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The group ring of an infinite cyclic group is the Laurent polynomial ring $R=\mathbb{Z}[t,t^{-1}]$, with augmentation $\epsilon (t)=1$. Multiply any element in the kernel by a power of $t$ to obtain a polynomial vanishing at one. The factor theorem makes it divisible by $t-1$. Thus the augmentation ideal is $(t-1)R$. Since $R$ is an integral domain, multiplication by $t-1$ is an injective module map, so this ideal is a free $R$-module of rank one, even though its additive group has infinite rank.

For the additive infinite cyclic group $\mathbb{Z}$, the generating set of a group $S=\{-1,1\}$ produces the infinite line: each integer $n$ is adjacent to $n-1$ and $n+1$. Adding the moves $\pm 2$ gives $T=\{-2,-1,1,2\}$ and additional edges skipping one vertex. Each skip still costs one step in the word metric.

For integers $a,b$, these word metrics areThe absolute value counts unit moves, and the ceiling function counts pairs plus a possible leftover unit. Every $T$-move changes the integer by at most two, while moves of size two followed by at most one move of size one attain this bound. For example, the graph path $0,1,2,3,4,5$ has five steps for $S$, while $0,2,4,5$ has three for $T$. This calculation gives the concrete comparison

The picture shows only a finite window of each infinite Cayley graph. By contrast, $U=\{-2,2\}$ alone fails to be a generating set of a group for $\mathbb{Z}$: it generates the subgroup $2\mathbb{Z}$. Its move graph has two connected components of a graph, the even integers and the odd integers. Within either component the graph distance is $|a-b|/2$; between different components no finite graph path exists. Adding a shortcut is different from deleting the moves needed for connectivity.

A cyclic group is a group that can be generated by repeating a single group element $g$ and its inverse element:For an additive group, the same expression is written $\{kg:k\in \mathbb{Z}\}$: $kg$ means repeated addition, including subtraction when $k$ is negative. A finite cyclic group has a repeating cycle, while an infinite cyclic group is isomorphic to the additive integers. The notation $C_n$ refers to a finite cyclic group of order $n$.

This definition compresses a whole composition rule into one repeatable operation. Once $g$ is chosen, $g^a{g}^b=g^{a+b}$, so working with group elements becomes working with integer exponents. In the finite case those exponents are reduced using modular arithmetic. It also proves that every cyclic group is an abelian group, because $a+b=b+a$. A finite cyclic group of order $n$ has exactly $d$ solutions of $x^d=e$ for every positive integer divisor $d$ of $n$: in exponents, $n\mid dk$ has exactly $d$ solutions modulo $n$. The standard finite and infinite models are described in Group Theory (v4.01) by James S. Milne (section 1.16).

A nonzero finitely generated subgroup of the additive rational numbers is an infinite cyclic group. Put the generators over a common positive denominator. Their integer numerators generate the ideal given by their greatest common divisor, and Bezout identity expresses that divisor as an integer combination of them. Dividing back by the denominator gives a single generator.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

A group packages reversible operations together with a rule for composing them. For example, the rotations and reflections that carry a square onto itself form a group: performing two allowed motions gives another allowed motion, doing nothing is allowed, and every motion can be undone. The elements are the motions themselves, rather than the vertices they move. Labels in the figure track those vertices; they are markers, rather than extra decorations that the motions must preserve.

Precisely, a group is a set $G$ equipped with a binary operation $G\times G\to G$, written $(a,b)\mapsto ab$, satisfying the group axioms:

The group operation need not be ordinary numerical multiplication. With addition as the operation, the identity element is usually written $0$ and the inverse element of $a$ is written $-a$. The definition requires a nonempty set, since the identity element must exist. See McKernan's MIT lecture 1, Definition 1.1 for the axioms.

Why insist on associativity? It makes a string of operations unambiguous without specifying parentheses. For function composition, both $(fg)h$ and $f(gh)$ send $x$ to $f(g(h(x)))$. This fixes the order of execution while allowing regrouping. Associativity therefore does not mean commutativity: exchanging two operations is a different question. Ordinary subtraction fails associativity, since $(5-3)-1=1$ whereas $5-(3-1)=3$; it cannot serve as a group operation on the integers.

