Group Generated by Reciprocal of z and 1 minus z

Theorem
Then $\left({S, \circ}\right)$ denote the group generated by $\dfrac 1 z$ and $1 - z$.

Then $\left({S, \circ}\right)$ is a finite group of order $6$.

Proof
By definition:
 * $S = \left\{ {f_1, f_2, f_3, f_4, f_5, f_6}\right\}$

where $f_1, f_2, \ldots, f_6$ are complex functions defined for all $z \in \C \setminus \left \{ {0, 1}\right\}$ as:


 * $f_1 \left({z}\right) = z$


 * $f_2 \left({z}\right) = \dfrac 1 {1 - z}$


 * $f_3 \left({z}\right) = \dfrac {z - 1} z$


 * $f_4 \left({z}\right) = \dfrac 1 z$


 * $f_5 \left({z}\right) = 1 - z$


 * $f_6 \left({z}\right) = \dfrac z {z - 1}$

Also by definition, $\circ$ denotes composition of functions.