# Is the class of inverse semigroups globally determined?

This question is a follow-up to this one I asked on math.stackexchange. I've decided to ask here because I believe this is a research-level question. I'm sorry if I'm wrong -- I'm not a researcher myself and may well be mistaken about it.

I will reitarate the definitions I posted in an answer to that question. For a semigroup $S,$ the set $\mathcal P (S)=2^S\setminus \lbrace \emptyset\rbrace$ with mutliplication given by $$AB=\lbrace ab\ |\ a\in A,b\in B\rbrace$$ is called the power semigroup of $S$. (I don't really know what removing the empty set from it makes any better, but this is what I found in most papers.) We say that two semigroups are globally isomorphic iff their power semigroups are isomorphic. We say that a class of semigroups $\mathscr C$ is globally determined iff any two globally isomorphic semigroups in $\mathscr C$ are automatically isomorphic, ie. for any $S,T\in\mathscr C$ we have $$\mathcal P(S)\cong\mathcal P(T)\implies S\cong T$$

The question I've linked to was about the class of all groups. As Steve D kindly pointed out in his answer, it is actually very simple to see that the class of all groups is globally determined. I started wondering if we can generalize this fact to all inverse semigroups, but in vain. There is little literature on this, even if I said otherwise previously. (I then meant that there was more than I'd expected.) Also, it seems to me that about 1990 papers on these things stopped appearing completely. Some that I've found are referenced on the linked page.

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You might add to the clariity of presentation (and utiility of your definition) by saying a class is globally determined if for S and T in the class S is glob. iso. to T implies S is iso. to T . The current presentation leads to a careless (on the part of the reader) but alternate reading involving cardinality. Gerhard "Ask Me About System Design" Paseman, 2012.04.30 –  Gerhard Paseman Apr 30 '12 at 15:58
Thank you. I've added some words to make it clearer. Is it OK now? –  Michał Masny Apr 30 '12 at 16:08
My former comments were wrong. I found a paper with google saying all finite semigroups are globally determined. –  Benjamin Steinberg Apr 30 '12 at 16:21
Is it a new paper? I know about a 1984 paper by Gould and Iskra in which they say that finite simple semigroups are globally determined. –  Michał Masny Apr 30 '12 at 16:29
Actually, there is an paper without proofs in a 1987 conference proceedings by Tamura. Google preview obscures the relevant pages and it doesn't seem a full journal version was ever published. –  Benjamin Steinberg Apr 30 '12 at 20:03

(comment) I do not know the answer to your question, but I do know the reason why sometimes it is convenient to remove the empty sets. This is because more properties of the semigroup are preserved in passing to its power. For example, if S is a rectangular band (that is, it satisfies xyz=xz) then $2^S \setminus \{ \emptyset\}$ is still a rectangular band, while $2^S$ is not (take $y= \emptyset$ and $x, z$ nonempty).

There are general results characterizing what is preserved in passing from some structure to its power, even in a general algebraic setting (structures with any number and any kind of operations, possibly also relations, even infinitary). If you disallow the empty set, satisfaction of linear equation is preserved ("linear" means that each variable appears at most once on each side). If you allow the empty set, you have to deal with balanced linear equations (this means that each variable appears exactly once on each side). For example, the above equation xyz=xz is linear but not balanced.

Probably, such arguments are not of much use in the case of inverse semigroups, since I do not see how to define them only in terms of linear equations (even allowing inverse as a unary operation). However, in case someone is interested, references for the general cases can be found in Brink, C.: Power structures. Algebra Universalis 30, 177–216 (1993) or Gratzer G., Whitney S., Infinitary varieties of structures closed under the formation of complex structures, Colloq. Math. 48 (1984) 1-5.

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Power semigroups of inverse semigroups contain all finite semigroup and so there is no identity satisfied even by power sets of finite semigroups v –  Benjamin Steinberg May 1 '12 at 20:57
I suspected something like this. Anyway, $2^S$ is determined by $2^S \setminus \{ \emptyset\}$, and conversely, hence, as far as the present question is concerned, there is no essential difference, whether we remove the empty set or not. Hence my "answer" is definitely only a remark (it seems I have not enough reputation to put remarks; if some admin wants to change my answer into a remark perhaps it appropriate) –  Paolo Lipparini May 9 '12 at 12:16

This is not a proof yet, but reduces the question to a simpler one. I think that inverse semigroups are not determined up to isomorphism by their power semigroups. Indeed, consider any countable linear order $L$. It can be viewed as a semilatice (=idempotent inverse semigroup) $S(L)$ with $ab=a$ iff $a\le b$. Order $L$ is isomorphic to $L'$ iff the semigroups $S(L), S(L')$ are isomorphic. Every subset $S$ of $L$ satisfies $SS=S$. So $2^L$ is a semilattice too. The product of two subsets $XY$ consists of all elements of $X$ which are not bigger than some element of $Y$ and all elements of $Y$ which are not bigger than some element of $X$. In other words, let $a$ be the supremum of elements of $X$, $b$ be the supremum of elements of $Y$. Assume that $a\ge b$. Then $XY$ consists of all elements of $X\cup Y$ which do not exceed $b$.

Now let $L$ be the order $\mathbb{Q}$ and $L'$ be $L$ without 0. Are $2^L$ and $2^{L'}$ isomorphic? My conjecture is that the answer is yes" and this is an example of two non-isomorphic inverse semigroups with isomorphic power semigroups. A possibly easier example would be $L=\{-1/n, 0, 1/n\mid n=1,2,...\}$, $L'=\{1/n,-1/n\mid n=1,2,...\}$.

Update. No, for linear orders, the rigidity is true (if $2^L=2^{L'}$, then $L=L'$). Indeed, first the element $L\in 2^L$ can be described as the only element $x$ such that $\forall y (xy=x \& \forall t (yt=t\to xt=t) ) \to y=x$. Then let $U=L2^L$, i.e. the ideal generated by $L$ in $2^L$. It consists of all ideals of the order $L$. An element $x\in U$ has the maximal element in $L$ if and only if there exists $m=m(x)\in 2^L$ such that

• $Lm=x$, and

• for every $n\in 2^L$ such that $Ln=x$ we have $an=n$.

Now 1-element subsets in $2^L$ are those $m\in 2^L$ for which there exists $x\in U$ such that $m=m(x)$.

The order on 1-element subsets is defined as $m\le m'$ iff $mm'=m$. Thus if $2^L$ and $2^{L'}$ are isomorphic, then $L$ and $L'$ are isomorphic (since $L$ is elementary determined inside $2^L$).

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@Mark, semilattices are globally determined. In fact any isomorphism of 2^E with 2^{E'} for semilattices E and E' preserves singletons. See springerlink.com/content/0w638t6921834144 –  Benjamin Steinberg May 2 '12 at 16:41
@Ben: Right. I just rediscovered that fact (my proof can be adapted to all semilattices). I guess then the expected answer for all inverse semigroups should be yes". –  Mark Sapir May 2 '12 at 17:01
I expect so, at least if on has an isomorphism as involutive semigroups. I think one can recover the Schein completion, which determines the semigroup. –  Benjamin Steinberg May 2 '12 at 17:49
@Ben: What is Schein completion? Does Schein know that he has a completion named after him? –  Mark Sapir May 2 '12 at 22:16