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Choose a local coordinate $w$ with $w(p)=0$. Differentiation at $p$ gives a homomorphismIt is injective. Indeed, a nonidentity Möbius transformation fixing $p$ and having derivative one there is parabolic, hence conjugate to a nonzero translation and of infinite order; no such element can belong to the finite group $H$. Every finite subgroup of ${\mathbb{C}}^{\times }$ is a cyclic group of roots of unity, so $H$ is cyclic. Write $n=|H|$.

The averaged coordinatehas derivative one at $p$, so it is a coordinate on a smaller neighbourhood. For $g\in H$, reindexing by $k=hg$ givesAfter shrinking to an $H$-invariant disc $U$, the group therefore acts as all rotations $z\mapsto {\zeta }^{j}z$, where $\zeta$ is a primitive $n$th root of unity. The invariant coordinateidentifies $H\backslash U$ with a disc and gives it a Riemann-surface chart. In these coordinates the quotient map is exactlyThis is the local cyclic quotient of a Riemann surface.

Existence follows from the three-transitivity of Möbius transformations: one can send three distinct source points to $0,1,\infty$ and then send those to the three distinct target points.

For uniqueness, suppose $S$ and $T$ have the same values at three distinct points. Then $M=T^{-1}S$ fixes those points. Ifthe finite fixed points satisfy $c{z}^2+(d-a)z-b=0$. A nonidentity Möbius map therefore has at most two fixed points on the Riemann sphere. Since $M$ has three, it is the identity, and $S=T$.

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

Letbe reflection in the unit circle, and represent a Möbius transformation $g$ byA direct calculation shows that $JgJ$ is represented byThus $gJ=Jg$ exactly when $M$ and $\tau (M)$ differ by a nonzero scalar. Applying the conjugate-linear involution $\tau$ twice shows that this scalar has modulus one. Rescaling $M$ by a suitable complex scalar then makes $M=\tau (M)$. Consequently all commuting maps, and only those maps, have the formThis is the Möbius maps commuting with reflection in the unit circle classification.

For such a map,It follows that $|g(z)|<1$ whenever $|z|<1$ precisely whenThe inverse has the same property, so these and only these maps preserve the unit disc.

The domains are conformally equivalent. The Möbius transformationmaps the upper half of the unit disc onto the first quadrant: its diameter $(-1,1)$ maps to the positive real axis and its upper semicircle maps to the positive imaginary axis. Squaring maps the first quadrant bijectively and conformally onto the upper half-plane. Thereforeis the required equivalence, as recorded by the conformal equivalence between the upper half-disc and upper half-plane.

Write $\tau =x+iy\in \mathfrak{h}$. Forthe imaginary part of the Möbius transformation isAmong the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau +d|$. Such a minimum exists because only finitely many lattice points lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Then $\gamma \tau$ has maximal imaginary part in its modular group orbit.

Applying an integral translation does not change that imaginary part, so arrangeIf $\mathrm{Im}(\gamma \tau )<\sqrt{3}/2$, then $|\gamma \tau |<1$. The modular inversion $S:z\mapsto -1/z$ would givecontradicting maximality. Hence every orbit meets the stated region. This is the reduction to the standard modular region argument.

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.

For the Möbius transformation $z^{\prime }=(az+b)/(cz+d)$ with $ad-bc=1$,The power of $|c{z}_i+d|$ contributed by every pair containing $i$ iswhere momentum conservation and the mass-shell condition $p_i^2=4/{\alpha }^{\prime }$ were used. The Koba-Nielsen factor therefore contributes $|c{z}_i+d{|}^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb{C})$ invariant, and division by its volume removes the residual conformal-gauge redundancy.

The valency theorem states that for a nonconstant analytic map between compact connected Riemann surfaces,for every target point $w$.

For a rational map $f=p/q$ on the Riemann sphere, first cancel common factors; then the degree of a rational map of the Riemann sphere isThe analytic isomorphisms are exactly the degree-one maps, namely the Möbius transformations.

The required transformation isIt sends $\infty \leftrightarrow 1$ and $0\leftrightarrow -1$. Its fixed points satisfy

The octahedral rotation orbits on the Riemann sphere have possible sizescorresponding respectively to vertices, face centres, edge centres, and generic points.

In the displayed $F$, no numerator factor vanishes at $0$ or at a fourth root of one. Hence $0,\pm 1,\pm i$ are poles of order four. The numerator and denominator have degrees $24$ and $20$, so $F(z)\sim z^4$ at infinity; infinity is also a pole of order four. Thusand

Finally let $z$ have stabilizer of order $e$. A local coordinate turns its stabilizer action into rotation by $e$th roots of unity. Since $F$ is invariant, the first nonconstant term of its local expansion has exponent divisible by $e$, so $m_F(z)\ge e$. The orbit has $24/e$ points and therefore contributes at least $24$ to the fibre through $F(z)$. The valency theorem and $\mathrm{deg}F=24$ leave no room for any further point. The degree-sized invariant separates finite-group orbits, so $F(z)=F(w)$ implies that $z$ and $w$ lie in the same orbit.

