Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 3 / 7B (Algebra and Geometry)

If all three points are finite, define the Möbius transformationIts numerator vanishes at $\alpha$, its pole is at $\gamma$, and substitution of $\beta$ gives $1$. The three points are distinct, so the numerator and denominator are not proportional and the associated $2\times 2$ matrix has nonzero determinant.

If one point is infinite, the limiting formulas areIn each case the finite values and the limit at infinity give $g(\alpha )=0$, $g(\beta )=1$, $g(\gamma )=\infty$, and the corresponding determinant is nonzero. Distinctness permits at most one of the points to be infinite. Hence every ordered triple of distinct points on the Riemann sphere can be normalized to $(0,1,\infty )$.

Each element of $H$ permutes the three-point set, giving a group homomorphism $\rho :H\to S_3$. To prove injectivity, let an element fix all three points. Conjugate it by the normalizing Möbius transformation from part (i). The conjugate fixes $0,1,\infty$. A Möbius transformation $z\mapsto (az+b)/(cz+d)$ fixing $0$ has $b=0$, and fixing infinity has $c=0$. It is then $z\mapsto kz$; fixing $1$ forces $k=1$. Thus the conjugate, and hence the original transformation, is the identity. The kernel of a group homomorphism is trivial.

For surjectivity, let $(x_1,x_2,x_3)=(\alpha ,\beta ,\gamma )$ and choose any permutation $\sigma$. Let $g$ normalize the ordered triple $(x_1,x_2,x_3)$ and let $g_{\sigma }$ normalize $(x_{\sigma (1)},x_{\sigma (2)},x_{\sigma (3)})$. Then $g_{\sigma }^{-1}g$ sends $x_j$ to $x_{\sigma (j)}$, lies in $H$, and induces the chosen permutation. Thus $\rho$ is onto andFor the normalized set, the six Möbius transformations permuting three points are $z$, $1-z$, $1/z$, $1/(1-z)$, $z/(z-1)$, and $(z-1)/z$. Conjugating by $g$ gives the corresponding six maps for the original set.