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The fundamental group of the figure-eight knot complement has the displayed two-generator one-relator presentation. It follows from the Artin automorphism of a braid for $({\sigma }_1{\sigma }_2^{-1}{)}^2$ by eliminating the third meridian from its fixed-generator presentation.

Use the braid $B=({\sigma }_1{\sigma }_2^{-1}{)}^2$ from part (c). Put meridians $a,b,c$ around its three strands at one end. At a positive crossing, the Wirtinger presentation relations induce the Artin automorphism of a braidFor the inverse crossing on strands two and three, they giveThe first formula is the usual crossing conjugation, and the second is its inverse on the relevant pair. Following one block gives $h=f\circ g$:The four-crossing braid induces $h^2$. Closing it identifies each ending meridian with its starting meridian. Applying the Seifert-van Kampen theorem to the crossing regions and the closing arcs therefore giveswhereBoth elementary crossing automorphisms preserve $abc$. Hence $h^2(abc)=abc$, and the first two closure relations imply the third: after substitution $abc=ab{h}^2(c)$, so cancellation gives $c=h^2(c)$.

The first remaining relation can be solved without adding any relation:Substitute this expression into $b=c^{-1}ac$. Free cancellation and multiplying the two sides by inverse words giveThese are Tietze transformations, eliminating a generator defined by one relation and retaining the other. Thus the figure-eight knot group has the requested one-relator presentation