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The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would obtain a positive (albeit partial) answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into the direct product of a family $(G_i)_{i \in I}$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, every subdirectly irreducible group embeds into a Prüfer group (and each non-empty subsemigroup of a Prüfer group is still a group).

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  • $\begingroup$ I don't know much about subdirect products, but is the semigroup $\langle a, b \mid ab = baa \rangle$ isomorphic to a subdirect product of groups? What happens when we push this through the machinery of Birkhoff's theorem? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Oct 8 at 6:19
  • $\begingroup$ @Carl-FredrikNybergBrodda Let's call S the semigroup you are suggesting to consider. By Adian's embedding theorem, S embeds into a group. What about the intersection of all proper congruences on S? If it's proper, then S is subdirectly irreducible and we have an example proving that the answer to the OP is no. Here (following Birkhoff), a congruence is proper if it's not the discrete congruence $\Delta := \{(x,x): x \in S\}$. $\endgroup$
    – Salvo Tringali
    Commented Oct 8 at 6:47
  • $\begingroup$ By definition, S is the quotient of the free semigroup on the two-element set $\{a, b\}$ by the smallest congruence $\theta$ containing the pair $(ab, baa)$. Let me ask a naive question: is there any "direct relation" between $\theta$ and the intersection of all proper congruences on S? Are they one and the same thing (up to the replacing of a and b with their $\theta$-classes)? If yes, this is probably a basic property of presentations that I should know but I don't. $\endgroup$
    – Salvo Tringali
    Commented Oct 8 at 7:10
  • $\begingroup$ I know nothing about semigroup, but… if I understand rightly, Kearnes’ proof goes through when the universal group $U$ of $S$ is residually finite. (Such a $U$ embeds into the product of all its finite quotients, and $S$ will surject each finite quotient.) $\endgroup$
    – HJRW
    Commented Oct 8 at 7:15
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    $\begingroup$ To cover the residually finite and abelian case, observe that if a semigroup $S$ is embeddable in a residually torsion group, then it is a subdirect product of groups. $\endgroup$
    – YCor
    Commented Oct 8 at 18:10

2 Answers 2

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The answer is no. There indeed exists a nonabelian subsemigroup of a group all of whose quotient groups are abelian.

Lemma. Let $G$ be a bi-ordered group and $G_+$ its subsemigroup of positive elements. Then every homomorphism $G_+\to L$, $L$ a group, uniquely extends to $G$.

Proposition. Under the assumptions of the lemma, suppose that every proper quotient group of $G$ is abelian. Then every quotient group of $G_+$ is abelian.

Proof of the proposition: let $L$ be a group and $G_+\to L$ be surjective. So its extension to $G$, given by the lemma, is surjective but not injective. So $L$ is abelian.

Example. Let $F$ be Thompson's group of the interval (piecewise affine increasing self-homeos of $[0,1]$ whose breakpoints are dyadic and slopes are integral powers of $2$.). Then every proper quotient of $F$ is abelian, and $F$ is bi-orderable (e.g. $F_+$ can be defined as the set of nonidentity elements of $F$ whose first nontrivial slope is $>1$).

Proof of the lemma. Let $f:G_+\to L$ be a homomorphism, $L$ a group.

Claim a: $f(g^{-1}hg)=f(g)^{-1}f(h)f(g)$ for all $g,h\in G_+$ [note that for this to make sense, we need bi-orderability, i.e., invariance of $G_+$ under conjugation]. Indeed, $f(h)f(g)=f(hg)=f(gg^{-1}hg)=f(g)f(g^{-1}hg)$ which proves the claim.

Claim b: if $g_1,h_1,g_2,h_2\in G_+$ with $g_1h_1^{-1}=g_2h_2^{-1}$ then $f(g_1)f(h_1)^{-1}=f(g_2)f(h_2)^{-1}$.

Indeed, write the assumption as $g_1 h_1^{-1}h_2h_1=g_2h_1 $. Then $$f(g_2)f(h_1)=f(g_2h_1)=f(g_1 h_1^{-1}h_2h_1)=f(g_1)f(h_1^{-1}h_2h_1)=f(g_1)f(h_1)^{-1}f(h_2)f(h_1),$$ which is what is desired (the last equality used Claim a).

Claim b makes it valid to define $f(gh^{-1})=f(g)f(h)^{-1}$ for $g,h\in G_+$, so $f$ is now a well-defined map $G\to L$.

