Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,y\in G$ the intersection $xA\cap yB$ is finite. Let $\Phi:G\to 2^G$ be a function assigning to each element $x\in G$ some finite set $\Phi(x)\subset G$.



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> **Question.** Is it true that there exist elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus\Phi(y)$ such that $xa=yb$?
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**Comment.** The answer is affirmative if $ab=ba$ for any $a\in A$ and $b\in B$. In this case we can choose two elements $a\in A\setminus\Phi(b)$ and $b\in B\setminus \Phi(a)$ and put $x=b$ and $y=a$.



A combinatorial property of uncountable groups
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $A\cup B$. Let $\Phi:G\to 2^G$ be the function assigning to each $g\in G$ the set of letters in the irreducible representation of $g$.


Now assume that there exist elements $x,y\in G$ and $a\in G\setminus\Phi(x)$ and $b\in G\setminus \Phi(y)$ such that $xa=yb$. Since $a\notin\Phi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $\Phi(y)=\Phi(xab^{-1})=\Phi(x)\cup\{a,b\}$, which contradicts the choice of $b\notin \Phi(y)$. 




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By the way, my purpose was to resolve Question 2.2 from [this survey of Protasov](https://arxiv.org/pdf/1807.03028.pdf). This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.


