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Let $F$ be a free group generated by two elements. Let $\{a,b\}$ and $\{c,d\}$ are two different generating set.

Q:Prove that $[a,b]$ is either conjugate to $[c,d]$ or its inverse.

Here $[a,b]=aba^{-1}b^{-1}$ is the commutator.

If we consider $F$ to be the fundamental group of torus with one puncture then I have a geometric proof. It will be better if I can have an algebraic proof for this complete algebraic fact.

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What have you tried? Have you tried, for example, just coming up with two simple distinct generating sets and seeing if it actually works out? –  zibadawa timmy Nov 4 '13 at 18:49
    
It's not even true that $[a,b]$ is conjugate to $[b,a]$. –  Derek Holt Nov 4 '13 at 22:05
    
@Zibadawa I have tried with simple examples. For example F is the fundamental group of torus with one puncture, hence the generators will be simple closed curves there. Now you cut along the curves you will get the square with hole and the boundary will be the two generator. Hence I can get the above result. What I actually want is some algebraic proof. –  tessellation Nov 5 '13 at 4:12
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What can we assume? If you are happy to use Nielsen's generating set for the automorphism group of a free group then plug each automorphism in and verify it that way (this uses the fact that every generating pair $(A, B)$ of $F(a, b)$ is in the automorphic orbit of the pair $(a, b)$). The result you are talking about is in Magnus, Karrass and Solitar's book Combinatorial group theory, which is an entirely algebraic text. I cannot find it in it though - but it is there somewhere! The proof in their book will be complete, but they might use Nielsen's generating set for $Aut(F_2)$ to do it. –  user1729 Nov 5 '13 at 14:17
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Theorem 3.9 of Magnus, Karrass and Solitar is precisely this question. Proof: Plug the generators for $\operatorname{Aut}(F_2)$ into $[a, b]$ and see that it is sent to a conjugate of $[a, b]$ or of $[a, b]^{-1}$. –  user1729 Nov 5 '13 at 14:56

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