Group Generated by Reciprocal of z and 1 minus z/Cayley Table

Cayley Table for Group Generated by Reciprocal of $1 / z$ and $1 - z$
We have:
 * $f_1 \left({z}\right) = z$


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


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


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


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


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

Hence from Group Generated by Reciprocal of z and 1 minus z:
 * $\begin{array}{r|rrrrrr}

\circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ f_2 & f_2 & f_3 & f_1 & f_6 & f_4 & f_5 \\ f_3 & f_3 & f_1 & f_2 & f_5 & f_6 & f_4 \\ f_4 & f_4 & f_5 & f_6 & f_1 & f_2 & f_3 \\ f_5 & f_5 & f_6 & f_4 & f_3 & f_1 & f_2 \\ f_6 & f_6 & f_4 & f_5 & f_2 & f_3 & f_1 \\ \end{array}$

Expressing the elements in full:


 * $\begin{array}{c|cccccc}

\circ & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \hline z & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \dfrac 1 {1 - z} & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z & \dfrac z {z - 1} & \dfrac 1 z & 1 - z \\ \dfrac {z - 1} z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} & 1 - z & \dfrac z {z - 1} & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z \\ 1 - z & 1 - z & \dfrac z {z - 1} & \dfrac 1 z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} \\ \dfrac z {z - 1} & \dfrac z {z - 1} & \dfrac 1 z & 1 - z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z \\ \end{array}$