty if and only if $\theta_I$ is trivial, that is, $\theta_I = \Delta_H := \{(x,x): x \in H\}$. Since a semigroup is subdirectly irreducible if and only if the intersection of all its non-trivial congruences is non-trivial (see, for instance, p. 765 of the 1944 BAMS paper by Birkhoff cited in the OP), it follows that $\theta_M \ne \Delta_H$ and hence $M \ne \emptyset$. []

We are finally ready to prove the main result:

Proof of Theorem. Suppose for the sake of contradiction that the semigroup $S$ defined at the beginning of this post has a subdirect representation $p \colon S \to \prod_{j \in J} S_j$, where all the $S_j$'s are subdirectly irreducible cancellative semigroups. Each of the $S_j$'s is then a homomorphic image of $S$; so, by Lemmas 1 and 2, $S_j$ is a (cancellative) duo semigroup. It follows, by Lemmas 3 and 4, that the $S_j$'s are groups. This is however impossible, for YCor showed here that $S$ is not a subdirect product of groups. []