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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.

One standard five-operation version of the Tietze transformations is: invert a relator; conjugate a relator by a word; multiply a relator on either side by another relator; insert or delete the empty relator; and introduce or eliminate a generator together with a defining relation $y=w$, where $w$ uses the other generators and an eliminated generator occurs nowhere else. The elementary relator operations are reversible. Renaming and reordering the generators and relators is harmless. The Tietze theorem states that two finite presentations define isomorphic groups exactly when they are related by a finite sequence of these transformations.

For a connected cell complex with one vertex, take one generator for each oriented one-cell and one relator from the attaching loop of each two-cell. Higher-dimensional cells do not change the fundamental group. This is the cellular presentation theorem, obtained by applying van Kampen to the two-skeleton. Conversely a finite presentation is realized by a bouquet of its generator circles with one disk attached along each relator word, giving a finite connected complex with the required fundamental group.

Triangulate each relator polygon by adding diagonals from one boundary vertex. Each diagonal is a new one-cell, and every triangular two-cell has an attaching word of length at most three. Algebraically, for a word $s_1\cdots s_{\ell }$ with $\ell \ge 4$, introduce successive partial-product generators $y_2,\dots ,y_{\ell -2}$. Use relatorsEliminating the new generators recovers the original word; words already of length at most three need no change. This proves the triangular-presentation claim geometrically and algebraically.

For the free abelian group of rank two, useIts two relations say $c=ab=ba$. For the requested transformation sequence, replace the second relator by the product of the first and second, freely reducing it to $ab{a}^{-1}{b}^{-1}$. Invert the first relator, giving $c{b}^{-1}{a}^{-1}$, the definition $c=ab$. Eliminate $c$ together with this defining relation. The remaining presentation is exactly the stated two-generator commutator presentation.

The construction is a triangular group presentation.

A triangular group presentation has relators of word length at most three. Every finitely presented group admits one. For a relator $s_1\cdots s_{\ell }$ of length $\ell \ge 4$, introduce generators $y_2,\dots ,y_{\ell -2}$ and replace it byThe defining relations say $y_j=s_1\cdots s_j$; eliminating those generators by Tietze transformations recovers the original relation. Every new relator has length three. Geometrically these operations triangulate each relator disk in its presentation complex by inserting diagonals, so they preserve the fundamental group.