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?)



Cardinality of factors of infinite non-abelian groups
Let $G=C\_2\ast C\_3$ be the free product of cyclic groups $C\_2=\langle a\rangle$ and $C\_3=\langle b\rangle$.


Let $A$ be the subset of $G$ consisting of
$$b,bab,babab,\dots$$
together with all reduced words ending with $a$ except for
$$ba,baba,bababa,\dots.$$


Then $A$ is a set of left coset representatives for both $B\_1=\langle a\rangle$ and $B\_2=\langle b\rangle$, so $G=A\cdot B\_1=A\cdot B\_2$.


