Thread: Proposition Regarding Cosets and Orders

1. Proposition Regarding Cosets and Orders

Let $A$ be a finite subset of a group $G$ (not necessarily a subgroup). Denote by $A^2$ the set $\{a_1 a_2 | a_1, a_2 \in A\}$.

Prove that $|A^2| = |A| \iff$ the following is true:
$A$ equals a left coset $aH$ for some subgroup $H \leq G$ and some
element $a \in G$, and $A$ also equals some right coset $Hb$, $b \in G$.

The backward implication is easy enough since all that's required is a simple bijection, but I'm having problems proving the forward implication.

2. I also realize that the professor went through quite a bit about group actions, would they be required to give a proof of this?

3. I haven't worked this out completely, but I'll throw out some ideas.

Let $A=\{a_1, a_2,\ldots, a_n\}$ and write $A^2=\bigcup_{a\in A} aA$. Since $a\in G$, the operation $x\mapsto ax$ is injective. Therefore, $|aA|=|A|$ for every $a\in A$. Now, we are given that $|A^2|=|A|$. This implies that every term in the union is equal; that is, $a_1A=a_2A=\ldots=a_nA$.

This seems to be coset-like behaviour, but I don't quite see to how get the conclusion.

4. Solved it -- took much more effort than I thought was initially required, presumably because this question was adapted from a conference paper that's currently investigating groups with such properties.

Basically the proof shows that the identity element is in $a^{-1}A$ for any $a \in A$. Then $A \subset a^{-1}A^{2}$. From there, a few further arguments will show that $a^{-1}A$ is a subgroup, and it is clear that $A = a.a^{-1}A$.