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Problem 1. Is it true that each uncountable group $G$ contains two subsets $A,B\subset G$ such that

1) for any $x,y\in G$ the intersection $xA\cap yB$ is finite and

2) for any function $\Phi:G\to 2^G$ assigning to each element $g\in G$ a finite subset $\Phi(g)\subset G$ there are two elements $x,y\in G$ and points $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$.

Remark. Such sets $A,B$ do exist if $G$ contains a subgroup $H$ that admits a homomorphism onto a group $\Gamma$ that contains an uncountable subset $U$ with infinite centralizer $C(U)=\bigcap_{u\in U}\{x\in\Gamma:xu=ux\}$. This means that a counterexample if exists, should be very non-commutative, like a Jonsson group, constructed by Shelah. We recall that a group $G$ is Jonsson if it is uncountable but all proper subgroups of $G$ are countable.

Problem 2. What is the answer to the Problem for simple Jonsson groups (constructed by Shelah)?

Comment. Problem 1 is a combinatorial reformulation of Question 2.2 from this survey of Protasov. Question 2.2 asks if the countability of a group is equivalent to the normality of its finitary ballean. This question was also repeated (as Problem 12.6) in the paper "The normality and bounded growth of balleans" of Banakh and Protasov.

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  • $\begingroup$ Say that $(A,B)$ is a good pair in $G$. So if I understand correctly, there's no good pair if $G$ is countable. Is there an immediate way to see that if $(A,B)$ is a good pair then both $A,B$ are uncountable? $\endgroup$ – YCor Oct 18 '18 at 11:22
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    $\begingroup$ Basic remark: if $G\subset H$ and $G$ has a good pair then so does $H$. As mentioned in Protasov, uncountable free groups and uncountable abelian groups admit good pairs. So a group without good pair has no uncountable abelian or free subgroup. $\endgroup$ – YCor Oct 18 '18 at 11:25
  • $\begingroup$ @YCor If $(A,B)$ is a good pair, then we only can assert that one of the sets $A$ or $B$ is uncountable. Moreover, for a countable group $A$ and an uncountable group $B$ the pair $(A,B)$ is good in the group $G=A\oplus B$. $\endgroup$ – Taras Banakh Oct 18 '18 at 12:13
  • $\begingroup$ There's no "the Jonsson group constructed by Shelah", as Shelah makes high use of the the axiom of choice and those various choices most likely produce many non-isomorphic Jonsson groups. $\endgroup$ – YCor Oct 18 '18 at 12:13
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    $\begingroup$ Problem 2 has a negative answer: any center-free $\aleph_1$-Jonsson group (i.e. of cardinal $\aleph_1$) is a counterexample. (It is not hard to check that any $\aleph_1$-Jonsson group has a centerfree $\aleph_1$-Jonsson quotient.) $\endgroup$ – YCor Oct 18 '18 at 23:14
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Problems 1 and 2 both have affirmative answers (implying that the finitary ballean of any uncountable group is normal).

Two cases are possible:

I. There exists a countable subgroup $A\subset G$ and an uncountable subset $B\subset G$ such that $bAb^{-1}\cap A$ is infinite for all $b\in B$. Replacing $B$ by a smaller uncountable set, we can assume that the family $(bA)_{b\in B}$ is disjoint. The latter condition can be used to show that the sets $A,B$ satisfy the condition 1 of the Problem.

We claim that for any function $\Phi:G\to [G]^{<\omega}$ there are elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$. Since $B$ is uncountable, there exists an element $b\in B\setminus\bigcup_{a\in H}\Phi(a)$. The set $bAb^{-1}\cap A$ is infinite and hence contains some element $a\notin\Phi(b)$. Put $x=b$ and $y=bab^{-1}\in A$. Observe that $xa=ba=yb$, $a\notin\Phi(b)=\Phi(x)$ and $b\notin\Phi(y)$.

II. There exists a countable infinite subgroup $A\subset G$ and an uncountable set $B'\subset G$ such that $bAb^{-1}\cap A$ is finite for every $b\in B'$. Replacing $B'$ by a smaller uncountable subset, we can aditionally assume that the family $(AbA)_{b\in B'}$ is disjoint.

We claim that the sets $A$ and $B=\{aba^{-1}:a\in A,\;b\in B'\}$ satisfy the conditions 1 and 2. The condition 1 will follow as soon as we check that for every $x\in G$ the intersection $xA\cap B$ is finite. Assuming that this intersection is not empty, we can find elemente $b\in B'$ and $a,\alpha\in A$ such that $x\alpha=aba^{-1}$. Taking into account that the family $(AvA)_{v\in B}$ is disjoint, we conclude that $xA\cap B=abA\cap B\subset abA\cap b^A$ where $b^A=\{gbg:g\in A\}$.

Given any element $z\in abA\cap b^A$, we can find elements $\alpha,g\in A$ with $ab\alpha=z=gbg^{-1}$ and conclude that $a^{-1}g=b\alpha gb^{-1}\in A\cap bAb^{-1}$. So, the element $g$ belongs to the finite set $F=a(A\cap gAg^{-1})$ and then $z=gbg^{-1}\in b^F:=\{fbf^{-1}:f\in F\}$. Therefore, $xA\cap B=abA\cap b^A$ is contained in the finite set $b^F$, which means that the sets $A,B$ satisfy the condition 1.

Now given any function $\Phi:G\to [G]^{<\omega}$, we shall find elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$.

Using the uncountability of the set $B'$, choose an element $u\in B'\setminus \bigcup_{a\in A}a\Phi(a)a^{-1}$. Find $a\in A\setminus \Phi(u)$. Put $y=a\in A$, $x=u$ and $b=a^{-1}ua\in B$. It follows that $xa=ua=aa^{-1}ua=yb$. Also $a\notin\Phi(u)=\Phi(x)$ and $b=a^{-1}ua\notin \Phi(a)$ (as $u\notin a\Phi(a)a^{-1}$).

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