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: === $\text G 0$: Closure === By inspection it can be seen that $\struct {S, \circ}$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === From Composition of Mappings is Associative, it follows that $\struct {S, \circ}$ is associative. {{qed|lemma}} === $\text G 2$: Identity === We have that: :$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$ Thus $\struct {S, \circ}$ has an identity element. {{qed|lemma}} === $\text G 3$: Inverses === From the above analysis: {{begin-eqn}} {{eqn | l = f_1 \circ f_1       | r = f_1 }} {{eqn | l = f_2 \circ f_3       | r = f_1 }} {{eqn | l = f_3 \circ f_2       | r = f_1 }} {{eqn | l = f_4 \circ f_4       | r = f_1 }} {{eqn | l = f_5 \circ f_5       | r = f_1 }} {{eqn | l = f_6 \circ f_6       | r = f_1 }} {{end-eqn}} Thus every element of $\struct {S, \circ}$ has an inverse. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group. $\struct {S, \circ}$ has $6$ elements and so is a finite group of order $6$. {{qed}}