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 $\Phi(x)\subset G$.

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 $xa=yb$?

**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 $y=a$.