Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Neither universal algebra nor semigroup theory is something I really know much about, so this question might not be appropriate for MO; if so, I'll move it to MSE.

Recently, I've been playing around with some semigroups in the context of universal algebra and I've run across a class of semigroups with a particularly nice-seeming (at least, to me) property. I've put my motivation for considering this property at the end of the question.

The property of semigroups I'm interested in is:

Suppose $S=(S, *)$ is a semigroup (I'll omit the "$*$" where not confusing), and $\Sigma, A\subseteq S$ are disjoint such that $\Sigma\sqcup A$ generates $S$. There is a unique minimal congruence (=equivalence class respecting $*$) $E$ on $S$ containing the relations $$\{(u, v): u, v\in \Sigma\} \text{ and }\{(a, b): a, b\in A\};$$ basically, we have $sEs'$ iff $s, s'$ can be represented by words $w, w'\in (\Sigma\sqcup A)^*$ such that $w'$ is gotten from $w$ by replacting some symbols from $\Sigma$ occurring in $w$ with other symbols from $\Sigma$, and some symbols from $A$ occurring in $w$ with other symbols from $A$.

Then say:

$(\Sigma, A)$ generates $S$ nicely if

  • (1) $A$ is an $E$-equivalence class;

  • (2) for each $s\in S$ there is a unique $z=z_s\in\mathbb{Z}$ such that for any words $\sigma\in \Sigma^*$, $a\in A^*$ with $\vert\sigma\vert-\vert a\vert=z-1$, we have $\hat{\sigma}s\hat{a}\in A$ (where, pedantically, $\hat{\sigma}$ denotes the element of $S$ corresponding to the word $\sigma$, etc.); and

  • (3) $z_{st}=z_s+z_t$.

Note that for $a\in A$, $\sigma\in \Sigma$, we have $z_a=1$ and $z_\sigma=-1$. So $z_s$ is just counting how many elements of $A$ versus $\Sigma$ are needed to describe $s$. (This number, $z_s$, is what I called a "weight" in the title. "Weight" doesn't seem like a good term for it, but I couldn't think of a better one.)


A semigroup $S$ is nicely generated if it is nicely generated by some pair of disjoin $\Sigma, A\subseteq S$.

My question is twofold:

(1) What semigroups are nicely generated? More broadly, are there natural necessary/sufficient conditions for being nicely generated?


(2) Where can I find out more about nicely generated semigroups (or anything related)?

I suspect this is something very basic, but I haven't been able to find it; partly because I don't really know what to search for. Towards (1), the only thing I know is the obvious fact that if $S$ is nicely generated then $S$ has $(\mathbb{Z}, +)$ as a semigroup quotient.

Now the "why would anyone care?" part:

Example: Consider the semigroup $S$ generated by $\{\sigma, a\}$ subject to the single relation $\sigma aa=a$. Then $(\{\sigma\}, \{a\})$ generates $S$ nicely.

This is the semigroup induced (in a weird way) by the trivial algebra with domain $\{a\}$ in the language $\{\sigma\}$ consisting of a single binary operation.

More generally, suppose $\mathcal{A}=(\Sigma, A)$ is an algebra (in the sense of universal algbera) with domain $A$ and language $\Sigma$, where each element of $\Sigma$ is a binary operation. Then consider the semigroup $Sem(\mathcal{A})$ generated by $\Sigma\sqcup A$, subjected to all relations of the form $$ \sigma a b=c$$ for $\mathcal{A}\models$"$\sigma(a, b)=c$".

Now, unless I've made a mistake (which I may have), $Sem(\mathcal{A})$ is always nicely generated by $(\Sigma, A)$; and conversely, given any semigroup nicely generated by $(\Sigma, A)$, we can pull back to an algebra. This leads to a bunch of potentially interesting questions: for example, what does it mean if $Sem(\mathcal{A})=Sem(\mathcal{B})$? And how should we think about a semigroup homomorphism $Sem(\mathcal{A})\rightarrow Sem(\mathcal{B})$? Etc.

(Of course, this can be generalized to contexts beyond just allowing binary operations in $\Sigma$; but that gets more tedious - in particular, we have to assign elements of $\Sigma$ weights, corresponding to their arities, and sum these when computing $z_s$ - and I'm not sure it's much more interesting.)

share|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.