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The integer matrices $A=(\begin{array}{cc}1& 2\\ 0& 1\end{array})$ and $B=(\begin{array}{cc}1& 0\\ 2& 1\end{array})$ generate a discrete subgroup. For every nonzero integer $m$, $A^m$ sends $|x|<1$ into $|x|>1$, while $B^m$ sends $|x|>1$, including infinity, into $|x|<1$. The ping-pong lemma identifies their subgroup as the free product of two infinite cyclic groups. This separates the proof of discreteness from the proof of freeness.

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.

Suppose a connected subset $C\subset F$ contains distinct points $a,b$. The distance function $x\mapsto |x-a|$ is $1$-Lipschitz, and its image on $C$ is connected and contains $0$ and $|b-a|$. It therefore contains the whole interval between those values. A Lipschitz map cannot increase Hausdorff dimension: taking images of arbitrarily fine covers multiplies their diameter powers by at most a fixed constant. Since a nondegenerate interval has Hausdorff dimension one, ${\dim }_{H}C\ge 1$, contradicting ${\dim }_{H}F<1$. Thus every connected subset is a singleton, so $F$ is totally disconnected.

A Möbius transformation pairs two disks when it maps their boundary circles to one another and takes the exterior of the first disk onto the interior of the second. Two pairs of disks with disjoint interiors give the ping-pong lemma construction of a free two-generator group. Tangencies are allowed in the convention suggested by the question; this is often called a Tangent-disk Schottky group, rather than a strictly separated Schottky group.

Here is a concrete configuration that also makes the dimension argument precise. Take $A(z)=z+2$ and $B(z)=z/(iz+1)$. The first pairs the half-plane disks $\Re z<-1$ and $\Re z>1$, which are disks on the sphere tangent at infinity. The second pairs $|z-i|<1$ and $|z+i|<1$, tangent at zero; their interiors are also disjoint from the half-plane disks. These pairings give a free discrete Kleinian group by ping-pong lemma, with a nonempty open fundamental region.

Set $C=AB{A}^{-1}{B}^{-1}$. Conjugating the subgroup generated by $B,C$ by $M(z)=iz/(2-z)$ gives matricesThe product of the second matrix with the first is $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. These are the standard parabolic generators of the principal level-two modular group, whose ideal-quadrilateral fundamental region tiles the upper half-plane and whose Kleinian limit set is the entire extended real line. Consequently the original group's Kleinian limit set contains the circle $|z-1|=1$, as well as its image under $A$, the distinct tangent circle $|z-3|=1$.

It follows that the limit set has Hausdorff dimension at least one and is not homeomorphic to a circle. Indeed a proper subset of a circle cannot itself be a circle: after deleting a missing point it lies in a line, which cannot contain an embedded circle. A space containing two distinct complete circles thus cannot be homeomorphic to a single circle. This explains why dimension at least one does not force circle topology. The tangent-circle terminology is consistent with the primary discussion of T-Schottky configurations in arxiv.org/pdf/math/0701572 .

A representative of an element of $G$ is a determinant-one matrix $\left(\begin{array}{cc}a& b\\ \overline{b}& \overline{a}\end{array}\right)$ over the Gaussian integers, considered up to its common sign. It acts by a Möbius transformation of the disk. Since $T(0)=b/\overline{a}$ and $|a{|}^2-|b{|}^2=1$, we haveOnly finitely many Gaussian-integer pairs $(a,b)$ can have $T(0)$ in a fixed compact subdisk. More generally, if $T$ carries some point of a compact set $K$ into $K$, and $K$ lies in the hyperbolic ball of radius $R$ about zero, the triangle inequality gives $\rho (0,T0)\le 2R$. Thus there are only finitely many such $T$. This proves the properly discontinuous group action, and in particular discreteness. The specified $A,B$ belong to $G$, so their subgroup $H$ is discrete as well.

The fixed-point equations reduce to $-i(z-1{)}^2=0$ for $A$ and $i(z+1{)}^2=0$ for $B$. Thus their unique fixed points are respectively $1$ and $-1$ on the ideal boundary. Neither transformation is the identity; their displayed determinant-one representatives have trace two. They are therefore parabolic Möbius transformations.

