A Shelah group in ZFC?

In his famous paper "On a problem of Kurosh, Jonsson groups, and applications" of 1980, Shelah constructed a CH-example of an uncountable group $$G$$ equal to $$A^{6641}$$ for any uncountable subset $$A\subset G$$.

Let us call a group $$G$$

$$\bullet$$ $$n$$-Shelah if $$G=A^n$$ for each subset $$A\subset G$$ of cardinality $$|A|=|G|$$;

$$\bullet$$ Shelah if $$G$$ is $$n$$-Shelah for some $$n\in\mathbb N$$;

$$\bullet$$ almost Shelah if for each subset $$A\subset G$$ of cardinality $$|A|=|G|$$ there exists $$n\in\mathbb N$$ such that $$A^n=G$$;

$$\bullet$$ Jonsson if each subsemigroup $$A\subset G$$ of cardinality $$|A|=|G|$$ coincides with $$G$$.

$$\bullet$$ Kurosh if each subgroup $$A\subset G$$ of cardinality $$|A|=|G|$$ coincides with $$G$$.

It is clear that for any group $$G$$ the following implications hold:

finite $$\Leftrightarrow$$ 1-Shelah $$\Rightarrow$$ $$n$$-Shelah $$\Rightarrow$$ Shelah $$\Rightarrow$$ almost Shelah $$\Rightarrow$$ Jonsson $$\Rightarrow$$ Kurosh.

In the mentioned paper, Shelah constructed a ZFC-example of an uncountable Jonsson group and also a CH-example of an uncountable 6641-Shelah group.

Problem 1. Can an infinite (almost) Shelah group be constructed in ZFC?

Problem 2. Find the largest possible $$n$$ (which will be smaller than 6640) such that each $$n$$-Shelah group is finite.

This result of Protasov implies

Theorem (Protasov). Each countable Shelah group is finite.

It is easy to show that each 2-Shelah group is finite.

Problem 3. Is each 3-Shelah group finite?

• @NateEldredge Thank you for the comment. I have corrected the title. – Taras Banakh Oct 23 '18 at 0:09
• When creating names for new mathematical objects I would propose to refrain from using weak + name :-) – Jan_Ch. Oct 23 '18 at 5:44
• I think that history will tell us, but there is nothing which is "almost Shelah". You're either Shelah or not at all Shelah. :-) – Asaf Karagila Oct 23 '18 at 12:58
• @PaulPlummer It is the construction: for every cardinal $\lambda$ with $\lambda^+=2^\lambda$ Shelah constructs a 6643-Shelah group of cardinality $\lambda^+$. But by the Easton's Theorem cardinals $\lambda$ with $\lambda^+=2^\lambda$ need not exist in ZFC. On the other hand, such a cardinal $\lambda$ (namely $\lambda=\aleph_0$ exists under CH. – Taras Banakh Oct 23 '18 at 13:21
• Easton's Theorem only applies to regular cardinals. To have $2^\lambda>\lambda^+$ everywhere, including at singular $\lambda$, is much more difficult and requires some very large cardinal hypotheses: jstor.org/stable/2944324 – François G. Dorais Oct 24 '18 at 0:26

I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)

Anyway:

there's no "3-Shelah" group. That is, every infinite group admits a subset $$W$$ such that $$W^3\neq G$$ and $$|W|=G$$. (Actually one can arrange $$W\cup W^2\cup W^3\neq G$$.)

Let $$G$$ be an infinite group. Let $$A$$, by Zorn, be a maximal subset such that $$1\notin A\cup A^2\cup A^3$$. Denote by $$\langle A\rangle$$ the subgroup generated by $$A$$, and $$G^{(6)}$$ the subgroup of $$G$$ generated by $$\{g^6:g\in G\}$$; clearly $$G^{(6)}$$ is normal in $$G$$.

For every $$g\in G\smallsetminus A$$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $$1\notin A\cup A^2\cup A^3$$, this means that $$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ Hence one of the following holds: $$g=1$$ or $$g^2=1$$ $$g^3=1$$ or $$g\in A^{-1}$$, or $$g^2\in A^{-1}$$ or $$g\in (A^2)^{-1}$$.

Hence, $$g^6\in \langle A\rangle$$ for all $$g\in G$$; equivalently, $$G^{(6)}\subset \langle A\rangle$$. In $$G/G^{(6)}$$, every element satisfies $$x^6=1$$. Since groups of exponent 6 are solvable, it follows that either $$G=G^{(6)}$$, or $$G$$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $$W$$. Otherwise, $$\langle A\rangle=G$$, that is, $$A$$ generates $$G$$. In particular, since $$G$$ is infinite, $$|A|=|G|$$, so we put $$A=W$$.

• What about a set $A=A^{-1}$ with $A^3\ne G$? – Taras Banakh Oct 23 '18 at 13:23
• Can you please explain why the subgroup $B$ is normal? – Yair Hayut Oct 23 '18 at 15:49
• $B$ is not always normal, it's fixed now. – YCor Oct 23 '18 at 17:14
• About multiple questions in a single thread, see this discussion: meta.mathoverflow.net/questions/3458/…. – YCor Oct 23 '18 at 17:14
• The argument does not adapt to the additional requirements $A=A^{-1}$, or $1\in A$. – YCor Oct 23 '18 at 17:20