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For nontrivial groups $A,B$, the free product $A\ast B$ is amenable exactly in the exceptional case $A\cong B\cong C_2$. In that case it is the infinite dihedral group, hence a virtually cyclic group and amenable. In every other case the normal form theorem for a free product and the ping-pong lemma produce a subgroup isomorphic to $F_2$; the fact that a subgroup of an amenable group is amenable then obstructs amenability.

A discrete subgroup of Euclidean isometries is a frieze group when its translation subgroup is infinite cyclic and has finite index. Translations along a line alone give $\mathbb{Z}$; adding perpendicular mirrors gives the infinite dihedral group, while adding a mirror along the translation direction gives $\mathbb{Z}\times C_2$. These groups describe patterns periodic in one planar direction.

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.

The Hilbert basis theorem says that if $B$ is a commutative Noetherian ring, then $B[t]$ is Noetherian; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an ideal $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are ideals of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The ascending chain condition makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\dots ,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min (m,N)$ and express its leading coefficient as ${\sum }_j{a}_{dj}{b}_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-{\sum }_j{b}_j{t}^{m-d}{f}_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient rings with zero divisors.

Put $A=\mathbb{C}G$ and $L=\mathbb{C}[X,X^{-1}]$. The infinite dihedral group has normal forms $X^m$ and $X^{m}Y$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto -n$ on $\mathbb{Z}$ shows that these forms are distinct. ThusThe Hilbert basis theorem makes $\mathbb{C}[X]$ Noetherian, and its localization $L$ is Noetherian: an ideal of a localization is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a Noetherian module over $L$. Every left ideal of $A$ is an $L$-submodule, so all its ascending chains stabilize. Therefore $\mathbb{C}G$ is left Noetherian.

For the center of the infinite dihedral group algebra, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent integral domain, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric Laurent polynomial is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge 1$. With $z=X+X^{-1}$, the recursionexpresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. HenceThe identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequentlyis a free $R$-module of rank four.

Let $M$ be a nonzero simple module on the left. For any $m\ne 0$, $M=Am$, so it is finite over $R$. Every central $r\notin P={\mathrm{Ann}}_R(M)$ acts bijectively: its kernel and image are $A$-submodules, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-module. For $0\ne b\in B$, $bM=M$. If $m_1,\dots ,m_d$ generate $M$, write $m_i=b{\sum }_j{c}_{ij}{m}_j$. The adjugate matrix identity for $I-bC$ shows that $\det (I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a field, and $P$ is a maximal ideal of $R$. This proves the general principle for simple modules over an algebra finite over its center in this instance.

Every maximal ideal of $\mathbb{C}[z]$ is $(z-\lambda )$ for some $\lambda \in \mathbb{C}$, by factorization of complex polynomials. The module $M=Am$ is therefore a quotient of the four-dimensional algebra $A/(z-\lambda )A$. In particular ${\dim }_{\mathbb{C}}M\le 4<\infty$, as required.

The actual simple dimensions are at most two. If $\lambda \ne \pm 2$, choose $u\ne \pm 1$ with $u+u^{-1}=\lambda$. The matrices $X=\mathrm{diag}(u,u^{-1})$ and $Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ give a representation of $A/(z-\lambda )A$ onto $M_2(\mathbb{C})$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both algebras have dimension four, so this is an isomorphism. If $\lambda =\pm 2$, the element $X\mp 1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided ideal is square zero. It annihilates every simple module, since its action on a simple module is either zero or the whole module, and the latter contradicts its square being zero. The remaining quotient is $\mathbb{C}[C_2]$, whose two simple modules have $Y=\pm 1$. Thus the generic simple modules have dimension two, and the four exceptional one-dimensional characters have $X=\pm 1$, $Y=\pm 1$.

A discrete subgroup of the Euclidean isometry group has the identity isolated in the group topology. Equivalently, no sequence of distinct group elements converges to the identity; translating neighbourhoods then isolates every element. Such a subgroup is closed, and its intersection with any compact subset is finite: otherwise two elements of a convergent sequence would have quotient tending to the identity.