An inverse element makes reversal exact and equations solvable. In any group, $ax=b$ has the unique solution $x=a^{-1}b$, since $a^{-1}(ax)=(a^{-1}a)x=x$. Similarly, $xa=b$ has solution $x=b{a}^{-1}$. The side matters in a non-abelian group. Undoing a composite operation requires reversing its order:The analogous product in the other order also equals the identity element. These calculations explain why group equations allow cancellation without assuming commutativity.

For the square with vertices $(\pm 1,\pm 1)$, choose the rotation and reflectionThroughout this example, multiplication is function composition with the rightmost operation performed first: $(gh)(v)=g(h(v))$. Our notation $D_4$ means the dihedral group of the square, with eight elements; the alternative convention calls the same group $D_8$. Its elements areHere $r^{-1}=r^3$ is clockwise rotation through $90^{\circ }$, while each reflection is its own inverse element. There are exactly eight motions: the image of one vertex has four choices, and the image of an adjacent vertex has two choices; together these determine the motion. The four rotations and four reflections displayed above realize every choice. The vertex permutations give a concrete group action. The general polygon construction is discussed in Group Theory (v4.01) by James S. Milne (section 1.17).

Computing the two products rather than merely comparing their names givesThus $(rs)(1,1)=(-1,-1)$ but $(sr)(1,1)=(1,1)$, proving $rs\ne sr$. In fact $rs$ is reflection in the line $y=-x$, whereas $sr$ is reflection in $y=x$. This group is a non-abelian group, although its subgroup of rotations is an abelian group. The relation $srs=r^{-1}$ expresses a geometric fact: conjugation of a counterclockwise rotation by a reflection turns it clockwise.

A contrasting example is the cyclic group $\mathbb{Z}/4\mathbb{Z}$ under addition in modular arithmetic. Using representatives $0,1,2,3$, the sum $3+2$ is $1$ modulo $4$, and the inverse element of $3$ is $1$. Repeated addition of $1$ runs through $0,1,2,3,0$. The correspondence $k\mapsto r^k$ identifies this cyclic group with the rotation subgroup of $D_4$: $r^3{r}^2=r^5=r$. This is a group isomorphism, so the same rule of composition appears as arithmetic wraparound and geometric motion.

By comparison, $(\mathbb{Z},+)$ is an infinite cyclic group. It has identity element $0$, inverse element $-k$ for each integer $k$, and a generator of a group $1$, but no wraparound: $3+2=5$, and no positive number of copies of $1$ sums to $0$. Both examples are abelian groups. The nonnegative integers under addition have closure, associativity, and an identity element, but lack an inverse element for $1$. Reversibility is precisely what fails.

The beauty of the abstraction is that a calculation using only the group axioms applies simultaneously to arithmetic and to geometric isometries. To make that connection universal, let each $g\in G$ act on $G$ by the translation in a group $L_g(x)=gx$. Each $L_g$ is a bijection, with inverse $L_{g^{-1}}$, and $L_g{L}_h=L_{gh}$. Distinct elements give distinct maps because $L_g(e)=g$. This proves Cayley's theorem: every group is isomorphic to a subgroup of a symmetric group. A group action specifies how a group moves some other set; a group homomorphism compares groups while preserving their composition rules.

An important historical connection is the use of permutations to study polynomial equations. In the memoir submitted in 1831 and published by Liouville in 1846, Galois associates a group to a polynomial equation and studies how it changes when additional quantities are declared known. Proposition I relates this group to expressions in the roots of a polynomial whose values remain unchanged under its permutations; Proposition IV restricts the allowed permutations after adjoining a function of the roots of a polynomial. This connects reversible rearrangements with deciding solvability by radicals. See Galois's original memoir, pp. 417–433, especially Propositions I and IV; Liouville's introduction, p. 382.

