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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.

Let $T(z)=(az+b)/(cz+d)$, with $ad-bc\ne 0$. If $c=0$, then $d\ne 0$ anda composition of a nonzero scaling and a translation. If $c\ne 0$, rearrange the fraction asThus every Möbius transformation is a composition, in order, of translation by $d/c$, reciprocal inversion, nonzero scaling by $(bc-ad)/c^2$, and translation by $a/c$. All formulas act on the Riemann sphere, including zero and infinity. The scaling parameter must be nonzero for that factor to be a Möbius transformation.

For three distinct finite points, define the cross-ratio normalizationIt sends $z_1,z_2,z_3$ to $0,1,\infty$ respectively. If $z_3=\infty$, use $(\zeta -z_1)/(z_2-z_1)$; if $z_1=\infty$, use $(z_2-z_3)/(\zeta -z_3)$; if $z_2=\infty$, use $(\zeta -z_1)/(\zeta -z_3)$. These are the corresponding limits of the same formula and cover all possible triples. Form the analogous normalization $F_w$ of the target triple. Thenhas all three specified images, proving existence. For uniqueness, the quotient of any two such maps fixes the three source points. Conjugating by $F_z$ gives a Möbius transformation fixing $0,1,\infty$. Fixing zero and infinity forces it to be $\lambda \zeta$, and fixing one forces $\lambda =1$. The quotient is identity, so the two maps are equal.

The Möbius transformations permuting three points are therefore in one-to-one correspondence with permutations of the triple. For the normalized set the six maps and their ordered images of $(0,1,\infty )$ areThey realize all six permutations. The action map to the symmetric group is a homomorphism, is injective by the three-point uniqueness proof, and is onto by this list. Hence