12
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @NateEldredge Thank you for the comment. I have corrected the title. $\endgroup$ – Taras Banakh 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @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. $\endgroup$ – Taras Banakh Oct 23 '18 at 13:21
  • 2
    $\begingroup$ 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 $\endgroup$ – François G. Dorais Oct 24 '18 at 0:26
11
$\begingroup$

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 $A=W$.

$\endgroup$
  • $\begingroup$ What about a set $A=A^{-1}$ with $A^3\ne G$? $\endgroup$ – Taras Banakh 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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