The undirected infinite line is both the Cayley graph of the infinite cyclic group $\mathbb{Z}$ with generating set $\{1\}$ and the Cayley graph of the infinite dihedral group $C_2\ast C_2$ with its two involutory generators. Thus an unlabelled Cayley graph need not determine its group.

Use integral coefficients throughout, allowing a nontrivial orientation character. A dimension-$d$ Poincare duality group has a finite resolution by finitely generated projective group-ring modules andas abelian groups. The action on the top copy of $\mathbb{Z}$ may be by signs. This group-ring criterion is particularly suited to the Lyndon–Hochschild–Serre spectral sequence.

Apply that sequence to the extension with coefficient module $\mathbb{Z}G$:As an $N$-module, $\mathbb{Z}G$ is a direct sum of copies of $\mathbb{Z}N$ indexed by the cosets in $Q$. Cohomology of $N$ commutes with this direct sum because its projective resolution is finitely generated. Hence the inner cohomology vanishes except when $b=n$, where it is a direct sum of rank-one groups indexed by $Q$. The $Q$-action shifts these summands freely and transitively, with possible signs from the orientation module. Choosing a generator in one summand and transporting it by $Q$ identifies the resulting left $Q$-module with its regular module $\mathbb{Z}Q$. The residual right group action records the orientation twist; it need not be trivial.

Now the duality of $Q$ says that the only nonzero $E_2$ term is $(a,b)=(q,n)$ and that this term is $\mathbb{Z}$ as an abelian group. No differential can enter or leave it. Therefore $H^{\ast }(G;\mathbb{Z}G)$ has exactly the duality pattern in degree $n+q$. The usual extension-resolution construction also supplies a resolution by finitely generated projective modules: combine the finite kernel and quotient resolutions, lift the quotient differentials and add the necessary homotopy corrections. Its total length is at most $n+q$. Alternatively, the same spectral sequence for arbitrary coefficient modules gives the dimension bound, while the extension finiteness lemma supplies finite generation of the resolution. The surviving top term makes the bound sharp. This proves the extension rule for Poincare duality groups:Requiring a trivial top orientation action would need an additional orientation condition; the untwisted conclusion is not implicit in the hypotheses.

We next need two precise auxiliary dimension results. The Strebel infinite-index subgroup theorem says that an infinite-index subgroup of an integral $P{D}^d$ group has integral cohomological dimension at most $d-1$. The Stallings-Swan theorem says that a group of integral cohomological dimension at most one is free, including the trivial group. Since $G/N$ contains an element of infinite order, $N$ has infinite index in $G$, so $\mathrm{cd}N\le 2$. If its dimension is at most one, the required free-group conclusion follows, with finite rank because $N$ is finitely presented.

Suppose instead that $\mathrm{cd}N=2$. Finite presentation gives a partial free resolution with finitely generated modules through degree two. The kernel in degree one is finitely generated and, because the integral cohomological dimension is two, projective. Truncating at this kernel gives a finite resolution by finitely generated projective modules. Thus $N$ has the finiteness needed for group-ring cohomology to commute with direct sums.

Choose an infinite-order element $\bar{t}\in G/N$, lift it to $t\in G$, and let $H$ be the preimage of $⟨\bar{t}⟩$. Then $H=N⋊⟨t⟩$. Restriction of $\mathbb{Z}H$ to $N$ yields copies of $\mathbb{Z}N$ indexed by all powers of $t$. For every $j$, the quotient generator acts on $H^j(N;\mathbb{Z}H)$ by shifting this direct sum, possibly with an automorphism on each copy. Its invariants vanish, since an invariant finite-support sequence must be zero. Its coinvariants are one copy of $H^j(N;\mathbb{Z}N)$: the shift relations identify successive copies, and any twists can be removed by transporting the identifications along the integer index.

