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The free product $A\ast B$ is the amalgamated free product over the trivial group. Its nonidentity elements have unique reduced alternating normal forms.

Orient $S^3$ as the boundary of $B^4$. Push the interior of an oriented Seifert surface $F$ slightly into $B^4$, and let $X_F$ be the double branched covering of $B^4$ over this pushed-in surface. Its boundary is the two-fold branched cover of a knot. Define the knot signature byThe signature of a possibly degenerate intersection form means the number of positive eigenvalues minus the number of negative eigenvalues; its radical contributes zero.

Here is why the definition is independent of $F$. Given two choices $F_0,F_1$, glue their branched covers along the common boundary to obtainThis is the double branched covering of $S^4$ over the closed oriented surface obtained by gluing the pushed-in $F_0$ to $-F_1$. Every closed oriented surface in $S^4$ admits a Seifert hypersurface: its complement has an integral meridional class, represent it by a map to the circle, and take a regular value. Near the surface choose the angular map in its trivial normal disk bundle. The closure of a suitable regular fiber is a compact oriented three-manifold bounding the surface. Its trivial normal framing can be chosen to match this construction.

Push the interior of this Seifert hypersurface into $B^5$. The double branched covering of $B^5$ over the resulting properly embedded three-manifold has boundary $X$. The existence of the cover follows from the meridian homomorphism to $\mathbb{Z}/2$; smoothness near the branch set is the local map $(z,u)\mapsto (z^2,u)$. By the hypothesis about boundaries of five-manifolds, $\mathrm{sign}X=0$. Novikov additivity for gluing along an entire closed three-dimensional boundary givesThus the knot signature is independent of the Seifert surface. An isotopy of the knot carries the construction to an equivalent one, so it is also a knot invariant.

To calculate its intersection form, express $F$ as a disk with $2g$ bands. The double branched covering of $B^4$ over the pushed-in disk is again $B^4$. Each band lifts to a two-handle. The capped lifted cores give a basis of $H_2(X_F;\mathbb{Z})$ corresponding to the band-core basis of $H_1(F;\mathbb{Z})$. The framing of the $i$th lifted attaching circle is $2\mathrm{lk}(a_i^{+},a_i)=2V_{ii}$. For distinct bands, the two sheets contribute the two push-off linkings $V_{ij}$ and $V_{ji}$, so the mutual linking number is $V_{ij}+V_{ji}$. HenceA different integral basis changes this matrix by matrix congruence and leaves its signature unchanged.

In Figure 7 the upper knot is a connected sum of knots $T\#T$, while the lower one is $T\#\overline{T}$, where $T$ is the positive trefoil knot. Both knot groups have the same presentationTo see this, the Seifert-van Kampen theorem expresses the knot group of a connected sum of knots as the amalgamated free product of the two summand knot groups, amalgamating their meridians of a knot. A trefoil knot has presentation $⟨a,b\mid aba=bab⟩$ with $a$ a meridian. Reflection gives the same abstract group for the mirror. If it reverses the chosen meridian, send both generators to their inverses; the inverse braid relation is again the same relation. Thus the meridian amalgamation yields the displayed group presentation in both cases.

A band basis for the positive trefoil knot giveswhose symmetrization is negative definite. Therefore $\sigma (T)=-2$ and $\sigma (\overline{T})=2$. Boundary-connected-summing Seifert surfaces makes the Seifert matrix block diagonal, so the knot signature is additive under connected sum of knots. ConsequentlyThe upper knot and lower knot are therefore not isotopic, even though their knot groups are isomorphic. With the opposite global knot signature convention the first value is $+4$, and the distinction is unchanged.

Let $n=[G:H]<\infty$. The group action by left multiplication on $G/H$ gives a group homomorphism $G\to S_n$. Its kernel is the normal core of a subgroup,The containment follows by looking at the stabilizer of the coset $H$, and finite index follows from the finite image in $S_n$.

The Higman group isIt is visibly a finitely presented group. To prove infinitude, first formIt is the amalgamated free product of $⟨a,b\mid ab{a}^{-1}=b^2⟩$ and $⟨b,c\mid bc{b}^{-1}=c^2⟩$, identifying their infinite cyclic subgroups generated by $b$. Each factor is an HNN extension of an infinite cyclic group, so its base and stable letter both have infinite order. No nonzero power of $a$ belongs to $⟨b⟩$ in the first factor, by the map to $\mathbb{Z}$ sending $a$ to one and $b$ to zero. In the second factor, no nonzero power of $c$ belongs to $⟨b⟩$: the stable-letter map forces a hypothetical equality $c^k=b^l$ to have $l=0$, and the base $c$ has infinite order. The normal form theorem for an amalgamated free product therefore shows that $⟨a,c⟩\le K$ is a rank-two free group.

