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This is extracted from this question following Benjamin Steinberg's suggestion.

For a semigroup $S,$ let $P(S)$ denote the power semigroup of $S,$ which is made up of all non-empty subsets of $S$ with the operation $AB=\{ab\,|\,a\in A,b\in B\}.$

I'm thining about the following conditions $(P1)$ and $(P2)$ on $S:$

$(P1)$ Every left-cancellable element of $P(S)$ is a singleton.

$(P2)$ Every cancellable element of $P(S)$ is a singleton.

1) Are the two conditions equivalent? Are they equivalent for left-cancellative semigroups? Are they equivalent for cancellative semigroups?

2) Can we characterize semigroups satisfying these conditions? Left-cancellative ones? Cancellative ones?

The questions in bold interest me the most. The class of cancellative semigroups satisfying $(P1)$ contains groups and commutative semigroups. From above, any left-cancellative semigroup satisfying $(P1)$ will have to satisfy the right Ore condition.

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  • $\begingroup$ I would look at a cancellative semigroup which satisfies the left Ore condition but not the right one. This could potentially gave (P2) but not (P1). Maybe look at baumslag-solitar semigroups. They have Ore condition on only one side. $\endgroup$
    – Benjamin Steinberg
    Feb 3, 2015 at 3:56
  • $\begingroup$ If $S$ is a left-cancellative finite sgrp (either commutative or not), then $SX=SS=S$ for every $X\in P(S)$, with the result that no element of $P(S)$ is left-cancellative, unless $S$ is empty or a singleton, in which case (P1) and (P2) are trivially equivalent. This means that, as far as the focus is on left-cancellative sgrps, you're left (excuse the unintended pun!) with the case where no element of $S$, other than the identity if an identity exists, has finite order. $\endgroup$
    – Salvo Tringali
    Feb 3, 2015 at 8:19
  • $\begingroup$ I think from previous questions the op is interested in the infinite case $\endgroup$
    – Benjamin Steinberg
    Feb 3, 2015 at 14:50
  • $\begingroup$ @Salvo Thank you for your comment, but I think there's some confusion here. $SX=SS$ for all $X$ follows from the right cancellation law on the elements of $S,$ i.e. that the multiplication by $x$ on the right side is injective in $S$. But more importantly, what $SX=SS$ says is not that $X$ is not right-cancellable, but that $S$ is not left-cancellable, which has no bearing on the question I'm asking since that not the same side that the cancellative property that holds in $S$. $\endgroup$
    – Michał Masny
    Feb 3, 2015 at 17:19
  • $\begingroup$ Actually, a right-zero semigroup is left-cancellative and finite, and since its power semigroup is again a right-zero semigroup, it's also left-cancellative. $\endgroup$
    – Michał Masny
    Feb 3, 2015 at 17:19

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