Claim c: $f$ is a homomorphism $G\to L$. Indeed, for $g_1,h_1,g_2,h_2\in G_+$, we have $$f(g_1h_1^{-1}g_2h_2^{-1})=f(g_1h_1^{-1}g_2h_1(h_2h_1)^{-1})=f(g_1h_1^{-1}g_2h_1)f(h_2h_1)^{-1}$$ $$=f(g_1)f(h_1^{-1}g_2h_1)f(h_2h_1)^{-1}=f(g_1)f(h_1)^{-1}f(g_2)f(h_1)f(h_1)^{-1}f(h_2)^{-1}$$ $$=f(g_1h_1^{-1})f(g_2h_2^{-1}).$$ (Claim a is used in the 4th equality.) So the lemma is proved.

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    $\begingroup$ PS in categorical terms, the lemma precisely says that for a bi-ordered group $G$, the embedding $G_+\to G$ is the universal group embedding. $\endgroup$
    – YCor
    Commented Oct 8 at 18:08
  • $\begingroup$ Very nice. Just let me make one point slightly more explicit: If a sgrp S is a subdirect prod of a family $(G_i)_{i\in I}$ of groups, then, by def., S embeds into the direct prod of the $G_i$'s and, for each $j\in I$, there is a surjective sgrp hom $f_i:S\to G_i$. It follows, by the 1st isomorphism thm for sgrps, that $G_i$ is iso to the quotient sgrp $S/\ker(f_i)$, where $\ker(f_i)$ is the kernel of $f_i$. If S is not a group and each of its proper quotients is commutative, then $G_i$ is abelian and hence $S$ is commutative (as it's a subsgrp of the direct product of abelian groups). $\endgroup$
    – Salvo Tringali
    Commented Oct 9 at 12:10
  • $\begingroup$ @SalvoTringali I don't think of this in this way (I'm not sure how to define kernel of a semigroup homomorphism, so I wouldn't invoke kernels — quotient semigroups, in general, are by congruences). How I'd phrase it: the surjective homomorphism $S\to G_i$ is surjective, $S$ is not a group while $G_i$ is a group, so it cannot be bijective, hence $G_i$ is, by the assumption, abelian. $\endgroup$
    – YCor
    Commented Oct 9 at 12:14
  • $\begingroup$ Concerning "I'm not sure how to define kernel of a semigroup homomorphism [...] — quotient semigroups, in general, are by congruences": the kernel of a sgrp homomorphism is a congruence (on the domain of the homomorphism); see, for instance, Prop 3.3 in the 1995 edition of Grillet's book on sgrps. That being said: with my previous msg, I mainly wanted to emphasize that each of the $G_i$'s is indeed a quotient of S, which is of course true of any factor in any subdirect rep of any algebra (by the general formulation of the 1st isomorphism thm for algebras.) $\endgroup$
    – Salvo Tringali
    Commented Oct 9 at 12:38
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    $\begingroup$ By the way, YCor's answer can be used to prove that a cancellative semigroup need not be a subdirect product of subdirectly irreducible cancellative semigroups, see mathoverflow.net/questions/480364#480364 $\endgroup$
    – Salvo Tringali
    Commented Oct 10 at 7:46
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I seem to be late to this game, but since you asked for a reference request here is one which also gives a more elementary example which has interesting properties in its own right. The reference is P. Trotter, Cancellative simple semigroups and groups, Semigroup Forum Volume 14, pages 189–198, (1977). Let $\mathbb Q_+$ be the set of positive rational numbers. Consider the semidirect product $S=\mathbb Q_+\rtimes \mathbb Q_+$ of the multiplicative group $\mathbb Q_+$ with the additive semigroup $\mathbb Q_+$. This is clearly group embeddable into the affine group $\mathbb Q\rtimes \mathbb Q^\times$. The semigroup $S$ is ideal simple (it has no proper two-sided ideals) and it has no idempotents. Trotter shows that the unique minimal congruence on $S$ giving a group is the quotient map $S\to \mathbb Q_+$. Thus $S$ is not a subdirect product of groups.

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  • $\begingroup$ Thanks for the reference (and +1). I'm accepting YCor's answer because (i) it came first and (ii) I believe his construction can be used to give a negative answer to mathoverflow.net/questions/480202 $\endgroup$
    – Salvo Tringali
    Commented Oct 9 at 12:14
  • $\begingroup$ I thought about this example for your question. Both examples have a unique minimal group congruence. I'm not sure whether either answer your first question. I ran into this example a few days ago thinking about that question. Notice trotters example has the property that any cancellative quotient that has an idempotent is a group do it should have to be a subdirect product of cancellative semigroups without idempotents. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 9 at 15:51
  • $\begingroup$ I posted an answer: mathoverflow.net/questions/480364#480364 I hope I haven't overlooked anything irreparable. $\endgroup$
    – Salvo Tringali
    Commented Oct 9 at 16:21

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