A semigroup $S$ is called


$\bullet$ *$n$-Shelah* for a positive integer $n$ if $S=A^n$ for any subset $A\subset S$ of cardinality $|A|=|S|$;


$\bullet$ *Shelah* if $S$ is $n$-Shelah for some $n\in\mathbb N$.



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> **Question.** Can an infinite Shelah semigroup be commutative?
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This problem was motivated by the following results:


**Theorem ([Shelah, 1980](https://www.sciencedirect.com/science/article/pii/S0049237X08713466)).** For any infinite cardinal $\lambda$ with $\lambda^+=2^\lambda$ there exists a group $G$ of cardinality $|G|=\lambda^+$, which is a $6640$-Shelah semigroup.


**Theorem ([Banakh, 2022](https://arxiv.org/abs/2212.01750)).** For any cardinal $\lambda$ with $\lambda^+=2^\lambda$, every group $H$ of cardinality $|H|\le\lambda$ is a subgroup of a $36$-Shelah group $G$ of cardinality $|G|=\lambda^+$.


**Corollary.** Under CH there exists a Shelah semigroup of cardinality $\aleph\_1$.


**Theorem ([Protasov, 2010](https://link.springer.com/article/10.1007/s00012-010-0032-0)).** Each countable Shelah semigroup is finite.


**Theorem (folklore?).** A commutative group is finite iff it is a Shelah semigroup.


**Proposition ([@YCor, 2018](https://mathoverflow.net/questions/313516/a-shelah-group-in-zfc)).** A group is finite iff it is a 3-Shelah semigroup.


**Theorem ([Todorcevic, 1987](https://projecteuclid.org/download/pdf_1/euclid.acta/1485890491)).** There is a commutative binary operation $\cdot:X\times X\to X$ of a set $X$ of cardinality $|X|=\aleph\_1$ such that $X=A^2:=\{ab:a,b\in A\}$ for any uncountable subset $A\subset X$.




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**Added in Edit**, after reading the answer of Keith Kearnes who referred to [the paper](https://link.springer.com/article/10.1007%2FBF02944951) of Ralph McKenzie who studied Jonsson semigroups.


Let us recall that a semigroup $S$ is *Jonsson* if $S=\bigcup\_{n\in\mathbb N}A^n$ for any subset $A\subset S$ of cardinality $|A|=|S|$. It is clear that each Shelah semigroup is Jonsson.



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> **Theorem ([McKenzie, 1971](https://link.springer.com/article/10.1007%2FBF02944951)).** A Jonsson semigroup $S$ of infinite cardinality $\kappa$ is a non-commutative group if $\mathrm{cf}(\kappa)>\omega$ or $2^{<\kappa}\le\kappa$.
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McKenzie asked in [his paper](https://link.springer.com/article/10.1007%2FBF02944951) if this theorem remains true without set-theoretic assumptions.



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> **Problem (McKenzie, 1971).** Is each infinite Jonsson semigroup a group?
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Is this problem of McKenzie still open?



Can a Shelah semigroup be commutative?
An infinite Shelah semigroup must be a Jonsson semigroup (meaning that it is an infinite semigroup whose proper subsemigroups have lesser power). Therefore the following paper answers the question asked on this page:
McKenzie, Ralph  
 
On semigroups whose proper subsemigroups have lesser power.  
 
Algebra Universalis 1 (1971), no. 1, 21-25.
McKenzie proves that   

(1) the only commutative Jonsson semigroups are the generalized cyclic groups, and  

(2) under GCH every Jonsson semigroup is the underlying semigroup of a group.
To reiterate and overexplain: infinite + Shelah + commutative implies infinite + Jonsson + commutative implies generalized cyclic. But the generalized cyclic groups are not Shelah. Thus infinite + Shelah + commutative semigroups do not exist. 








