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If you take a generating pair for $F(a, b)$, $(a, B)$, then it is intuitively obvious that $B=a^ib^{\epsilon}a^j$ for $i, j\in \mathbb{Z}$, $\epsilon=\pm 1$. However, I cannot come up with a neat proof of this, and so was wondering if someone could either provide a reference or come up with a more elegant approach.

My proof is basically,

Assume $B\neq a^ib^{\epsilon}a^j$, so $B=a^{i_1}b^{i_2}a^{i_3}\bar{B}a^{j_3}b^{j_2}a^{j_1}$, so you end up with $\langle a, \hat{B}\rangle=F(a, b)$ where $\hat{B}=b^{i_2}a^{i_3}\bar{B}a^{j_3}b^{j_2}$ and no free cancellation happens between $a$ and $\hat{B}$, and $|\hat{B}|>1$. This means that $a\not\in \langle a, \hat{B}\rangle$, a contradiction.

I do not like this proof, but it seems to be the best I can come up any ideas would be appreciated!

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You don't specify, but I assume that $F(a,b)$ is the free group on two generators, and by a generating pair, you mean two elements of the group which generate the entire group? Also, what would constitute an elegant approach? We know exactly what the elements of $F(a,b)$ are, and your idea relies on nothing more than that. Would you like something that uses covering space theory? Or something that characterizes the automorphisms of $F(a,b)$? – Aaron Jul 16 '11 at 18:20
This is related to an old result of Magnus: if two words $r$ and $s$ in a free group have the same normal closure, then $r$ is conjugate to $s$ or $s^{-1}$. – user641 Jul 16 '11 at 20:21
@Steve D: I can see how that would make my approach neater - I end up with B=$B_0^{-1}b^{\epsilon}B_0$ freely reduced and just need to prove that $b$ is not a subword of $B_0$. Was this what you were getting at? – user1729 Jul 17 '11 at 11:45
@Aaron: I suppose my question is bad because "elegant" is subjective. I perhaps meant "elementary and short" or "can be obtained from another result in an elementary and short way". – user1729 Jul 17 '11 at 11:47
up vote 1 down vote accepted

The following works (and is much neater).

Basically, the abelianisation map induces a homomorphism from $\operatorname{Out}(G)$ to $\operatorname{Aut}(G^{\operatorname{ab}})$. In the free group on two generators this induced homomorphism is an isomorphism (a classic result of Nielsen). Thus, there exists a map $\phi: \operatorname{Aut}(F_2) \rightarrow \operatorname{Aut}(\mathbb{Z}\times \mathbb{Z})$, with $\ker(\phi)=\operatorname{Inn}(F_2)$.

So, if $\langle a, B\rangle = \langle a, b\rangle$ then $B\mapsto a^ib^{\epsilon}$ under the abelianisation map (because of the way $\mathbb{Z}\times \mathbb{Z}$ is generated). As the kernel of the induced map is $\operatorname{Inn}(F_2)$, we have that the pair $(a, B)$ is conjugate to the pair $(a, a^ib)$. That is, there exists $w\in F(A, b)$ such that $a^w=a$ and $B^w=a^ib$. As $a^w=a$, we have that $w=a^j$ for some $j\in \mathbb{Z}$, as required.

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