Möbius group ($\mathcal{M}\cong PS{L}_2(\mathbb{C})$)

The Möbius group is the group of all Möbius transformations of the Riemann sphere, under function composition. An invertible $2\times 2$ complex matrix represents $z\mapsto (az+b)/(cz+d)$, and nonzero scalar multiples represent the same transformation. Rescaling a representative to determinant one gives a surjective group homomorphism $S{L}_2(\mathbb{C})\to \mathcal{M}$ with kernel of a group homomorphism $\{I,-I\}$. Thus it is also the projective linear group $PG{L}_2(\mathbb{C})$ and the quotient $PS{L}_2(\mathbb{C})=S{L}_2(\mathbb{C})/\{\pm I\}$.

For two distinct points of the Riemann sphere, conjugate by a Möbius transformation sending them to zero and infinity. A transformation fixing those two points has the form $z\mapsto \lambda z$, $\lambda \ne 0$. Composition multiplies the parameter, so the pointwise stabilizer is isomorphic to the complex multiplicative group. For zero and one, conjugation by $z/(1-z)$ gives $z\mapsto \lambda z/[1+(\lambda -1)z]$.

The Möbius transformations preserving a three-point set act faithfully on it: one fixing all three is the identity. Every permutation is realized because a Möbius transformation can send any ordered distinct triple to any other, by composing the two normalizations to $0,1,\infty$. For that normalized set the six maps are $z$, $1-z$, $1/z$, $1/(1-z)$, $z/(z-1)$ and $(z-1)/z$. Thus its setwise stabilizer is the symmetric group $S_3$. Normalizing any other three-point set conjugates its stabilizer to this group.

The Möbius transformations generated by $D_{\lambda }(z)=\lambda z$ with $\lambda \ne 0$ and $I(z)=1/z$ have exactly the two displayed forms. The relations $I^2=1$ and $I{D}_{\lambda }=D_{{\lambda }^{-1}}I$ reduce every word to $D_{\lambda }{I}^{ϵ}$, $ϵ\in \{0,1\}$. This is a semidirect product of ${\mathbb{C}}^{\times }$ with a cyclic group of order two acting by multiplicative inversion. It preserves the unordered pair $\{0,\infty \}$, and only its scaling elements fix zero.

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.