Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
9  
Pick a finite group $\Gamma$ and two subgroups $A,B\subseteq\Gamma$ such that $AB\neq BA$. Pick any free group $F$ and any surjective morphim $f:F\to\Gamma$, and consider the subgroups $f^{-1}(A)$ and $f^{-1}(B)$. –  Mariano Suárez-Alvarez Feb 1 '13 at 19:56

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
The sentence «So for every pair of subgroups $A$, $B$ of any finite group we have $AB=BA$.» is plainly false. Maybe you could rephrase this somehow? –  Mariano Suárez-Alvarez Feb 1 '13 at 21:30
2  
@Mariano: I think that the intended logic of the proof is: if the OP's statement were true, then <<...>> would be true. But (as he and you agree) it is plainly false, it is easy to find a counterexample. –  Lee Mosher Feb 2 '13 at 1:18
    
@Lee, I know :-) –  Mariano Suárez-Alvarez Feb 2 '13 at 19:14
    
@Mariano, I did so. –  doug Feb 3 '13 at 16:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.