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 $y=a$.

  • $\begingroup$ (in commutative case) "we can choose $a\in A-\Phi(b)$ and $b\in B-\Phi(a)$": I can't make any sense of this: you need to have chosen $b$ to choose $a$ and vice versa. What you can do is to consider $A'$ infinite countable subset of $A$, define $W=\bigcup_{a\in A'}\Phi(a)$, then choose $b$ in $B-W$, and then choose $a$ in $A'-\Phi(b)$. $\endgroup$ – YCor Oct 18 '18 at 14:24

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 $b\notin \Phi(y)$.

By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. 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.

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    $\begingroup$ Just to mention, in this case, $xA\cap yB$ is at most a singleton for every pair $(x,y)$, namely if $x^{-1}y=a_0b_0^{-1}$ for some $(a_0,b_0)\in A\times B$, this singleton is equal to $\{xa_0\}(=\{yb_0\})$; otherwise it's empty. In general, this argument holds as soon as $\langle A\rangle\cap \langle B\rangle=\{1\}$. $\endgroup$ – YCor Oct 18 '18 at 9:53
  • $\begingroup$ Do you have a restatement of Protasov's question using the language of infinite combinatorics? $\endgroup$ – YCor Oct 18 '18 at 10:12
  • $\begingroup$ @YCor Yes, and I am thinking of posing this as a separate problem. $\endgroup$ – Taras Banakh Oct 18 '18 at 10:19
  • $\begingroup$ Link to the separated problem: mathoverflow.net/questions/313128/… $\endgroup$ – YCor Oct 18 '18 at 11:18

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