Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 3 / 7B / ii

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.