Similarly,contains $⟨c,a⟩$ as a rank-two free group. Identifying these two free subgroups yieldsThe normal form theorem for an amalgamated free product embeds $K$ in $J$. In particular, $J$ contains a free group of rank two and is infinite.

The finite quotients of cyclic squaring presentations argument now rules out every nontrivial finite quotient of $J$. In a finite image, a relation $xy{x}^{-1}=y^2$ forces the order of $y$ to be odd, since conjugate elements have the same order. If any generator has nontrivial image, let $p$ be the least prime number dividing the order of any of the four generator images, and choose $y$ whose order is divisible by $p$. Its predecessor $x$ conjugates it to its square. If $r$ is the order of $x$, iterating conjugation gives $2^r\equiv 1(\mathrm{mod}p)$. Hence the multiplicative order of $2$ modulo $p$ divides $r$. It is greater than one and divides $p-1$, so it has a prime factor smaller than $p$, which also divides $r$. This contradicts the minimal choice of $p$. Thus all four generator images are trivial. $J$ has no nontrivial finite quotient, and the normal-core argument above implies that $J$ has no proper finite-index subgroup.

For the final argument, Conjugation preserves the order of an element. Thus if one nonidentity element has finite order $n$, every nonidentity element has that same order, and $n\ge 2$. Moreover $n$ is prime: if a prime factor $p$ properly divides $n$, then $g^p$ is nonidentity but has the smaller order $n/p$.

When $n\ge 3$, the element $g^2$ is nonidentity, so choose $x$ with $xg{x}^{-1}=g^2$. The conjugator is not the identity, since $g^2\ne g$, and therefore $x^n=1$. Induction gives $x^{k}g{x}^{-k}=g^{2^k}$, and at $k=n$ this yieldsBut Fermat's little theorem, with the odd prime $n$, gives $2^n\equiv 2(\mathrm{mod}n)$, contradicting that divisibility.

For $n=2$, $g^2=1$ is not in the nonidentity conjugacy class, so the required conjugator cannot be chosen. Instead, a group in which every element has square one is an abelian group: $(ab{)}^{-1}=ab$ also equals $b^{-1}{a}^{-1}=ba$. In an abelian group every conjugacy class is a singleton, so one nonidentity class permits only one nonidentity element, giving a group of order two. This contradicts infinitude. Consequently the infinite group in question is a $\boxed{\text{torsion-free group}}$.

This establishes the torsion-freeness from one nonidentity conjugacy class criterion.

We construct the output presentation explicitly, using HNN extensions and an amalgamated free product. Throughout, $P$ also denotes the group given by the input finite group presentation, and $X=\{x_1,\dots ,x_n\}$. Every word introduced below is a literal computable word in the displayed generators; testing whether $w=1$ is never part of the construction.

First fix the auxiliary Higman groupWe need two facts: $⟨a_1,a_3⟩$ is a free group of rank two, and the normal closure of $a_1$ is $Q$. The latter follows immediately from the relations: setting $a_1=1$ forces successively $a_2=a_2^2$, $a_3=a_3^2$, and $a_4=a_4^2$, hence all three are $1$.

Here is a normal-form verification of the first fact. The three-generator groupis an amalgamated free product of two Baumslag-Solitar groups $BS(1,2)$ along their infinite cyclic subgroups generated by $a_2$. Their base cyclic groups embed by Britton's lemma, so this amalgamation is legitimate. Nonzero powers of $a_1$ in the first factor lie outside $⟨a_2⟩$, as shown by the homomorphism recording the exponent of its stable letter $a_1$. Nonzero powers of $a_3$ in the second factor also lie outside $⟨a_2⟩$: the stable-letter exponent map first forces a putative equality $a_3^r=a_2^s$ to have $s=0$, and then injectivity of the base cyclic group forces $r=0$. The normal form theorem for an amalgamated free product therefore makes $⟨a_1,a_3⟩$ a rank-two free group in $Q_{123}$.

The same argument applies to $Q_{341}$, formed from the other two cyclic conjugation relations. The full $Q$ is $Q_{123}{\ast }_{⟨a_1,a_3⟩}{Q}_{341}$, so that rank-two free group embeds in $Q$ as well. In particular $a_1$ has infinite order.

Now take the free productand setIntroduce $n+2$ letters $u_0,\dots ,u_{n+1}$, and define the finite presentationFinally impose the Higman relations and identifyThus the algorithm outputsThe symbols $c,C,g_j$ in this display are abbreviations for the explicit words above, not extra generators. Renaming the finitely many new generators to avoid $X$ makes this a uniform effective construction.

The pushout of groups isIts maps from $G_1,G_2$ agree on $H$, and every other pair of group homomorphisms agreeing on $H$ factors through one unique homomorphism from $G$. This is the universal property. When the $i_j$ are injective, it is the amalgamated free product, and the two factors and the identified copy of $H$ embed in it. Some conventions use this name for the quotient even without injectivity; the embedding assertion requires injectivity.

Given group homomorphisms $i_A:C\to A$ and $i_B:C\to B$, their amalgamated free product is a group $A{\ast }_{C}B$ with maps $j_A:A\to A{\ast }_{C}B$ and $j_B:B\to A{\ast }_{C}B$ satisfying $j_A{i}_A=j_B{i}_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_A{i}_A=f_B{i}_B$ factors through a unique homomorphism $A{\ast }_{C}B\to H$.

The Seifert-van Kampen theorem says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce

Here is a direct proof of the requested generation statement. For a based loop $\gamma$ in $X$, the Lebesgue number lemma supplies a subdivisionsuch that every arc $\gamma {|}_{[t_{j-1},t_j]}$ lies in one $U_{{\epsilon }_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma (t_j)$ lies in $U_1\cap U_2$. Since the intersection is path-connected, choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Thenis a loop in $U_{{\epsilon }_j}$. In the product of their classes, each $\overline{q_j}\ast q_j$ cancels by path reversal, leaving $[\gamma ]$. Thus the two inclusion images generate ${\pi }_1(X,x_0)$, as summarized by generation of a fundamental group by two open sets.

Now let $a,b$ be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

The two attaching classes $A=ab$ and $B=a^2{b}^3$ commute. Their exponent matrix has determinantso they form a basis of $⟨a,b⟩\cong {\mathbb{Z}}^2$, withSubstituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentationThis is the calculation in two Möbius bands attached to a torus.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the symmetric group $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective group homomorphism from ${\pi }_1(Y)$ onto the nonabelian group $S_3$. Therefore ${\pi }_1(Y)$ is nonabelian.

Direct multiplication gives $A^3=B^2=-I$, so $⟨A⟩$ has order $6$, $⟨B⟩$ has order $4$, and their intersection is $\{I,-I\}$ of order $2$. For the amalgamated free productthe Bass-Serre tree has verticesand edges $S{L}_2(\mathbb{Z})/\{\pm I\}$. Each edge $g\{\pm I\}$ joins $g⟨A⟩$ to $g⟨B⟩$. Thus it is an infinite bipartite tree in which the $⟨A⟩$-vertices have degree $3$ and the $⟨B⟩$-vertices have degree $2$:with a third branch leaving every ${\circ }_3$ vertex and the same pattern continuing on every branch.

Introduce $d=c^2$ and $e=a^2$. The two vertex groupsare Klein bottle groups. In $A$, the subgroup $⟨a^2,d⟩$ is ${\mathbb{Z}}^2$ of index two; in $B$, the subgroup $⟨e,c^2⟩$ is also ${\mathbb{Z}}^2$ of index two. IdentifyinggivesEliminating $d,e$ from this amalgamated free product recovers exactly the two given relators.

The Bass-Serre tree is bipartite with vertex sets $G/A$ and $G/B$ and edge set $G/⟨a^2,c^2⟩$. Both edge-group inclusions have index two, so every vertex has degree two. The tree is therefore a bi-infinite line, with $A$- and $B$-type vertices alternating.

For homomorphisms $i_j:H\to G_j$, the pushout of groups is $(G_1\ast G_2)/⟨⟨i_1(h)i_2(h{)}^{-1}:h\in H⟩⟩$. It has the universal property for homomorphisms out of $G_1,G_2$ agreeing on $H$. When the maps $i_j$ are injective, it is the amalgamated free product, with the factors and common subgroup embedded.