Ascending sequences of idempotents in inverse semigroups

Question

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.

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 $X_{0,0}\times\mathbf{Z}^2$.

Define

• $u$ 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?]