Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced by $S$: $$ (X, Y) \mapsto XY := \{xy \colon x \in X, \, y \in Y\}. $$ In the literature, $\mathcal P(S)$ is called the power semigroup (or global) of $S$. Two semigroups are then said to be globally isomorphic if their power semigroups are isomorphic.
QUESTION. Let $S$ and $T$ be semigroups and assume $S$ is cancellative and globally isomorphic to $T$. Is it known whether $S$ is necessarily isomorphic to $T$? And what about the case when $S$ is cancellative and commutative?
If any of the answers turns out to be yes, I would greatly appreciate a reference.
Update (Feb 25, 2024). On the 2nd page of [M. Gould, J.A. Iskra, and C. Tsinakis, Globals of completely regular periodic semigroups, Semigroup Forum 29 (1984), 365-374], one can read:
Tamura [15] surveyed the recent papers and added to the list the class of all semilattice-indecomposable cancellative semigroups".
Here, 'the list' is the list of semigroups for which it was known by the time of Tamura's survey that two semigroups $S$ and $T$ in a certain class are globally isomorphic if and only if they are isomorphic; and [15] is [T. Tamura, On the semigroups of subsets of semigroups, Proc. 6th Symposium on Semigroups, Ritsumeikan Univ. (1982), 11-18]. Unfortunately, I don't have a copy of Tamura's paper.
In the meanwhile, I've arXived a short note where, as a special case of a more general result, a positive answer to the above question is provided for the class of cancellative commutative semigroups. The (cancellative) non-commutative case remains largely open.
