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?

  • $\begingroup$ @NateEldredge Thank you for the comment. I have corrected the title. $\endgroup$ Oct 23 '18 at 0:09
  • 1
    $\begingroup$ When creating names for new mathematical objects I would propose to refrain from using weak + name :-) $\endgroup$
    – Jan_Ch.
    Oct 23 '18 at 5:44
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    $\begingroup$ I think that history will tell us, but there is nothing which is "almost Shelah". You're either Shelah or not at all Shelah. :-) $\endgroup$
    – Asaf Karagila
    Oct 23 '18 at 12:58
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    $\begingroup$ @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. $\endgroup$ Oct 23 '18 at 13:21
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    $\begingroup$ 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 $\endgroup$ 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 :)


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 $A=W$.

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

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