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.

First it is necessary to establish the Cayley table for $\left({S, \circ}\right)$.

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

Hence the Cayley table is established:

Taking the group axioms in turn:

G0: Closure
By inspection it can be seen that $\left({S, \circ}\right)$ is closed.

G1: Associativity
From Composition of Mappings is Associative, it follows that $\left({S, \circ}\right)$ is associative.

G2: Identity
We have that:
 * $\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$

Thus $\left({S, \circ}\right)$ has an identity element.

G3: Inverses
From the above analysis:

Thus every element of $\left({S, \circ}\right)$ has an inverse.

All the group axioms are thus seen to be fulfilled, and so $\left({S, \circ}\right)$ is a group.

$\left({S, \circ}\right)$ has $6$ elements and so is a finite group of order $6$.