# If $G=AB$ and $|G|=|A||B|$, then $G=A^{-1}B$.

Let $$A$$ and $$B$$ be subsets of a finite Abelian group $$G$$ such that $$G=AB$$ and $$|G|=|A||B|$$. Show that $$G=A^{-1}B$$.

My attempt: If $$A$$ is a subgroup, then $$A=A^{-1}$$ and we're done. If $$A$$ is not a subgroup, then there exists $$a_1,a_2\in A$$ such that $$a_1a_2^{-1}\notin A$$. Thus, there exists $$a\in A$$ and a non-trivial $$b\in B$$ such that $$a_1a_2^{-1}=ab$$.

How do I proceed from here? Any hints are appreciated.

• If $A$ is not necessarily a subgroup, how do you define $A^{-1}$? Jun 27 at 23:02
• @JohnDouma I assumed it is $\{a^{-1}:a\in A\}$. Jun 27 at 23:03

Maps between finite sets with the same cardinality are injective iff surjective. Because of $$|G|=|A||B|=|A\times B|$$, such a map is $$\phi\colon A\times B\rightarrow G,(a,b)\mapsto ab$$. It is surjective due to $$G=AB=\phi(A\times B)$$ and therefore injective, which means the representation of an element $$g\in G$$ as $$g=ab$$ with elements $$a\in A$$ and $$b\in B$$ is unique. Let $$\psi\colon A\times B\rightarrow G,(a,b)\mapsto a^{-1}b$$ and let $$a_1,a_2\in A$$ and $$b_1,b_2\in B$$ be elements with $$a_1^{-1}b_1=a_2^{-1}b_2\Leftrightarrow a_2\circ b_1=a_1\circ b_2$$ using the group $$G$$ is abelian. Because of the uniqueness of representation, we have $$a_1=a_2$$ and $$b_1=b_2$$. Therefore $$\psi$$ is injective, therefore $$\psi$$ is surjective and $$G=\psi(A\times B)=A^{-1}B$$.
Suppose $$f : X \to Y$$, where $$X$$ and $$Y$$ are finite with $$|X| = |Y|$$. Then $$f$$ is injective if and only if it is surjective.
Let us consider the group operation restricted to $$A^{-1} \times B$$. This is a function from $$A^{-1} \times B$$ to $$G$$. Note that $$|A^{-1} \times B| = |A^{-1}||B| = |A||B| = |G|.$$ The middle equality comes from the fact that inversion is a bijection on $$G$$, so it preserves the cardinality of sets.
Suppose $$A^{-1}B \neq G$$. Then, this restricted group operation is not surjective, and hence is not injective. In particular, this implies that there are $$a_1, a_2 \in A$$ and $$b_1, b_2 \in B$$, such that $$a_1 \neq a_2$$ or $$b_1 \neq b_2$$, and $$a_1^{-1} b_1 = a_2^{-1} b_2.$$ Given $$G$$ is Abelian, this is equivalent to $$a_2 b_1 = a_1 b_2,$$ such that $$a_2 \neq a_1$$ or $$b_1 \neq b_2$$.
If we consider the group operation restricted instead to $$A \times B$$, we now have shown that this map is not injective. It is therefore not surjective, and hence $$G \neq AB$$, establishing the contrapositive.