G$, the intersections $gF(a)g^{-1} \cap F(a)$ and $gF(a)g^{-1} \cap F(b)$. In the latter case, one appears to need, as part of the procedure, to be able to determine whether or not $g \in F(b)F(a)$. For this additional question, despite the fact that, in his solution to the word problem for one-relator groups, Magnus proved that one can always decide if a given element lies in a given Magnus subgroup, the usual inductive technique seems to run aground in the two-generator case when neither generator has exponent sum zero in the relator.

$\mspace{-16.5mu}$ [to [@C]]{}The reader of [@C] should note that although, in the definition of notation on page 273 of [@C], it is made clear that the sets $A^*_+$ and $A^*_-$ may both be empty and similarly for $C^*_+$ and $C^*_-$, there is no specific discussion in Sections 5–6 of [@C] of what happens when these possibilities arise. However, as we point out below, in practice the results in these sections and the similar results in are employed only in situations where all of $A^*_+, A^*_-, C^*_+, C^*_-$ are nonempty. This point is clarified in the introduction to . It is also worth pointing out that $B^*$ may be empty 