Let $\Sigma$ be a finite set. Let $F\_\Sigma$ be the free group over $\Sigma$. Let $G$ and $H$ be finite index subgroups of $F\_\Sigma$. Consider the sets $GH$ and $HG$. Is it always true that $GH=HG$? If not, could you provide a counter-example?


The motivation for this question is automata theory. The subgroups G and H each represents a finite deterministic permutation automata. If the proposition above is true, it says something about the structure of the product automata.



Let G and H be finite index subgroups of a free group. Does GH=HG?
No, ths is not true.
Let $E$ be a finite group and let $\phi:F\_\Sigma\to E$ be a surjective group homomorphism. Let $A,B\subset E$ any subgroups and let $G=\phi^{-1}(A)$ and $H=\phi^{-1}(B)$.
If $GH=HG$ holds, then by applying $\phi$ we get $AB=BA$.
So if the claim was true, then for every pair of subgroups $A,B$ of any finite group we would have $AB=BA$.
Now it's easy to find a counterexample.


