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A finite abelian subgroup of the Möbius group is cyclic or isomorphic to $C_2\times C_2$. If it contains an element of order greater than two, conjugate that element's two fixed points to $0,\infty$. Every commuting map must preserve these points individually: an interchange would conjugate the element to its inverse. All maps are then scalings, forming a cyclic finite subgroup of ${\mathbb{C}}^{\times }$. If every nonidentity element has order two, choose one with fixed points $0,\infty$; the subgroup acts on this pair with image of size at most two and kernel contained in $\{z,-z\}$. Its size is therefore at most four, giving the asserted possibilities.

Let $T_b(z)=z+b$ and $I(z)=1/z$ on the Riemann sphere. For $s\ne 0$, direct composition gives the displayed identity. Every nonzero complex scaling $z\mapsto \lambda z$ is obtained by choosing $s^2=-\lambda$. Since translations, scalings and reciprocal inversion generate the Möbius group, translations and reciprocal inversion alone also generate it.

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, acting on the Riemann sphere $\hat{\mathbb{C}}=\mathbb{C}\cup \{\infty \}$. Proportional matrices represent the same map. Thus it is ${\mathrm{PGL}}_2(\mathbb{C})$, equivalently ${\mathrm{PSL}}_2(\mathbb{C})$ after normalizing the determinant. A zero denominator gives the image $\infty$, while $\infty$ maps to $a/c$ when $c\ne 0$ and stays at infinity when $c=0$.

Fixing both zero and infinity forces $b=c=0$, leaving $z\mapsto \lambda z$ with $\lambda \ne 0$. Composition multiplies the parameters, giving a bijective group homomorphism from the multiplicative group of nonzero complex numbers to this pointwise stabilizer.

For the other pair, take $T(z)=z/(1-z)$, which sends zero to zero and one to infinity. Conjugate the scaling subgroup by this Möbius transformation:Every map fixing zero and one arises this way. Composition again multiplies $\lambda$, proving both subgroups are isomorphic to ${\mathbb{C}}^{\ast }$. This is the Möbius pointwise stabilizer of two points; conjugation changes the selected fixed points without changing its group structure.

Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point $o$, the limit set of a Kleinian group is the set of boundary accumulation points of $Go$. Its definition is independent of $o$: two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and $G$-invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.

The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms $f(z)=\alpha z$, with $|\alpha |>1$, and $g(z)=\beta z+b$, with $b\ne 0$. But $f^{-n}g{f}^n(z)=\beta z+b{\alpha }^{-n}$ is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.

Let $\xi$ be any point of the Kleinian limit set, and choose distinct $g_n$ with $g_{n}o\to \xi$. The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to $\xi$ uniformly away from one exceptional boundary point $\eta$. To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is $\eta$ and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, $g_{n}o$ has this same image limit, so the attracting point is indeed $\xi$.

Choose distinct ${\zeta }_1,{\zeta }_2$ in the Kleinian limit set other than $\eta$. Then $g_n{\zeta }_1$ and $g_n{\zeta }_2$ both tend to $\xi$, belong to the Kleinian limit set, and remain distinct because $g_n$ is injective. If $\xi$ were isolated, both would eventually equal $\xi$, an impossibility. The limit set has no isolated points.

The assertion is true over the complex numbers. It suffices to generate every nonzero scaling using translations and $I(z)=1/z$, since these together with scalings already generate the Möbius group.

For $s\ne 0$, compose the maps rightmost first. Away from the intermediate poles, the successive values areHenceFor every $\lambda \ne 0$, choose a complex square root $s$ with $s^2=-\lambda$. This gives $D_{\lambda }$ and proves Möbius scaling generated by translations and reciprocal inversion. Equality of the rational maps extends the identity to all points of the Riemann sphere, including the intermediate exceptional values.

A subgroup of the Möbius group ${\mathrm{PSL}}_2(\mathbb{C})$ is a discrete subgroup when the identity is isolated in its induced topology, equivalently every group element is isolated. We prove the compact-set criterion for a properly discontinuous group action: for every compact $K\subset {\mathbb{H}}^3$, only finitely many $g\in G$ satisfy $gK\cap K\ne \varnothing$.

Fix an origin $o$ and take $K\subseteq \overline{B}(o,R)$. If $x\in K$ and $gx\in K$, the triangle inequality and invariance of hyperbolic distance giveThe set of Möbius isometries with $d(o,go)\le 2R$ is compact. To see this explicitly, represent an isometry by a determinant-one complex matrix and use singular-value decomposition:The compact factors fix $o$, and the diagonal factor moves it hyperbolic distance $s$, as seen from the vertical dilation by $e^s$ in upper half-space. Thus the bounded-displacement set is the continuous image of $\mathrm{SU}(2)\times [0,2R]\times \mathrm{SU}(2)$, modulo central signs.

By the allowed closedness fact, its intersection with $G$ is a closed discrete subset of a compact space, hence finite. Otherwise an infinite sequence would accumulate in that intersection, contradicting isolation of its limit point. This proves every discrete Möbius subgroup acts properly discontinuously on hyperbolic three-space. Finite stabilizers are allowed; discreteness alone does not imply a free action.

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 ${\mathrm{PSL}}_2(\mathbb{C})$.

The Poincare ball model has ideal boundary the Riemann sphere. To construct the hyperbolic extension of a Möbius transformation, conjugate the ball to the upper half-space model with coordinates $(z,t)\in \mathbb{C}\times (0,\infty )$ and metric $(|dz{|}^2+d{t}^2)/t^2$. The generators of the Möbius group extend asandTranslations and similarities plainly preserve the metric; Euclidean inversion rescales both the squared Euclidean line element and $t^2$ by the same factor, proving the last case. Composition gives the unique orientation-preserving hyperbolic extension. Equivalently, for a determinant-one matrix its upper half-space formula is

An isometry fixing the ball origin acts on its tangent space by an orientation-preserving orthogonal transformation; radial geodesics then show that the whole map is that ordinary rotation. Conversely every rotation preserves the ball metric. On the sphere these are exactly the projective special-unitary mapsthe image of $SU(2)$ modulo its central signs.