Write each Euclidean isometry uniquely as $x\mapsto Qx+b$, $Q\in O(2)$. Its linear part gives the group homomorphism $G\to O(2)$, since the linear part of a composite is the product of the linear parts. Its kernel is exactly the translation subgroup $G_T$. The translation vectors form a discrete additive subgroup $L\subset {\mathbb{R}}^2$.

If $L\ne \{0\}$, choose a shortest nonzero vector $a$, possible by discreteness and finiteness in compact balls. On its line, Euclidean division of a scalar coefficient by one proves $L\cap \mathbb{R}a=\mathbb{Z}a$: a nonzero remainder would be shorter than $a$. If all vectors are collinear, this finishes the proof. Otherwise orient the perpendicular coordinate so some vector has positive height. Reduce every vector's component along $a$ modulo $a$ into a bounded interval. Among vectors with positive height at most that of a chosen vector, these reduced representatives lie in a compact rectangle and are finite. Choose $b$ with the smallest such positive height. Subtracting an integer multiple of $b$ from any vector of $L$ reduces its height into $[0,b_{\perp })$; minimality makes the remainder's height zero, so it is an integer multiple of $a$. ThusThis derives the rank-at-most-two translation lattice rather than assuming every discrete subgroup of the plane is a full lattice.

Two examples with $G_T\cong \mathbb{Z}$ are the group generated by $T(x,y)=(x+1,y)$ alone, and the infinite dihedral group generated by $T$ together with $R(x,y)=(-x,y)$. The latter has $RTR=T^{-1}$; its elements are translations $T^k$ and reflections $T^{k}R$, whose mirror lines are $x=k/2$. The first group preserves the orientation of every translated motif; the second also reverses it across these mirrors.

A frieze group is a discrete subgroup of Euclidean isometries whose translation subgroup is generated by one nonzero vector and has finite index. A plane crystallographic group is a discrete subgroup of Euclidean isometries with a rank-two translation lattice of finite index; equivalently its action has a compact fundamental region.

Here are three frieze groups, with explicit generators. The translation $T(x,y)=(x+1,y)$ alone gives $\mathbb{Z}$. Adding the vertical reflection in mathematics $S(x,y)=(-x,y)$ gives $\mathbb{Z}⋊C_2$, the infinite dihedral group, since $STS=T^{-1}$. Instead adding the horizontal reflection in mathematics $H(x,y)=(x,-y)$ gives $\mathbb{Z}\times C_2$, since $HT=TH$. Their translation parts are exactly $⟨T⟩$ and their point parts have respectively one, two and two elements.

For plane crystallographic groups, let $T_1(x,y)=(x+1,y)$ and $T_2(x,y)=(x,y+1)$. These generate ${\mathbb{Z}}^2$. Adding the half-turn $R_2(x,y)=(-x,-y)$ gives ${\mathbb{Z}}^2⋊C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives ${\mathbb{Z}}^2⋊C_4$, with $R_4{T}_1{R}_4^{-1}=T_2$ and $R_4{T}_2{R}_4^{-1}=T_1^{-1}$. These are the translation, half-turn and square-rotation wallpaper groups. Integer translation vectors and a finite point part make each group a discrete subgroup; the unit square meets every orbit, establishing a cocompact group action.

For the final claim, suppose a Euclidean isometry $R$ has order five. An orientation-reversing planar isometry has finite order only when it is a reflection in mathematics, so $R$ is a rotation through $\theta =2\pi j/5$, $j=1,2,3,4$, about some centre. Write $A$ for its linear rotation part. If $t_v\in G$ is a nonzero translation, conjugation gives $R{t}_v{R}^{-1}=t_{Av}$. Thus $G$ contains translations by $v$ and by $(A+A^{-1})v=cv$, where $c=2\cos \theta$.