The infinite cyclic group has integral cohomological dimension one, so only columns zero and one occur in the spectral sequence. Column zero vanishes and column one consists of these coinvariants. We obtain the cyclic-extension shift of group-ring cohomologyas abelian groups. Moreover $H^2(N;\mathbb{Z}N)\ne 0$. Here is a useful proof of that last point: if top cohomology of a finite projective resolution were zero, its last dual differential would be onto. Its target is projective, so that differential splits. Dualizing back splits the last injection of the original resolution, shortening it and contradicting integral cohomological dimension two.

It follows that $H^3(H;\mathbb{Z}H)\ne 0$, and hence $\mathrm{cd}H=3$. Strebel's theorem forces $H$ to have finite index in $G$. It is therefore itself a $P{D}^3$ group, by finite-index invariance of Poincare duality for torsion-free groups. The displayed shift now makes $H^j(N;\mathbb{Z}N)$ zero except for $j=2$, where it is $\mathbb{Z}$. Together with its finite projective resolution, this provesThe classification of two-dimensional Poincare duality groups identifies the latter with the fundamental group of a closed aspherical surface. This classification includes orientation-twisted surface groups; it does not include the two-sphere or real projective plane, whose fundamental groups do not have integral cohomological dimension two.

In this latter case we have also proved that $G/N$ is virtually infinite cyclic, since $H/N=⟨\bar{t}⟩$ has finite index. If necessary replace the surface group by its characteristic index-two orientable subgroup and take the subgroup generated by it and $t$. This still has finite index in $G$ and is a semidirect product of an orientable closed aspherical surface group by $\mathbb{Z}$. The Dehn-Nielsen-Baer theorem for closed orientable surfaces realizes the outer automorphism defined by conjugation by $t$ as a surface homeomorphism $f$. For the torus this is just realization of $G{L}_2(\mathbb{Z})$ by linear homeomorphisms. Changing the chosen lift accounts for an inner automorphism. The mapping torusis a surface bundle over $S^1$ with fundamental group ${\pi }_1(S){⋊}_{f_{\ast }}\mathbb{Z}$. Thus a finite-index subgroup of $G$ is the fundamental group of a surface fibration over the circle. Taking a further double cover of the base if required makes the monodromy orientation-preserving.

Write $\hat{X}=X_{/Y}$. The formal completion of a scheme restricts on each chart toThus $\hat{X}$ is a formal scheme over $\mathrm{Spf}\mathbb{Z}[[q]]$. The ordinary torus $D(q)$ has disappeared from its underlying space: every point of $\hat{X}$ lies on $Y$. In particular, only adjacent completed charts intersect, and their two possible branch overlaps inside any one chart are disjoint.

Construct the action using the integral lattice automorphismIt carries $v_i$ to $v_{i+3}$ and ${\sigma }_i$ to ${\sigma }_{i+3}$, so induces a toric morphism $g:X\to X$ that is an automorphism. It preserves the lattice projection to the second coordinate, hence preserves $q$ and $Y$. On the dense algebraic torus it isIn the local coordinates this sends $U_i$ to $U_{i+3}$ withThese identities show regularity on every chart without requiring $q$ to be invertible there. They also make the action continuous in the $q$-adic topology, so it extends to $\hat{X}$. Its powers shift $D_i$ to $D_{i+3k}$; no nonzero power is the identity or fixes a point of the formal support. Therefore $G=⟨g⟩\cong \mathbb{Z}$ is a nontrivial infinite cyclic group acting freely.

Construct the quotient explicitly, using three representative chartswith $q=X_i{Y}_i$. Glue their branch opens cyclically, taking subscripts modulo three, byThese are genuine isomorphisms of formal schemes. For example, the completed localization of an adic ring on the first open iswhere $X_i=q/Y_i$. On the second it is $\mathbb{Z}[X_{i+1},X_{i+1}^{-1}][[q]]$, and the gluing sends $X_{i+1}$ to $Y_i^{-1}$ and $Y_{i+1}$ to $q{Y}_i$. Thus it preserves $q$ and is continuous, with a continuous inverse.

