Theorem: Group Generated by Reciprocal of z and 1 minus z

Let $\struct {S, \circ}$ denote the '''group generated by $\dfrac 1 z$ and $1 - z$'''.
Then $\struct {S, \circ}$ is a finite group of order $6$.

Proof:
By definition:
:$S = \set {f_1, f_2, f_3, f_4, f_5, f_6}$
where $f_1, f_2, \ldots, f_6$ are complex functions defined for all $z \in \C \setminus \set {0, 1}$ as:
{{begin-eqn}}
{{eqn | l = \map {f_1} z
      | r = z
}}
{{eqn | l = \map {f_2} z
      | r = \dfrac 1 {1 - z}
}}
{{eqn | l = \map {f_3} z
      | r = \dfrac {z - 1} z
}}
{{eqn | l = \map {f_4} z
      | r = \dfrac 1 z
}}
{{eqn | l = \map {f_5} z
      | r = 1 - z
}}
{{eqn | l = \map {f_6} z
      | r = \dfrac z {z - 1}
}}
{{end-eqn}}
Also by definition, $\circ$ denotes composition of functions.
First it is necessary to establish the Cayley table for $\struct {S, \circ}$.
First note that it is apparent by definition of composition of functions that:
:$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$
It remains to establish the rest of the compositions
{{begin-eqn}}
{{eqn | l = f_2 \circ f_2
      | r = \dfrac 1 {1 - \dfrac 1 {1 - z} }
}}
{{eqn | r = \dfrac {1 - z} {1 - z - 1}
}}
{{eqn | r = \dfrac {z - 1} z
}}
{{eqn | r = f_3
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_2 \circ f_3
      | r = \dfrac 1 {1 - \dfrac {z - 1} z}
}}
{{eqn | r = \dfrac z {z - \paren {z - 1} }
}}
{{eqn | r = z
}}
{{eqn | r = f_1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_2 \circ f_4
      | r = \dfrac 1 {1 - \dfrac 1 z}
}}
{{eqn | r = \dfrac z {z - 1}
}}
{{eqn | r = f_6
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_2 \circ f_5
      | r = \dfrac 1 {1 - \paren {1 - z} }
}}
{{eqn | r = \dfrac 1 z
}}
{{eqn | r = f_4
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_2 \circ f_6
      | r = \dfrac 1 {1 - \dfrac z {z - 1} }
}}
{{eqn | r = \dfrac {z - 1} {z - 1 - z}
}}
{{eqn | r = 1 - z
}}
{{eqn | r = f_5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_3 \circ f_2
      | r = \dfrac {\dfrac 1 {1 - z} - 1} {\dfrac 1 {1 - z} }
}}
{{eqn | r = \paren {1 - z} \dfrac {1 - \paren {1 - z} } {1 - z}
}}
{{eqn | r = z
}}
{{eqn | r = f_1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_3 \circ f_3
      | r = \dfrac {\dfrac {z - 1} z - 1} {\dfrac {z - 1} z}
}}
{{eqn | r = \dfrac {z - 1 - z} z \dfrac z {z - 1}
}}
{{eqn | r = \dfrac 1 {1 - z}
}}
{{eqn | r = f_2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_3 \circ f_4
      | r = \dfrac {\dfrac 1 z - 1} {\dfrac 1 z}
}}
{{eqn | r = z \dfrac {1 - z} z
}}
{{eqn | r = 1 - z
}}
{{eqn | r = f_5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_3 \circ f_5
      | r = \dfrac {\paren {1 - z} - 1} {1 - z}
}}
{{eqn | r = \dfrac z {z - 1}
}}
{{eqn | r = f_6
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_3 \circ f_6
      | r = \dfrac {\dfrac z {z - 1} - 1} {\dfrac z {z - 1} }
}}
{{eqn | r = \dfrac {z - \paren {z - 1} } {z - 1} \dfrac {z - 1} z
}}
{{eqn | r = \dfrac 1 z
}}
{{eqn | r = f_4
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_4 \circ f_2
      | r = \dfrac 1 {\dfrac 1 {1 - z} }
}}
{{eqn | r = 1 - z
}}
{{eqn | r = f_5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_4 \circ f_3
      | r = \dfrac 1 {\dfrac {z - 1} z}
}}
{{eqn | r = \dfrac z {z - 1}
}}
{{eqn | r = f_6
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_4 \circ f_4
      | r = \dfrac 1 {1 / z}
}}
{{eqn | r = z
}}
{{eqn | r = f_1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_4 \circ f_5
      | r = \dfrac 1 {1 - z}
}}
{{eqn | r = f_2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_4 \circ f_6
      | r = \dfrac 1 {\dfrac z {z - 1} }
}}
{{eqn | r = \dfrac {z - 1} z
}}
{{eqn | r = f_3
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_5 \circ f_2
      | r = 1 - \dfrac 1 {1 - z}
}}
{{eqn | r = \dfrac {1 - z - 1} {1 - z}
}}
{{eqn | r = \dfrac z {z - 1}
}}
{{eqn | r = f_6
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_5 \circ f_3
      | r = 1 - \dfrac {z - 1} z
}}
{{eqn | r = \dfrac {z - \paren {z - 1} } z
}}
{{eqn | r = \dfrac 1 z
}}
{{eqn | r = f_4
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_5 \circ f_4
      | r = 1 - \dfrac 1 z
}}
{{eqn | r = \dfrac {z - 1} z
}}
{{eqn | r = f_3
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_5 \circ f_5
      | r = 1 - \paren {1 - z}
}}
{{eqn | r = z
}}
{{eqn | r = f_1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_5 \circ f_6
      | r = 1 - \dfrac z {z - 1}
}}
{{eqn | r = \dfrac {z - 1 - z} {z - 1}
}}
{{eqn | r = \dfrac 1 {1 - z}
}}
{{eqn | r = f_2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_6 \circ f_2
      | r = \dfrac {\dfrac 1 {1 - z} } {\dfrac 1 {1 - z} - 1}
}}
{{eqn | r = \dfrac 1 {1 - z} \dfrac {1 - z} {1 - \paren {1 - z} }
}}
{{eqn | r = \dfrac 1 z
}}
{{eqn | r = f_4
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_6 \circ f_3
      | r = \dfrac {\dfrac {z - 1} z} {\dfrac {z - 1} z - 1}
}}
{{eqn | r = \dfrac {z - 1} z \dfrac z {z - 1 - z}
}}
{{eqn | r = 1 - z
}}
{{eqn | r = f_5
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_6 \circ f_4
      | r = \dfrac {\dfrac 1 z} {\dfrac 1 z - 1}
}}
{{eqn | r = \dfrac 1 z \dfrac z {1 - z}
}}
{{eqn | r = \dfrac 1 {1 - z}
}}
{{eqn | r = f_2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_6 \circ f_5
      | r = \dfrac {1 - z} {1 - z - 1}
}}
{{eqn | r = \dfrac {z - 1} z
}}
{{eqn | r = f_3
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = f_6 \circ f_6
      | r = \dfrac {\dfrac z {z - 1} } {\dfrac z {z - 1} - 1}
}}
{{eqn | r = \dfrac z {z - 1} \dfrac {z - 1} {z - 1 - z}
}}
{{eqn | r = z
}}
{{eqn | r = f_1
}}
{{end-eqn}}
Hence the Cayley table is established:
{{:Group Generated by Reciprocal of z and 1 minus z/Cayley Table}}
Taking the group axioms in turn: