Let $A$ and $B$ be arbitrary nonempty subsets of a group $G$. Then the product $AB$ is called direct, and we denot it by $A \cdot B$, if the representation of every its element by $x=ab$ with $a\in A$, $b\in B$ is unique.
It is obvious that if $AB=A \cdot B$ then $|A B|=|A||B|$, and the converse is true if both are finite.
Now, let $A \cdot B_1=A \cdot B_2$. If $A$ is finite then $|B_1|=|B_2|$. Also, if $G$ is abelian then we can define an injective map between $B_1$ and $B_2$ (and also between $B_2$ and $B_1$). Hence, the question is:
- Let $G$ be an infinite non-abelian group, and $A,B_1,B_2$ their nonempty subsets. Is it true that if $A \cdot B_1=A \cdot B_2$ then $|B_1|=|B_2|$? (what about $A \cdot B_1=A \cdot B_2=G$, if the answer is negative?)