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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 with...so any ideas would be appreciated!

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    @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".2011-07-17

1 Answers 1

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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.