Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 6D (Groups)

$G{L}_2(\mathbb{C})$ consists of invertible $2\times 2$ complex matrices and $S{L}_2(\mathbb{C})=\ker \det$. It is normal, and the first isomorphism theorem gives $G{L}_2/S{L}_2\cong {\mathbb{C}}^{\ast }$.

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, including the usual action on $\infty$. Scalar matrices act trivially and are the whole kernel, so $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

The map $z\mapsto z-w$ sends $w$ to $0$, proving transitivity (with inversion handling $\infty$). Fixed points satisfy the quadratic $c{z}^2+(d-a)z-b=0$, with multiplicity and $\infty$ included, so a nonidentity has one or two. The stabiliser of $\infty$ is $S=\{z\mapsto az+b:a\ne 0\}$.

Choose a fixed point $p$ of a nonidentity transformation and conjugate by a Möbius map carrying $p$ to $\infty$; the conjugate lies in $S$. The identity already lies there.