To find the Dirichlet domains, use the Möbius transformationIt takes the disk to the upper half-plane, and direct substitution givesFor $w=x+iy$ and $v$ in the upper half-plane, the hyperbolic distance obeys $\cosh \rho (w,v)=1+|w-v{|}^2/(2y\mathrm{Im}v)$. Since $C(A^{k}0)=i-2k$, the distance inequalities are equivalent toThese reduce to $kx+k^2\ge 0$ for all integers $k$. The cases $k=1,-1$ imply $-1\le x\le 1$, which also suffices for every other $k$. Therefore the cyclic Dirichlet domain becomes the strip $|\mathrm{Re}w|\le 1$.

The two vertical hyperbolic geodesics pull back to arcs joining $1$ to $i$ and $1$ to $-i$. In disk coordinates they are the circles $|z-(1+i)|=1$ and $|z-(1-i)|=1$, restricted to the disk. ThusConjugating by $z\mapsto -z$ gives $B$, so its two hyperbolic geodesics join $-1$ to $i$ and $-1$ to $-i$, andThe requested drawing and their intersection are shown below.

For any $z_0$, the orbit points satisfying $\rho (0,T{z}_0)\le \rho (0,z_0)$ form a nonempty finite set, by proper discontinuity and compactness of closed hyperbolic balls. Choose a closest one, $q=T{z}_0$. For every integer $k$, its minimality givesand the same inequality with $B$. Hence $q$ lies in the ideal hyperbolic quadrilateralIts ideal vertices are $1,i,-1,-i$.

For the quotient identification, it remains to justify that this covering region is a fundamental polygon. The four open corner caps excluded by $P$ are pairwise disjoint inside the disk. The transformation $A$ carries the exterior of the lower-right cap into the upper-right cap; $A^{-1}$ reverses the pairing. Similarly $B$ carries the exterior of the upper-left cap into the lower-left cap, and $B^{-1}$ reverses this. These assertions follow either from the upper-half-plane formulas or by mapping their boundary circles and testing zero: $A0=(1+i)/2$ and $B0=(-1-i)/2$.

Apply a reduced word in $A^{\pm 1},B^{\pm 1}$ from right to left to a point in the interior of $P$. The first letter puts it into its target cap. Every subsequent letter can do the same, since its inverse cap is distinct from the preceding target cap unless the word cancels. Thus the final point lies in the target cap of the leftmost letter, never in the interior of $P$. This ping-pong lemma argument proves freeness and disjointness of distinct translated interiors. Together with orbit coverage it makes $P$ a fundamental polygon. The corresponding argument on closed caps shows that boundary identifications are generated by the stated side pairings.

The generator $A$ pairs the side from $1$ to $-i$ with the side from $1$ to $i$, fixing the ideal endpoint $1$ and sending $-i$ to $i$. The generator $B$ pairs the side from $-1$ to $i$ with the side from $-1$ to $-i$, fixing $-1$ and sending $i$ to $-i$. There are three ideal-vertex classes: $\{1\}$, $\{-1\}$ and $\{i,-i\}$. The first two give parabolic ends from $A,B$; the third does too, since $AB$ is a nonidentity parabolic transformation fixing $i$. Adding one point at each end produces an oriented closed surface with one face, two edges and three vertices. Its Euler characteristic is $1-2+3=2$, so it is a sphere. Removing the three added points gives a thrice-punctured sphere, proving

Consider the Möbius transformationsTheir group lies in ${\mathrm{PSL}}_2(\mathbb{Z})$, which is a discrete subgroup of the Möbius group: integer entries cannot approach the identity matrix except by eventually equalling it, and passing to the quotient by $\{I,-I\}$ preserves this property.

For freeness let $X=\{x\in \mathbb{R}:|x|>1\}$ and $Y=\{x\in \mathbb{R}:|x|<1\}$. For every nonzero integer $n$, $A^n(Y)=(2n-1,2n+1)\subset X$ andA reduced word involving both generators can be conjugated to a cyclically reduced word beginning with $A^n$ and ending with a nonzero power of $B$, after interchanging the generators if necessary. Repeated use of these inclusions shows that this word sends $X$ into the proper subset $A^n(Y)$ of $X$, so it cannot be the identity. Nonzero pure powers of either generator are also nonidentity. This is the ping-pong lemma in this particular action, and proves that $\boxed{⟨A,B⟩\cong F_2}$ is a free group of rank two as well as a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$.

