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

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Cayley Table for Group Generated by Reciprocal of $1 / z$ and $1 - z$

We have:

\(\ds \map {f_1} z\) \(=\) \(\ds z\)
\(\ds \map {f_2} z\) \(=\) \(\ds \dfrac 1 {1 - z}\)
\(\ds \map {f_3} z\) \(=\) \(\ds \dfrac {z - 1} z\)
\(\ds \map {f_4} z\) \(=\) \(\ds \dfrac 1 z\)
\(\ds \map {f_5} z\) \(=\) \(\ds 1 - z\)
\(\ds \map {f_6} z\) \(=\) \(\ds \dfrac z {z - 1}\)


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}$


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