# A combinatorial property of uncountable groups

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 $$y=a$$.

• (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 $A'-\Phi(b)$. – 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.
• 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 $\langle A\rangle\cap \langle B\rangle=\{1\}$. – YCor Oct 18 '18 at 9:53