For a primitive fifth root $\zeta$, dividing $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$ by ${\zeta }^2$ gives $c^2+c-1=0$. Hence $c$ is irrational. The pigeonhole principle, applied to $N+1$ fractional parts of $0,c,\dots ,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\le 1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the discrete subgroup hypothesis. There is no element of order five. This argument needs only the given nonzero translation, not a rank-two lattice.

Each Möbius strip deformation retracts onto its core circle. Its boundary travels twice around that circle. Thicken the two strips slightly across their common boundary to obtain an open cover whose intersection deformation retracts onto that boundary circle. The Seifert-van Kampen theorem therefore givesDepending on the boundary orientation the relation may first read $a^2=b^{-2}$; replacing $b$ by its inverse gives the displayed presentation.

The map $a,b\mapsto 1$ respects the relation and surjects onto the infinite cyclic group $\mathbb{Z}$, so the fundamental group is infinite. It also surjects onto the infinite dihedral group: map $a$ and $b$ to the reflections $s(x)=-x$ and $t(x)=2-x$ of the real line. Both squares are the identity, but $st(x)=x-2$ and $ts(x)=x+2$ differ. An abelian group cannot have this nonabelian quotient. The fundamental group is infinite and nonabelian.

Let $a,b$ be the generating involutions. If $a\ne b$, set $x=ab$ and $y=b$. Then $x\ne e$, so its order of a group element is at least two or infinite, andTherefore $x,y$ generate the group and satisfy the required dihedral inversion relation. If $a=b$, the group is $C_2$ and $x=y=a$ gives its permitted degenerate description. This proves the two-involution characterization of a dihedral group without excluding either $C_2$ or the infinite dihedral group.

The two-involution characterization of a dihedral group uses the extended convention that includes $C_2$ and the infinite dihedral group; the two generators need not be distinct. Put $a=xy$ and $b=y$. The inversion relation givesIf $a\ne e$, both $a,b$ are involutions and generate the original group. If $a=e$, then $x=y$ and the group is $C_2$; choose the same nonidentity element twice. Thus two elements of order two generate the group, including the degenerate case allowed here. The distinction between square equal to the identity and order of a group element exactly two is essential.

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 $D_{\infty }$.

Yes, but there is only one nontrivial isomorphism type that works. If $G=C_2$, thenThe infinite dihedral group contains its infinite cyclic rotation subgroup with index two, so it is amenable by virtually abelian groups are amenable.

Suppose now that $G\ncong C_2$. The action of $G\ast G$ on its Bass-Serre tree is non-elementary: one vertex degree is greater than two, so there are hyperbolic elements with disjoint pairs of endpoints. Suitable powers satisfy the ping-pong lemma on the boundary and generate a copy of $F_2$. Since a subgroup of an amenable group is amenable, an amenable group cannot contain this nonamenable subgroup. The amenability of a free product therefore yieldsfor nontrivial finitely generated $G$.

Maschke's theorem says that if $H$ is a finite group and the characteristic of a field $F$ does not divide $|H|$, then every finite-dimensional $F$-representation of $H$ is semisimple.

To prove it, let $W$ be an invariant subspace of a representation $V$, and choose any linear projection $P:V\to W$. Average its conjugates:The scalar $|H{|}^{-1}$ exists by the characteristic assumption. Reindexing the sum shows that $\overline{P}$ commutes with every $\rho (g)$. Since each summand restricts to the identity on $W$, so does $\overline{P}$; hence $\overline{P}$ is an equivariant projection onto $W$. Its kernel is an invariant complement, and induction on $\dim V$ decomposes $V$ into irreducible representations.

The isometry group of $\mathbb{Z}$ is the infinite dihedral groupLet $\rho$ be a nonfaithful finite-dimensional complex representation. Its nontrivial normal kernel contains $t^m$ for some $m>0$, by the stated hint, so $\rho$ factors through the finite quotientMaschke's theorem decomposes this quotient representation into irreducibles. To bound their dimensions, choose a $t$-eigenvector $v$ with eigenvalue $\lambda$ in an irreducible constituent. The relation givesso $\mathrm{span}\{v,sv\}$ is invariant under both generators. Irreducibility makes it the whole constituent, whose dimension is therefore at most two. This proves the result on nonfaithful representations of the infinite dihedral group.

