Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 2D (Algebra and Geometry)

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.