Let $H_k$ be the open hyperbolic half-plane beyond ${\gamma }_k$, opposite the central region $U$. These three half-planes have disjoint closures in the disk, and hyperbolic reflection $J_k$ exchanges $H_k$ and its complementary half-plane. In particular $J_k(U^{\circ })\subset H_k$ and $J_i(H_j)\subset H_i$ when $i\ne j$. A reduced nonempty reflection word therefore sends an interior point of $U$ into the half-plane corresponding to its first letter. It cannot be the identity. This ping-pong lemma argument proves that the reflection group isThe usual reflection tiling makes $U$ a fundamental domain for $W$: successive reflection across an offending side reduces a point to $U$, and the disjoint-side configuration yields locally finite tiles. Equivalently this is the disjoint-geodesic case of the reflection fundamental-domain theorem; there are no vertex relations because the sides never meet.

The orientation-preserving subgroup consists of even words. Every pair $J_i{J}_j$ can be expressed in $A=J_2{J}_1$ and $B=J_3{J}_2$: the other nontrivial pairs are their inverses and $J_3{J}_1=BA$, $J_1{J}_3=A^{-1}{B}^{-1}$. Thus it is exactly $G$. A fundamental set is $U\cup J_{2}U$, with one representative chosen on each paired side; its interior is a fundamental domain. To see there are no relations between $A,B$, the reflection-word tree has vertices the reduced words and edges joining $w$ to $w{J}_i$. The even subgroup acts freely on this tree; its quotient has two vertices joined by three edges. Choosing the $J_2$ edge as a spanning tree leaves two independent loops, represented by $A$ and $B$. Reduced loops in a graph form a free group, soThe quotient is the double of the central three-sided region along its geodesic sides: an orientable pair of pants, topologically a sphere with three points removed. Its three ends are hyperbolic funnels, not cusps, because the sides are ultraparallel and the peripheral transformations have positive translation lengths. The compact convex core has three geodesic boundary circles; the complete quotient itself has no boundary.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

Consider the matricesTheir projective action on the positive real line isThese disjoint images give the monoid version of the ping-pong lemma: after removing a common initial letter, two different positive words act differently. Thus $A,B$ generate a free monoid. There are $2^n$ distinct positive words of length $n$, all lying in the radius-$n$ word ball for any generating set enlarged to contain $A,B$. Therefore the Exponential growth of SL2 of Z gives

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 $G$.

PutFor every nonzero integer $n$,LetThese are disjoint nonempty subsets of ${\mathbb{R}}^2$. If $(x,y)\in B$, thenso $u^n(B)\subseteq A$. Similarly, if $(x,y)\in A$, thenso $v^n(A)\subseteq B$.

The ping-pong lemma now identifies the subgroup generated by $u$ and $v$ withBoth generators have infinite order, so this is a free group of rank two inside the special linear group ${\mathrm{SL}}_2(\mathbb{Z})$.

Every finitely generated free group is a residually finite group. For $F_2$, the ping-pong lemma embeds it into ${\mathrm{SL}}_2(\mathbb{Z})$, and reduction modulo a prime in an integral matrix group separates each nonidentity element in a finite quotient.

A Tangent-disk Schottky group uses side-pairing disks with disjoint interiors but permits boundaries to meet. This convention must be distinguished from the strictly separated, purely loxodromic Schottky group convention. For example, the maps $A(z)=z+2$ and $B(z)=z/(iz+1)$ pair the half-plane disks $\Re z<-1,\Re z>1$ and the disks $|z-i|<1,|z+i|<1$. The ping-pong lemma gives a free discrete group. With $C=AB{A}^{-1}{B}^{-1}$, the subgroup generated by $B,C$ is conjugate by $M(z)=iz/(2-z)$ to the principal level-two modular group, generated by $\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$ and $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. Thus its Kleinian limit set contains the circle $|z-1|=1$ and its translate $|z-3|=1$. Its Hausdorff dimension is at least one and it cannot be homeomorphic to a single circle.