For the requested representation of $(\mathbb{Z},+)$, let its generator act on the polynomials of degree at most one by translation:In the basis $(1,x)$,This is a nontrivial Jordan block, so it is not diagonalizable. A direct sum of one-dimensional subrepresentations would make $T$ diagonalizable, proving that this nonsemisimple translation representation of the infinite cyclic group has the required property.

Finally, on $V=\mathbb{C}[x{]}_{\le 2}$ defineThen $S^2=I$ and $STS=T^{-1}$, so these operators define a quadratic-polynomial representation of the infinite dihedral group. In the basis $(1,x,x^2)$ they areHere $(T-I{)}^2\ne 0$ but $(T-I{)}^3=0$, so $T$ consists of one size-three Jordan block. If $V$ were a direct sum of subrepresentations of dimensions at most two, then $T$ would be block diagonal with Jordan blocks of sizes at most two, a contradiction. Thus this representation has the required failure of decomposition.

The Poincaré half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

Write a point of the band $X$ as $(u,t)$ with $u\in S^2$, $|t|\le 1/2$, after rescaling the radius of the first three coordinates. The antipodal identification isThe homotopy $(u,t)\mapsto (u,(1-s)t)$ is equivariant under this identification, so it descends to a deformation retraction of $Y$ onto the central sliceThusThe two boundary spheres of $X$ are exchanged by the antipodal map, so $\partial Y\cong S^2$. This is the mapping-cylinder model of punctured real projective three-space.

The standard cellular chain complex for ${\mathbb{RP}}^3$, with one cell in each dimension zero through three, has boundary maps alternating between multiplication by two and zero:Therefore the integral homology of real projective three-space isIn particular,

Apply the Mayer–Vietoris theorem toThe relevant pieces of the long exact sequence areandThe last map is injective, so

Since $\partial Y\cong S^2$ is simply connected and ${\pi }_1(Y)\cong {\pi }_1({\mathbb{RP}}^2)\cong \mathbb{Z}/2$, the Seifert-van Kampen theorem givesthe infinite dihedral group. Topologically, $Z$ is the double of punctured real projective three-space, namely ${\mathbb{RP}}^3\#{\mathbb{RP}}^3$.

The universal cover of each copy of $Y$ is $S^3$ with two disjoint open balls removed, homeomorphic to $S^2\times [0,1]$. The universal cover of the double strings infinitely many such cylinders together according to the Cayley line of $D_{\infty }$. Hence the familiar covering space is

The infinite dihedral group is $D_{\infty }\cong C_2\ast C_2$, with factors $A=⟨a⟩$ and $B=⟨b⟩$. Its Bass-Serre tree has vertex setand one edge indexed by each $g\in D_{\infty }$, joining $gA$ to $gB$. Since both factors have order two, every vertex has degree two. The connected tree is therefore a bi-infinite line.

The action is cocompact, and its vertex stabilizers are the finite conjugates of $A$ and $B$, so it is proper. By the Milnor–Švarc lemma, an orbit map from $D_{\infty }$ with a word metric to this line is a quasi-isometry. A simplicial bi-infinite line is quasi-isometric to $\mathbb{R}$, hence so is $D_{\infty }$.

A group generated by two involutions is dihedral in the extended convention allowing $C_2$ and the infinite dihedral group. If $a\ne b$, put $x=ab$, $y=b$; then $yx{y}^{-1}=x^{-1}$. Conversely, from the inversion relation, $y$ and $xy$ square to the identity and generate. If $xy=e$, the group is $C_2$ and the same nonidentity element is used twice. Exact order two, rather than merely a square equal to the identity, requires this exceptional case. This convention extends the usual finite dihedral group definition.