I've enocuntered the following question in my current research, and I'd appreciate any help you could give me. This is probably well known to experts on the subject.

Let $S = \langle K \rangle$ be a finitely generated inverse semigroup. Recall that the set $E$ of idempotents (i.e. elements $e \in S$ such that $e^2 = e$) is partially ordered via $e \leq f$ when $ef = fe = e$ (idempotents always commute in inverse semigroups).

Question: Can $S$ have an infinite ascending sequence of idempotents? That is, are there $e_n \in E$ such that $e_1 < e_2 < \dots < e_n < \dots$?

By $e < f$ I mean that $e \leq f$ and $e \neq f$. The only instance of the latter behavior I've come accross is the semigroup $S = (\mathbb{N}, \min)$ (and its' relatives), where $n \cdot m := \min\{n, m\}$. In this case we have that $S$ is equal to its' semilattice of idempotents, and $1 < 2 < \dots$, but this semigroup is not finitely generated.


1 Answer 1


Yes. Indeed, for $X$ a set, let $G_X$ be the group of partial bijections of $X$, that are defined and identity outside a countable subset. I claim that, for $X$ uncountable, every countable subset of $G$ is contained in a (5-generator) finitely generated submonoid (and hence in a finitely generated inverse submonoid).

The claim being granted, and using that the power set of $\omega$ contains a chain isomorphic to $(\mathbf{Q},\le)$, one obtains such a chain of idempotents in a suitable inverse monoid.

Note: the same claim was proved by Sierpinski and Banach in the 1930's for the monoid of all self-maps of every set, and by Galvin (1995) for the group of all permutations of every set.

Now let me prove the claim, inspired by Galvin's proof. Let $(f_n)_{n\in\mathbf{Z}}$ be a sequence in $G_X$. So there exists an infinite countable subset $X_{0,0}$ such that for every $n$, each $f_n$ is defined and identity outside $X_{0,0}$. Choose for all other $(m,n)\in\mathbf{Z}^2$ an infinite countable susbet $X_{m,n}$, pairwise disjoint. Henceforth, all maps are assumed to be defined and identity outside $X'=\bigcup_{m,n}X_{m,n}$. Also fix a bijection $X_{0,0}\to X_{m,n}$ for all $(m,n)\neq (0,0)$, so that we identify $X'$ to $X_{0,0}\times\mathbf{Z}^2$.


  • $u$ as the permutation $(x,m,n)\mapsto (x,m+1,n)$;

  • $r$ as the permutation $(x,0,n)\mapsto (x,0,n+1)$, $(x,m,n)\mapsto (x,m,n)$ for $m\neq 0$;

  • $f$ as the permutation $(x,m,n)\mapsto (f_m(x),m,n)$ for $n\ge 0$ and $(x,m,n)\mapsto (x,m,n)$ for $n\ge 0$.

I claim that for every $m$ we have $f_m\in\langle u,u^{-1},r,r^{-1},f\rangle$, where $\langle\cdots\rangle$ means the submonoid generated (actually, it follows that $f_m\in\langle u,r,f\rangle_{\text{inverse-monoid}}$).

Indeed, write $g_m=u^mfu^{-m}$: then $g_m$ is like $f$, but shifted $m$ times to the right. Then one sees that $g_m(r^{-1}g_mr)^{-1}=f_m$, and the claim is proved.

[Note 1: observe that $f_m$ is written as a word of length $\le 2+2(2m+1)=4m+6$ with respect to the given generators: since this only depends on $m$, this shows that $G_X$ is "strongly distorted" (as monoid, and as inverse monoid) and in particular strongly bounded, a.k.a. Bergman's property.]

[Note 2: Probably it's also true for $X$ countable, with some further preliminary lemmas. Also with only two generators.]

[Note 3: from Vagner-Preston, every countable inverse monoid embeds into $G_{\aleph_1}$. As corollary, every countable inverse monoid embeds into a 3-generated one. This is probably well-known?]

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    $\begingroup$ Note 3: It is a result of Chris Ash from 1978 that every countable inverse semigroup embeds into a 2-generated inverse semigroup. $\endgroup$ Commented Mar 26, 2020 at 12:05
  • $\begingroup$ Thank you to both. @Ycor, what do you mean by "strongly distorted"? Would you have a reference where I could read about Bergman's Property? $\endgroup$ Commented Mar 26, 2020 at 14:04
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    $\begingroup$ @DiegoMartínez The claim appears in [Byleen, Karl, "Inverse semigroups with countable universal semilattices". Semigroups and their applications (Chico, Calif., 1986), 37–42, Reidel, Dordrecht, 1987]. There he attributes it to Ash, and says it appears in a survey article; I have not been able to find this survey. I do not know what the proof looks like. $\endgroup$ Commented Mar 26, 2020 at 14:17
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    $\begingroup$ It is shown in proposition 4.2 of cambridge.org/core/services/aop-cambridge-core/content/view/… that every countable subset of the symmetric inverse monoid is contained in a 2-generated inverse subsemigroup. $\endgroup$ Commented Mar 26, 2020 at 14:38
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    $\begingroup$ @DiegoMartínez given a variety of structures (e.g., inverse monoids), say with finite signature (= finitely many finitary laws), an object $G$ is strongly distorted if there exists a numerical sequence $(a_n)$ such that for every sequence $(g_n)$ in $G$ there exists a finite subset $F\subset G$ such that for every $n$, $g_n$ belongs to the substructure generated by $F$ and has length $\le a_n$ with respect to $F$ in some reasonable sense. $\endgroup$
    – YCor
    Commented Mar 26, 2020 at 15:08

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