Inside ${\mathfrak{V}}_i$ we haveConsequently there are no triple-overlap conditions left to check: the two branch identifications in a chart have disjoint domains. By gluing of schemes along open subschemes in its formal version, the three charts produce a formal scheme $\mathfrak{C}$ over $\mathrm{Spf}\mathbb{Z}[[q]]$. Equivalently, reduce the same gluing modulo $q^{n+1}$ for every $n$; this constructs compatible schemes $C_n$, all on the same underlying space, with $C_n=C_{n+1}\times \mathrm{Spec}\mathbb{Z}[q]/(q^{n+1})$. Their inverse system has precisely the displayed formal charts.

For every integer $i$, map ${\hat{U}}_i$ isomorphically to ${\mathfrak{V}}_{\bar{i}}$ by identifying its coordinates with the corresponding capital-letter coordinates. The adjacent-chart formulas agree with the cyclic identifications, so these maps glue toThey are invariant under $g$, and ${\pi }^{-1}({\mathfrak{V}}_i)$ is the disjoint union of the translated opens ${\hat{U}}_{i+3k}$, each mapping isomorphically onto ${\mathfrak{V}}_i$. The group permutes these copies transitively. Thus the fibres of $\pi$ are exactly the $G$-orbits, and invariant sections on this inverse image are exactly the sections on ${\mathfrak{V}}_i$.

To verify the quotient property also for morphisms, take a $G$-invariant morphism $h:\hat{X}\to \mathfrak{T}$. Its restrictions to ${\hat{U}}_0,{\hat{U}}_1,{\hat{U}}_2$ agree on the adjacent branch opens. On the closing overlap they agree because the translated chart ${\hat{U}}_3$ has the same restriction as ${\hat{U}}_0$ under $g$. Hence they descend to a unique morphism $\mathfrak{C}\to \mathfrak{T}$. This proves that the gluing is the cyclic formal quotient of the infinite toric chain:Its fibre modulo $q$ is a triangle of three projective lines over $\mathbb{Z}$, each meeting its two neighbours at its two boundary sections. More generally, a shift by any $m\ge 3$ gives an $m$-gon. Completion is important here: translates of a formal chart are disjoint, whereas all ordinary toric charts still share the generic torus.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.

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.

The first Baumslag-Solitar group iswhere the second infinite cyclic group acts on the first by inversion. The presentation of a semidirect product proves this identification; both copies of $\mathbb{Z}$ are residually finite groups, since reduction modulo a suitable positive integer separates any nonzero integer. The normal factor is finitely generated, so the residual finiteness of semidirect products theorem applies. $\boxed{BS(1,-1)\text{ is residually finite}.}$

The two commutation relations make $c$ central, and the remaining relation is $ab=cba$. Repeatedly move $a$ past $b$, keeping the central powers in front. Induction first on $s$ and then on $r$ givesThereforeThis is the commutator identity in the integer Heisenberg group. The order of the factors fixes the positive sign of the exponent.

To check the claimed infinite order, map the generators to matricesTheir commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+m{E}_{13}$; hence $c$ has infinite order. This realizes the presentation in the Integer Heisenberg group and is sufficient to certify the cyclic subgroup, without needing to prove faithfulness of the whole representation.

With the ambient word metric from $\{a,b,c{\}}^{\pm 1}$, the displayed commutator givesFor any finite generating set $T$ of $⟨c⟩$, write its elements as powers of $c$ and set $B=\max \{|j|:c^j\in T\}$, with inverses included. Since $T$ generates an infinite cyclic group, $B\ge 1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, soA quasi-isometric embedding of the inclusion would require $|c^{k^2}{|}_H\ge {\lambda }^{-1}|c^{k^2}{|}_T-\epsilon$. Combining these bounds would give $4k\ge k^2/(B\lambda )-\epsilon$ for every $k$, which is impossible. Thus the inclusion of the central infinite cyclic subgroup is not a quasi-isometric embedding. Changing the ambient finite generating set only changes multiplicative constants, by equivalence of finite word metrics. This is the quadratic distortion of the Heisenberg center.

For a finite generating set $S$ of $G$, the Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and an edge from $g$ to $gs$ for every $g\in G$ and $s\in S\cup S^{-1}$; one may retain orientations and labels, or forget them.

