# Can a Shelah semigroup be commutative?

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?

• My easy construction is not worth being called a theorem. Also Shelah produced Shelah semigroups (that are groups) of cardinal $\aleph_1$ in ZFC. – YCor Oct 30 '18 at 10:17
• @YCor Ok. Then how to call your result? It is interesting anyway and should be mentioned in this context. – Taras Banakh Oct 30 '18 at 10:25
• @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? – Taras Banakh Oct 30 '18 at 10:28
• 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. – YCor Oct 30 '18 at 10:37
• @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). – Taras Banakh Oct 30 '18 at 10:40