8
$\begingroup$

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.

Theorem (Banakh, 2022). 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). 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?

$\endgroup$
9
  • $\begingroup$ 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, 2018 at 10:17
  • $\begingroup$ @YCor Ok. Then how to call your result? It is interesting anyway and should be mentioned in this context. $\endgroup$ Oct 30, 2018 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? $\endgroup$ Oct 30, 2018 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, 2018 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). $\endgroup$ Oct 30, 2018 at 10:40

1 Answer 1

13
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.