Editing Möbius transformation (section)

==== Mapping first to 0, 1, {{math|∞}} ====
It is easy to check that the Möbius transformation
<math display="block">f_1(z)= \frac {(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}</math>
with matrix
<math display="block">\mathfrak{H}_1 = \begin{pmatrix}
z_2 - z_3 & -z_1 (z_2 - z_3)\\
z_2-z_1 & -z_3 (z_2-z_1)
\end{pmatrix}</math>
maps <math>z_1,z_2 \text{ and } z_3</math> to {{tmath|1= 0,1,\ \text{and}\ \infty}}, respectively. If one of the ''<math>z_j</math>'' is <math>\infty</math>, then the proper formula for <math>\mathfrak{H}_1</math> is obtained from the above one by first dividing all entries by ''<math>z_j</math>'' and then taking the limit {{tmath|1= z_j\to\infty }}.

If <math>\mathfrak{H}_2</math> is similarly defined to map <math>w_1,w_2,w_3</math> to <math>0,1,\ \text{and}\ \infty,</math> then the matrix <math>\mathfrak{H}</math> which maps <math>z_{1,2,3}</math> to <math>w_{1,2,3}</math> becomes
<math display="block">\mathfrak{H} = \mathfrak{H}_2^{-1} \mathfrak{H}_1.</math>

The stabilizer of <math>\{0,1,\infty\}</math> (as an unordered set) is a subgroup known as the [[anharmonic group]].