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> **Problem 1.** Is it true that each uncountable group $G$ contains two subsets $A,B\subset G$ such that
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> 1) for any $x,y\in G$ the intersection $xA\cap yB$ is finite and 
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> 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 $xa=yb$.
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**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](https://www.sciencedirect.com/science/article/pii/S0049237X08713466).
We recall that a group $G$ is *Jonsson* if it is uncountable but all proper subgroups of $G$ are countable.



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> **Problem 2.** What is the answer to the Problem for simple Jonsson groups (constructed by [Shelah](https://www.sciencedirect.com/science/article/pii/S0049237X08713466))?
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**Comment.** Problem 1 is a combinatorial reformulation of Question 2.2 from [this survey of Protasov](https://arxiv.org/pdf/1807.03028.pdf). 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"](https://arxiv.org/abs/1810.07979) of Banakh and Protasov. 



A combinatorial property of uncountable groups, II
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 $u\notin a\Phi(a)a^{-1}$).


