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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

Suppose $G=⟨g_1,\dots ,g_n\mid R⟩$ has a finite group presentation. Work in the free product $B=G\ast F(a,b)$ and putThe $u_i$ freely generate a subgroup of $F(a,b)$. Indeed, in a reduced word in these conjugates, combine adjacent powers with the same index. Expanding the remaining word leaves nonzero powers of $a$ separated by nonzero powers of $b$, so the normal form theorem for a free product makes it nonidentity. This proves the free conjugate bases in a rank-two free group fact we need.

The group homomorphism $B\to F(a,b)$ killing $G$ sends $v_i$ to $a^{-i}b{a}^i$, including $v_0=b$. These are a free conjugate basis by the same argument with $a,b$ interchanged. A nonempty reduced word in the $v_i$ therefore has nontrivial image and cannot be trivial in $B$. Thus $u_i\mapsto v_i$ extends to an isomorphism between the two rank-$(n+1)$ free groups embedded in $B$.

Adjoin one stable letter implementing this isomorphism:We assume Britton's lemma, including its assertion that the base embeds in this HNN extension, and the factor-embedding assertion of the normal form theorem for a free product. Hence $G\hookrightarrow E$. The presentation is finite, and its relations express every generator in $a,t$:Eliminating $b,g_i$ by Tietze transformations therefore gives a two-generator finite presentation with $\boxed{G\hookrightarrow E=⟨a,t⟩}$. This explicitly constructs the two-generator HNN embedding; it does not rely on an arbitrary two-generator embedding preserving finite presentability.