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.



Ascending sequences of idempotents in inverse semigroups
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?]


