Does $A^{-1}A=G$ imply that $AA^{-1}=G$?

Question

Let $G$ be a group, $A\subseteq G$ and put $A^{-1}=\{ a^{-1}:a\in A\}$.

Is it true that if $A^{-1}A=G$ then $AA^{-1}=G$ (and vice versa)?

Answer 1

I think I could construct a random counter-example with GAP ($t$ is the set of the products $A^{-1}A$ forming $G$ , $w$ is the set of the products $AA^{-1}$)

The following GAP session:

gap> A:=[ (), (3,4), (2,4,3), (2,4), (1,2), (1,2,3), (1,2,3,4), (1,3,2,4), (1,4,2,3) ];;
gap> t:=[];;
gap> w:=[];;
gap> for u in A do 
>   for v in A do 
>     t:=Union(t,List([u^(-1)*v])); 
>     w:=Union(w,List([u*v^(-1)])); 
>   od;
> od;
gap> t;
[ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), 
  (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), 
  (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ]
gap> w;
[ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), 
  (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), 
  (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ]
gap> Length(t);
24
gap> Length(w);
23
gap> G:=AsGroup(t);
Group([ (3,4), (2,3), (1,2) ])
gap> StructureDescription(G);
"S4"
gap> AsGroup(w);
fail

demonstrates the subset $A \subseteq G = S_4$ with the required properties:

$$[ (), (3,4), (2,4,3), (2,4), (1,2), (1,2,3), (1,2,3,4), (1,3,2,4), (1,4,2,3)]$$

The products forming the group $G$ are :

$$A = [ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ] $$

The permutations of the product $AA^{-1}$ are

$$[ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ]$$

$A$ is a subset of $G$, as required. The permutation $(13)$ is missing in the last set.