Cardinality of factors of infinite non-abelian groups

Question

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?)

Answer 1

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 $G=A\cdot B_1=A\cdot B_2$.

Answer 2

Here's an example with $|B_1|=1, |B_2|=2$: Let $G=\mathbb Z*\mathbb Z$ (free product). Write non-identity elements of $G$ multiplicatively as reduced words with $X^{\mathbb Z-\{0\}}, Y^{\mathbb Z-\{0\}} $(with symbols $X,Y$). Let $A_0=X^{\mathbb Z - 2\mathbb Z}, A_1 = \{$reduced words with $X^{2\mathbb N},Y^{2\mathbb N}\}, A=A_0A_1, B_1=\{1\}, B_2=\{X^2,Y^2\}$. Note we can't have $A \cdot B_1=A \cdot B_2=G$ with $|B_1|=1, |B_2|>1$.