I want to prove the following result
Let $F=\{A_1,\dots,A_k\}\subset\mathbb{C}^{n\times n}$ be a family of matrices and let $G=\{A_iA_j:A_i,A_j\in F\}$ the family of all pairwise products of members in $F$. Show that if $F$ is a commuting family then so is $G$ but the converse may not be true.
The forward part is trivial by definition of a commuting family (a family of matrices in which every pair of matrices commutes) i.e. if $F$ is a commuting family ($A_iA_j=A_jA_i, \forall i,j$) then all pairwise products also commute i.e. \begin{eqnarray*} A_iA_jA_{i'}A_{j'}&=&A_iA_{i'}A_jA_{j'}\\ &=&A_{i'}A_iA_jA_{j'}\\ &=&A_{i'}A_iA_{j'}A_j\\ &=&A_{i'}A_{j'}A_iA_j, \forall i,j,i',j' \end{eqnarray*} Thus $G$ is a commuting family.
I am trying to find a counterexample to disprove the converse part. Let $A\in\mathbb{C}^{n\times n}$ nonsingular and some $B\in\mathbb{C}^{n\times n}$ s.t. $F=\{A,B,A^{-1}\}$. Then $G=\{AB,BA,I,BA^{-1},A^{-1}B\}$. Assume $G$ is a commuting family i.e. the following relations hold $$AB^2A=BA^2B$$ $$AB^2A^{-1}=B^2$$ $$ABA^{-1}B=A^{-1}BAB$$ $$BABA^{-1}=BA^{-1}BA$$ $$B^2=A^{-1}B^2A$$ $$A(A^{-1})^2B=A^{-1}B^2A^{-1}$$ Now if we assume that $F$ is a commuting family then $AB=BA$ and I was hoping this to be contradictory to some of the above equations but I cannot substantiate this. Maybe my example won't work.
Suppose $A^2=B^2=I$. Then $(AB)(BA)=AB^2A=A^2=I$; likewise $(BA)(AB)=I$. So $\{A^2,AB,BA,B^2\}=\{I,AB,BA\}$ is commuting. But it is easy to find $A$ and $B$ with $AB\ne BA$ (for instance reflection matrices).