As unlabelled undirected graphs, the Cayley graph of the infinite cyclic group $\mathbb{Z}$ with generator $1$ and the Cayley graph of the infinite dihedral groupwith generators $a,b$ are both the two-way infinite line. The first group is an abelian group and the second is not, so they are not isomorphic. Thus the requested example is the infinite line as a Cayley graph of $\mathbb{Z}$ and $D_{\infty }$.

The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

Keep $b=t^{-1}at$. Repeatedly applying $ba{b}^{-1}=a^2$ givesIn the word metric on $G$ from the finite generating set $\{a,t\}$, the right-hand side therefore has length at most $6n+1$, because $b$ is represented by the word $t^{-1}at$. In the intrinsic word metric on the infinite cyclic group $⟨a⟩$ generated by $a$, however,A quasi-isometric embedding would bound the latter by an affine function of the former. The exponential sequence above violates every such bound, so the inclusion $⟨a⟩\hookrightarrow G$ is not a quasi-isometric embedding.

A poly-infinite cyclic group has a finite subnormal series all of whose nontrivial factors are infinite cyclic groups. Intersecting its series with a subgroup again gives trivial or infinite cyclic factors; repetitions may be deleted. Splicing a kernel series with the preimage of a quotient series proves closure under extensions. Such groups are finitely presented groups: induct on the series, split the last extension by an infinite cyclic quotient, and use finite presentation of a semidirect product.

The trefoil is the torus knot $T_{2,3}$. It is the simplest nontrivial knot by crossing number of a knot: $c(T_{2,3})=3$. A standard three-crossing alternating knot diagram is reduced, so the Tait crossing-number theorem establishes minimality. The four-crossing knot diagram shown below also represents the trefoil knot, illustrating that a particular projection need not achieve that minimum. Its Seifert genus is one.

Here is a small algebraic certificate that a trefoil knot cannot be deformed into an unknot. One Seifert matrix for a trefoil knot isTaking the determinant gives the Alexander polynomial of a knotwhere $\doteq$ allows multiplication by a Laurent unit $\pm t^n$ in the Laurent polynomial ring. No such multiplication turns this three-term Alexander polynomial of a knot into the unknot's value $1$. Since the Alexander polynomial of a knot is a knot invariant, this computation rules out every possible ambient isotopy between the two knots; it does not merely rule out the deformations we happened to try.

The knot group supplies another certificate. The presentation $⟨a,b\mid a^2=b^3⟩$ maps onto the symmetric group $S_3$ by $a\mapsto (12)$ and $b\mapsto (123)$, since both sides of the relation map to the identity. These images generate a non-abelian group, so the knot group cannot be the unknot's infinite cyclic group. However, a trefoil knot and its mirror of a link have isomorphic knot groups: the abstract knot group alone does not distinguish all knots.

The unknot $U$ is the knot type represented by the planar circleEquivalently, an unknot is a knot that bounds a smoothly embedded disk in the three-sphere. The disk must itself be embedded; merely filling a loop with a map that is allowed to intersect itself does not establish that the knot is an unknot. Flattening a neighborhood of an embedded spanning disk explains why this criterion recovers the planar circle.

A complicated drawing can still represent an unknot. Starting with a crossing-free knot diagram of $U$, insert a small kink by a type I Reidemeister move. The resulting knot diagram has one crossing, but removing the kink exhibits the same unknot. Repeating in separate small regions gives knot diagrams of the unknot with arbitrarily many crossings, although the crossing number of a knot is $c(U)=0$. A drawing's apparent complexity is therefore not a certificate of knotting.

The normalized Alexander polynomial of a knot and Jones polynomial of the unknot are both $1$, and its knot group is the infinite cyclic group $\mathbb{Z}$, because the knot exterior is a solid torus. These are useful comparison values. In particular, a nonconstant Alexander polynomial of a knot certifies that a knot is not an unknot, but the converse fails: nontrivial untwisted Whitehead doubles also have Alexander polynomial of a knot $1$.