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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 $n\in\mathbb N$.

Question. Can an infinite Shelah semigroup be commutative?

This problem was motivated by the following results:

Theorem (Shelah, 1980). For any infinite cardinal $\lambda$ with $\lambda^+=2^\lambda$ there exists a group $G$ of cardinality $|G|=\lambda^+$, which is a $6640$-Shelah semigroup.

Corollary. Under CH there exists a Shelah semigroup of cardinality $\aleph_1$.

Theorem (Protasov, 2010). Each countable Shelah semigroup is finite.

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

Proposition (@YCor, 2018). A group is finite iff it is a 3-Shelah semigroup.

Theorem (Todorcevic, 1987). 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 $A\subset X$.


Added in Edit, after reading the answer of Keith Kearnes who referred to the paper 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.

Theorem (McKenzie, 1971). A Jonsson semigroup $S$ of infinite cardinality $\kappa$ is a non-commutative group if $\mathrm{cf}(\kappa)>\omega$ or $2^{<\kappa}\le\kappa$.

McKenzie asked in his paper if this theorem remains true without set-theoretic assumptions.

Problem (McKenzie, 1971). Is each infinite Jonsson semigroup a group?

Is this problem of McKenzie still open?

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  • $\begingroup$ My easy construction is not worth being called a theorem. Also Shelah produced Shelah semigroups (that are groups) of cardinal $\aleph_1$ in ZFC. $\endgroup$ – YCor Oct 30 '18 at 10:17
  • $\begingroup$ @YCor Ok. Then how to call your result? It is interesting anyway and should be mentioned in this context. $\endgroup$ – Taras Banakh Oct 30 '18 at 10:25
  • $\begingroup$ @YCor Are you sure that Shelah constructed a Shelah semigroup in ZFC? It is not stated this way in his paper. His ZFC-Theorem 2.9 is formulated differently and does not mention any $n_0$ which appear explicitly in the formulation of his CH-Theorem 2.1 Moreover, proving his Theorem C on non-topologizable groups, Shelah proves this theorem only under CH. Why? If he would produce a Shelah group in ZFC, why CH-appears in the application of this theorem to the problem of topologizability? $\endgroup$ – Taras Banakh Oct 30 '18 at 10:28
  • $\begingroup$ From the MR review of Shelah's paper: The author constructs some remarkable infinite groups, notably "Jónsson groups'', that is, groups of uncountable cardinality containing no proper subgroups (or better: no proper subsemigroup) of the same cardinality. One construction works for all successor cardinals if we assume the generalized continuum hypothesis; a variant works for $\aleph_1$ with no set-theoretic assumptions. mathscinet.ams.org/mathscinet-getitem?mr=579953 But maybe this refers to being Jonsson, and not the more precise exponent fact? I don't have the paper actually. $\endgroup$ – YCor Oct 30 '18 at 10:37
  • $\begingroup$ @YCor I read the ``variant" mentioned in MR-review as the fact that the ZFC-result of Shelah yields only a Jonsson semigroup, but not a Shelah semigroup (whose construction is different and do required additional assumptions like CH). $\endgroup$ – Taras Banakh Oct 30 '18 at 10:40
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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.

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