For $|a|<1$, the Möbius transformation $f_a(z)=(z-a)/(1-\overline{a}z)$ is an isometry of the Poincare disc model sending $a$ to zero. Rotations combined with these transformations describe its orientation-preserving isometries.

The group $PG{L}_2(F)=G{L}_2(F)/Z$ acts faithfully onby Möbius transformations. For $|F|=q$, this embeds $PG{L}_2(F)$ into the symmetric group $S_{q+1}$.

Normalize three distinct points of the Riemann sphere by a Möbius transformation $f$ sending them to $\infty ,0,1$. A fourth point lies on their generalized circle exactly when $f$ takes a real value there: the image of that generalized circle is the extended real line. Other standard cross-ratio ordering conventions differ by real fractional-linear transformations and preserve this reality criterion.

The special linear group $S{L}_2(\mathbb{R})$ acting by Möbius transformations has exactly three group orbits on the Riemann sphere: the extended real line and the two open half-planes. The identity $\mathrm{Im}((az+b)/(cz+d))=\mathrm{Im}z/|cz+d{|}^2$ preserves these sets. Upper triangular determinant-one matrices send $i$ to any prescribed point of the complex upper half-plane, and similarly send $-i$ through the lower half-plane. The stabilizer subgroups of $i$ and $-i$ are both $SO(2)$.

Rotations of the domain and target Riemann spheres act as Möbius transformations represented by SU(2) matrices. A combined symmetry requires the equivariance identity $R(gz)=hR(z)$ for the corresponding transformations. Symmetry of the Wronskian of a rational map alone is insufficient: it tests the ramification directions, not the entire map. Target rotations preserve the angular Jacobian of a rational map, so an equivariant map has a domain-invariant angular density.

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on the projective line by the Möbius transformation $x\mapsto (ax+b)/(cx+d)$, with the pole sent to infinity and infinity sent to $a/c$ when $c\ne 0$. The action is transitive because $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any finite $t$. The stabilizer subgroup of infinity consists of $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ with $a\ne 0$. The orbit-stabilizer theorem gives $|{\mathrm{SL}}_2({\mathbb{F}}_q)|=q(q^2-1)$. Its kernel is the scalar matrices of determinant one: a transformation fixing infinity has $c=0$, fixing zero then forces $b=0$, and fixing one gives $a=d$. Thus the kernel is $\{I,-I\}$, with these matrices coinciding in characteristic two.

At genus zero the residual Möbius transformation group fixes three marked insertion points, leaving one complex position to integrate. The corresponding worldsheet ghost fields give $|z_{ij}{z}_{jk}{z}_{ki}{|}^2$ for the three fixed points. For on-shell weight-$(1,1)$ string vertex operators, these ghost factors cancel the coordinate dependence of the fixed-point normalization, including the limit at infinity. The remaining Koba-Nielsen factor for four tachyons is the integrand of the complex beta integral, producing the Virasoro–Shapiro amplitude.

Under the stereographic projection $w=(x+iy)/(1-z)$, rotations about the third axis have representatives $\mathrm{diag}(e^{i\theta /2},e^{-i\theta /2})$, and rotations about the second axis have representativesThese belong to the special unitary group $SU(2)$. Euler-angle generation of the special orthogonal group $SO(3)$ shows that every sphere rotation becomes a Möbius transformation represented in $SU(2)$. Representatives $U$ and $-U$ induce the same transformation.

If a simple Steiner chain closes after $n\ge 3$ circles, rotating the starting circle between concentric boundaries gives another closed $n$-circle chain. Applying a Möbius transformation proves the corresponding statement for arbitrary boundary circles. Construction must stay in the same intervening family and proceed without reversing the chosen direction. Merely choosing arbitrary circles tangent to the preceding circle permits backtracking and does not assert closure.

An SU(2) matrix is a two-by-two complex unitary matrix with determinant one. It has the form $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Its action on the Riemann sphere is a rotational Möbius transformation; opposite matrices induce the same sphere rotation.

For a Steiner chain between concentric circles, tangencies of neighbouring circles are midpoints of their centres and lie on the circle of radius $\sqrt{R\rho }$. Under a Möbius transformation, their locus is a generalized circle, which may be a line. An ordinary-circle conclusion requires the pole of the inverse transformation to avoid the original locus.