Can a semigroup with zero be globally isomorphic to a semigroup without zero?

This is not a great question for sure and it may even be trivial for all I know, but a couple of years ago, when I still thought I'd be a mathematician, I spent quite a lot of time thinking about it and my advisor wasn't able to help so I'd like to ask it here, even if only for the feeling that my effort wasn't completely wasted if anybody reads it. Hopefully that's OK.

Two semigroups are called globally isomorphic if their power semigroups are isomorphic. The power semigroup $P(S)$ of a semigroup $S$ is the semigroup of all non-empty subsets of $S$ with the natural multiplication.

Mogiljanskaja gave examples of semigroups which are globally isomorphic but not isomorphic, and more examples have been produced since. All of those examples, at least the ones I've seen, are semigroups with zero which differ on their annihilators, say, $\mathrm{Ann}(S) = \{s\in S\,|\,(\forall r\in S)\, rs=0\}$ or ones built from such semigroups. In general, there doesn't seem to be a great variety of examples. Mogiljanskaja asked what other kinds of examples could be found, and the general question seems to be: "what of the semigroup structure does a power isomorphism have to preserve?"

The first question that came to my head was if it was possible to have a semigroup with zero globally isomorphic to a semigroup without zero. Obviously, the power semigroup of a semigroup with zero has a zero as well, so we would have to have a semigroup without zero whose power semigroup has a zero. It's easy to construct such semigroups, and they have been studied. A semigroup whose power semigroup has a zero is called a homogroup and there's a paper by Clifford and Miller from 1948 about them, giving a general construction: Semigroups Having Zeroid Elements, A. H. Clifford and D. D. Miller, American Journal of Mathematics, Vol. 70, No. 1 (Jan., 1948), pp. 117-125. (JSTOR)

They are exactly the semigroups which have a two-sided ideal that's a group. (I know that thanks to this question.) That has to be the least ideal of the semigroup and so its kernel. Any semigroup with zero is a homogroup and $\{0\}$ is that ideal in those.

So the question is whether a semigroup without zero can be globally isomorphic to a homogroup with a bigger kernel. And another question that immediately comes to mind is whether, simply, two globally isomorphic homogroups can have non-isomorphic kernels.

So these are the questions I'd like to ask. If there are such pairs, I think that would be genuinely interesting. Probably there aren't though and the question is boring, but I wasn't able to prove it.

• It's a nice question, which I've upvoted, but ... I'm in the mood to rant today ... and, although I'm sure it's not your fault, ... "globally isomorphic" is terrible terminology! – Jeremy Rickard May 14 '16 at 13:45
• @JeremyRickard I think so too :) – Michał Masny May 14 '16 at 14:06
• Is "power isomorphic" already used for something else? – Jeremy Rickard May 14 '16 at 14:28
• @Jeremy I don't know. The terminology comes from the power semigroup being called a "global" in some papers. I think the word "global" was originally used in Russian papers. As far as what I've read goes, "power" stuck to the semigroup and "global" stuck to the isomorphism. I'm not really sure why, and I could also simply be wrong. – Michał Masny May 14 '16 at 14:44

Here is an idea to prove the minimal ideal can be recovered. Let $G$ be the minimal ideal of $S$ and let $e$ be its identity; it is a central idempotent. Then $E=\{e\}$ is a central idempotent of $P(S)$ with $EP(S)$ isomorphic to $P(G)$. Since groups are globally determined you can recover $G$ (as the group of units of $P(G)$). So it remains to show the idempotent $E$ can be recovered from the semigroup structure of $P(S)$. My feeling is that it is the unique maximal central idempotent $F$ of $P(S)$ in the usual order on idempotents such that $FP(S)$ is isomorphic to the power set of a group.
• Hello Benjamin, thanks. On a tangent, I wanted to say that soon after your and Mark Sapir's comments on the linked question, I understood why a group $G$ cannot be globally isomorphic to a non-group $S$. If $P(G)$ is isomorphic to $P(S)$ then $S$ has to be a homogroup with a group of units $G'$ isomorphic to $G$ and kernel $I$. And so if $S$ is not a group, then $P(S)$ has a at least two elements: $G'$ and $I$ that absorb every unit in $P(S)$. Only $G$ absorbs every unit of $P(G)$. – Michał Masny May 14 '16 at 13:22