Let $G$ be a group and $\varnothing\neq A\subseteq G$ be finite. Suppose that $|AA|=|A|$; then there is $H\le G$ such that $A$ is a left coset of $H$. I'm having some trouble with the following:
Suppose $G$ is a group and $A \subseteq G$ is a finite, non-empty subset. Suppose further that $|AA| = |A|$; then there is a subgroup $H$ of $G$ such that $A$ is a left coset of $H$.
I have scribbles but nothing substantial. To me, the fact that $|AA| = |A|$ suggests that $H$ is a normal subgroup and $AA$ is also a left coset (i.e, if $A = gH$, then $AA = g^2H$); I've thus played around with the fact that $xA = AA = Ax$ to try and obtain the $g$ so that $g^{-1}A = H$ but no luck, and I now think that's a dead end anyway.
I considered group actions on $A$ by left multiplication but again I'm unable to extract anything concrete out of it.
I would appreciate help.
 A: Let $a\in A$. If $A$ is really a left coset of some subgroup $H$, then $A=aH$, and then $a^{-1}A$ will be the subgroup you want.  So that suggests trying to see whether that is in fact a subgroup or not.
So, let $H=a^{-1}A$.
Note that for each $x\in A$, $xA\subseteq AA$; since $|xA|=|A|=|AA|$ and these are all finite quantities, we have that $xA=AA$. Thus, for every $z,w\in A$ there exists a unique $y$ such that $xy=zw$. Similarly, $Ax=AA$ so there exists a unique $t$ such that $zw=tx$.
In particular, given $x\in A$ there exists $y_x$ such that $xa = ay_x$. Thus, $a^{-1}x = y_xa^{-1}$.  Symmetrically, given $x\in A$ there exists $z_x\in A$ such that $ax = z_xa$, and therefore $xa^{-1}=a^{-1}z_x$.
Clearly, $H$ contains $e$.
Let $a^{-1}r,a^{-1}s\in H$. Then there exists $z\in A$ such that $ry_s=za$ (using that $AA=Aa$), so we have:
$$\begin{align*}
(a^{-1}r)(a^{-1}s) &= a^{-1}r(y_sa^{-1})\\
&= a^{-1}(ry_s)a^{-1}\\
&= a^{-1}(za)a^{-1}\\
&= a^{-1}z \in H.
\end{align*}$$
So $H$ is closed under products.
Now, $H$ is finite. If $h=a^{-1}x\in H$, then $h^n\in H$ for all positive integers $n$. There must exist positive integers $r,s$, $r\lt s$, such that $h^r=h^s$. But then $h^{s-r}=e$, so $h^{s-r-1}=h^{-1}\in H$. This proves that $H$ is closed under inverses, and we are done.
A: Alright, I think I got it based on the approach @ArturoMagidin suggested with a bit of a tweak.
I consider $H = a^{-2}AA$ instead where $a \in A$ is arbitrary.
Using the fact that $|AA| = |A|$ one can come to the conclusion that for any $a' \in A$ we have $AA = Aa' = a'A$ and similarly that $a'^{-1}AA = A$, etc. This is just a bit of cardinality manipulation- for instance $a'A \subseteq AA$ but $|a'A| = |A| = |AA|$ and so $a'A = AA$.
It's easy to show that $1 \in a^{-2}AA$ of course.
Now suppose that $a^{-2}bc$, $a^{-2}df \in a^{-2}AA$; we'd like to show that $a^{-1}a^{-1}bcd^{-1}f^{-1}aa \in a^{-2}AA$.
Firstly, $f^{-1}aa \in f^{-1}AA = A$ since $AA = fA$. So $f^{-1}aa = a' \in A$. The expression now is  $a^{-1}a^{-1}bcd^{-1}f^{-1}aa  = a^{-2}bcd^{-1}a'$. Next $bcd^{-1} \in AAd^{-1} = A$ once again since $AA = Ad$ and so $bcd^{-1} = a'' \in A$ and thus now the expression is $a^{-2}bcd^{-1}a' = a^{-2}a''a' \in a^{-2}AA$ as desired. So $a^{-2}AA$ is a subgroup, and certainly $aH = a\cdot a^{-2}AA = a^{-1}AA = A$ so it's a coset as desired.
