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)?
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Sign up to join this communityLet $G$ 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)?
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.