Group Generated by Reciprocal of z and 1 minus z
Theorem
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:
| \(\ds \map {f_1} z\) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \map {f_2} z\) | \(=\) | \(\ds \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \map {f_3} z\) | \(=\) | \(\ds \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \map {f_4} z\) | \(=\) | \(\ds \dfrac 1 z\) | ||||||||||||
| \(\ds \map {f_5} z\) | \(=\) | \(\ds 1 - z\) | ||||||||||||
| \(\ds \map {f_6} z\) | \(=\) | \(\ds \dfrac z {z - 1}\) |
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
| \(\ds f_2 \circ f_2\) | \(=\) | \(\ds \dfrac 1 {1 - \dfrac 1 {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {1 - z} {1 - z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_3\) |
| \(\ds f_2 \circ f_3\) | \(=\) | \(\ds \dfrac 1 {1 - \dfrac {z - 1} z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - \paren {z - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_1\) |
| \(\ds f_2 \circ f_4\) | \(=\) | \(\ds \dfrac 1 {1 - \dfrac 1 z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_6\) |
| \(\ds f_2 \circ f_5\) | \(=\) | \(\ds \dfrac 1 {1 - \paren {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_4\) |
| \(\ds f_2 \circ f_6\) | \(=\) | \(\ds \dfrac 1 {1 - \dfrac z {z - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} {z - 1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_5\) |
| \(\ds f_3 \circ f_2\) | \(=\) | \(\ds \dfrac {\dfrac 1 {1 - z} - 1} {\dfrac 1 {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {1 - z} \dfrac {1 - \paren {1 - z} } {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_1\) |
| \(\ds f_3 \circ f_3\) | \(=\) | \(\ds \dfrac {\dfrac {z - 1} z - 1} {\dfrac {z - 1} z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1 - z} z \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_2\) |
| \(\ds f_3 \circ f_4\) | \(=\) | \(\ds \dfrac {\dfrac 1 z - 1} {\dfrac 1 z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z \dfrac {1 - z} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_5\) |
| \(\ds f_3 \circ f_5\) | \(=\) | \(\ds \dfrac {\paren {1 - z} - 1} {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_6\) |
| \(\ds f_3 \circ f_6\) | \(=\) | \(\ds \dfrac {\dfrac z {z - 1} - 1} {\dfrac z {z - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - \paren {z - 1} } {z - 1} \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_4\) |
| \(\ds f_4 \circ f_2\) | \(=\) | \(\ds \dfrac 1 {\dfrac 1 {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_5\) |
| \(\ds f_4 \circ f_3\) | \(=\) | \(\ds \dfrac 1 {\dfrac {z - 1} z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_6\) |
| \(\ds f_4 \circ f_4\) | \(=\) | \(\ds \dfrac 1 {1 / z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_1\) |
| \(\ds f_4 \circ f_5\) | \(=\) | \(\ds \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_2\) |
| \(\ds f_4 \circ f_6\) | \(=\) | \(\ds \dfrac 1 {\dfrac z {z - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_3\) |
| \(\ds f_5 \circ f_2\) | \(=\) | \(\ds 1 - \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {1 - z - 1} {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_6\) |
| \(\ds f_5 \circ f_3\) | \(=\) | \(\ds 1 - \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - \paren {z - 1} } z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_4\) |
| \(\ds f_5 \circ f_4\) | \(=\) | \(\ds 1 - \dfrac 1 z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_3\) |
| \(\ds f_5 \circ f_5\) | \(=\) | \(\ds 1 - \paren {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_1\) |
| \(\ds f_5 \circ f_6\) | \(=\) | \(\ds 1 - \dfrac z {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1 - z} {z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_2\) |
| \(\ds f_6 \circ f_2\) | \(=\) | \(\ds \dfrac {\dfrac 1 {1 - z} } {\dfrac 1 {1 - z} - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - z} \dfrac {1 - z} {1 - \paren {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_4\) |
| \(\ds f_6 \circ f_3\) | \(=\) | \(\ds \dfrac {\dfrac {z - 1} z} {\dfrac {z - 1} z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} z \dfrac z {z - 1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_5\) |
| \(\ds f_6 \circ f_4\) | \(=\) | \(\ds \dfrac {\dfrac 1 z} {\dfrac 1 z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 z \dfrac z {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_2\) |
| \(\ds f_6 \circ f_5\) | \(=\) | \(\ds \dfrac {1 - z} {1 - z - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z - 1} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_3\) |
| \(\ds f_6 \circ f_6\) | \(=\) | \(\ds \dfrac {\dfrac z {z - 1} } {\dfrac z {z - 1} - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac z {z - 1} \dfrac {z - 1} {z - 1 - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds f_1\) |
Hence the Cayley table is established:
- $\begin{array}{r|rrrrrr}
\circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ f_2 & f_2 & f_3 & f_1 & f_6 & f_4 & f_5 \\ f_3 & f_3 & f_1 & f_2 & f_5 & f_6 & f_4 \\ f_4 & f_4 & f_5 & f_6 & f_1 & f_2 & f_3 \\ f_5 & f_5 & f_6 & f_4 & f_3 & f_1 & f_2 \\ f_6 & f_6 & f_4 & f_5 & f_2 & f_3 & f_1 \\ \end{array}$
Expressing the elements in full:
- $\begin{array}{c|cccccc}
\circ & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \hline z & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \dfrac 1 {1 - z} & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z & \dfrac z {z - 1} & \dfrac 1 z & 1 - z \\ \dfrac {z - 1} z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} & 1 - z & \dfrac z {z - 1} & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z \\ 1 - z & 1 - z & \dfrac z {z - 1} & \dfrac 1 z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} \\ \dfrac z {z - 1} & \dfrac z {z - 1} & \dfrac 1 z & 1 - z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z \\ \end{array}$
Taking the group axioms in turn:
Group Axiom $\text G 0$: Closure
By inspection it can be seen that $\struct {S, \circ}$ is closed.
$\Box$
Group Axiom $\text G 1$: Associativity
From Composition of Mappings is Associative, it follows that $\struct {S, \circ}$ is associative.
$\Box$
Group Axiom $\text G 2$: Existence of Identity Element
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.
$\Box$
Group Axiom $\text G 3$: Existence of Inverse Element
From the above analysis:
| \(\ds f_1 \circ f_1\) | \(=\) | \(\ds f_1\) | ||||||||||||
| \(\ds f_2 \circ f_3\) | \(=\) | \(\ds f_1\) | ||||||||||||
| \(\ds f_3 \circ f_2\) | \(=\) | \(\ds f_1\) | ||||||||||||
| \(\ds f_4 \circ f_4\) | \(=\) | \(\ds f_1\) | ||||||||||||
| \(\ds f_5 \circ f_5\) | \(=\) | \(\ds f_1\) | ||||||||||||
| \(\ds f_6 \circ f_6\) | \(=\) | \(\ds f_1\) |
Thus every element of $\struct {S, \circ}$ has an inverse.
$\Box$
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 $6$.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(ii)}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Introduction