most recent 30 from http://mathoverflow.net 2013-05-19T13:11:16Z http://mathoverflow.net/feeds/question/120539 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120539/let-g-and-h-be-finite-index-subgroups-of-a-free-group-does-ghhg Yonatan 2013-02-01T19:47:24Z 2013-02-03T16:20:16Z

<p>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?</p> <p>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.</p>

http://mathoverflow.net/questions/120539/let-g-and-h-be-finite-index-subgroups-of-a-free-group-does-ghhg/120540#120540 anton 2013-02-01T20:02:24Z 2013-02-03T16:20:16Z

<p>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.</p>