= Solution
The <Möbius group> consists of maps $z\mapsto(az+b)/(cz+d)$ with $ad-bc\ne0$, acting on the <Riemann sphere> $\widehat{\mathbb C}=\mathbb C\cup\{\infty\}$. 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 $c\ne0$ and stays at infinity when $c=0$.
Fixing both zero and infinity forces $b=c=0$, leaving $z\mapsto\lambda z$ with $\lambda\ne0$. 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>:
$$
\boxed{T^{-1}(\lambda T(z))=\frac{\lambda z}{1+(\lambda-1)z},\qquad \lambda\in\mathbb C^*.}
$$
Every map fixing zero and one arises this way. Composition again multiplies $\lambda$, proving \b[both subgroups are isomorphic to $\mathbb C^*$]. This is the <Möbius pointwise stabilizer of two points>; conjugation changes the selected fixed points without changing its group structure.
Back to article page