A hyperbolic line is determined by its unordered pair of ideal endpoints. It is invariant precisely when that pair is preserved as a set. A nonidentity parabolic map has one fixed point, and its square also has one, so it cannot fix or interchange two distinct endpoints. Every nonparabolic nonidentity map has two fixed points, and their joining line is invariant.

More precisely, normalize the latter map to $z\mapsto \lambda z$. Unless $\lambda =-1$, any invariant pair must consist of zero and infinity: a swapped pair would be fixed pointwise by the square, whose only fixed points remain zero and infinity when ${\lambda }^2\ne 1$. This proves uniqueness for every nonidentity map except an elliptic half-turn. For $\lambda =-1$, every pair $\{z,-z\}$ is also invariant, giving infinitely many additional lines, besides the fixed-point axis. The identity preserves every line. ThusHere the uniqueness statement includes hyperbolic translations, genuine loxodromic maps and elliptic rotations through angles other than a half-turn.

Identify the Möbius group with ${\mathrm{PSL}}_2(\mathbb{C})$, the orientation-preserving isometry group of three-dimensional hyperbolic space. Here a properly discontinuous group action permits finite stabilizers: for each compact set $K$, only finitely many $g$ satisfy $gK\cap K\ne \varnothing$.

Fix $o\in {\mathbb{H}}^3$. The set of isometries with $d(o,go)\le R$ is compact. To see this, an orientation-preserving isometry is uniquely determined by the image of an oriented orthonormal frame at $o$. The image point lies in a compact closed hyperbolic ball and its frame lies in a compact rotation-group fiber. These choices parametrize the indicated set of isometries continuously.

A discrete subgroup of a topological group is closed: otherwise distinct subgroup elements accumulating at the same point would have quotients tending to the identity, contradicting isolation of the identity. Its intersection with a compact set is therefore finite. If $K\subseteq B(o,R)$ and $gK\cap K\ne \varnothing$, choosing $x,gx\in K$ gives $d(o,go)\le d(o,gx)+d(gx,go)\le 2R$. Hence discreteness implies a properly discontinuous group action. Conversely, if the subgroup is not discrete, distinct $g_n$ tend to the identity. For a compact ball $K$ with $o$ in its interior, eventually $g_{n}o\in K$, so infinitely many $g_{n}K$ meet $K$, contradicting proper discontinuity. This proves the equivalence.

Matrices of determinant one over the Gaussian integers are closed under multiplication and inversion, with inverse $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Quotienting by $\{\pm I\}$ gives the specified group of Möbius transformations. The Gaussian integers form a discrete set in the complex plane, so these matrices form a discrete subgroup of ${\mathrm{SL}}_2(\mathbb{C})$, and the finite quotient remains discrete. Thus this group acts properly discontinuously on ${\mathbb{H}}^3$.

Explicit representatives of all four types areThe first is $z\mapsto -1/z$ and has order two; the second is the nonidentity translation $z\mapsto z+1$. The third has positive real reciprocal eigenvalues $(3\pm \sqrt{5})/2$, giving translation along an axis without rotation. For the last, a nonreal trace excludes both unit-modulus and real eigenvalues. Its multiplier therefore has modulus different from one and a nonzero rotation angle: it is a loxodromic Möbius transformation.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

A group homomorphism preserves multiplication: $\Phi (gh)=\Phi (g)\Phi (h)$. Its group kernel $K=\{g:\Phi (g)=e_H\}$ is a subgroup, and $\Phi (gk{g}^{-1})=e_H$ for $k\in K$. Thus $gK{g}^{-1}=K$, proving it is a normal subgroup. The map $gK\mapsto \Phi (g)$ is well-defined because two representatives of a coset differ by an element of $K$. It is a homomorphism, has trivial kernel, and is surjective when $\Phi$ is. This proves the first isomorphism theorem, $G/K\cong H$. This proof does not use finiteness and therefore also applies below.

The identity matrix belongs to $S{L}_2(\mathbb{C})$. The determinant is multiplicative, giving closure, and matrix multiplication is associative. The inverse of a determinant-one matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, also of determinant one. Thus this special linear group is a group.

The fractional linear maps must be interpreted on the Riemann sphere, including infinity: a finite pole is sent to infinity and infinity is sent to $a/c$ when $c\ne 0$. As literal everywhere-defined maps of the complex plane they would not form the asserted group. Substitution into the two rational expressions gives $f_A\circ f_B=f_{AB}$. Matrix identities and inverses therefore give the identity and inverse transformations, proving that the Möbius group is a group and that $\Phi :A\mapsto f_A$ is a surjective homomorphism.

For $f_A$ to be the identity, $(az+b)/(cz+d)=z$ on all finite points where defined. Hence $c{z}^2+(d-a)z-b$ is the zero polynomial, so $b=c=0$, $d=a$, and $a^2=1$ from the determinant. ThereforeThe kernel is central and hence normal. In particular $\Phi$ is not an isomorphism, since $A$ and $-A$ represent the same transformation. The quotient is the projective special linear group in this complex two-dimensional case.

The tree-level Polyakov path integral for closed-string tachyons reduces in conformal gauge to their vertex-position integral on the Riemann sphere. Three positions are fixed by the residual Möbius group; the other $N-3$ complex positions are integrated. The embedding-field Gaussian integral produces momentum conservation and the Koba-Nielsen factor.