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!