Set of Affine Mappings on Real Line under Composition forms Group
Theorem
Let $S$ be the set of all real functions $f: \R \to \R$ of the form:
- $\forall x \in \R: \map f x = r x + s$
where $r \in \R_{\ne 0}$ and $s \in \R$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
Proof
We note that $S$ is a subset of the set of all real functions on $\R$.
From Set of all Self-Maps under Composition forms Semigroup, we have that $\circ$ is associative.
Consider the real function $I: \R \to \R$ defined as:
- $\forall x \in \R: \map I x = 1 \times x + 0$
We have that:
- $I \in S$
- $I$ is the identity mapping.
So $S$ is not empty.
Then we note:
Let $f, g \in S$ such that:
- $\map f x = r_1 x + s_1$
- $\map g x = r_2 x + s_2$
For all $x \in \R$, we have:
\(\ds \map {g \circ f} x\) | \(=\) | \(\ds \map g {\map f x}\) | Definition of Composition of Mappings | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g {r_1 x + s_1}\) | Definition of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds r_2 {r_1 x + s_1} + s_2\) | Definition of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {r_2 r_1} x + \paren {r_2 s_1 + s_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds r x + s\) | where $r = r_2 r_2$ and $s = r_2 s_1 + s_2$ |
This demonstrates that $\struct {S, \circ}$ is closed.
Let $\phi: \R \to \R$ be the real function defined as:
- $\forall x \in \R: \map \phi x = p x + q$
where $p \in \R_{\ne 0}$ and $q \in \R$.
Then:
\(\ds y\) | \(=\) | \(\ds p x + q\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {y - q} p\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac 1 p y + \paren {-\dfrac q p}\) |
Thus we define $\psi: \R \to \R$ be the real function defined as:
- $\forall y \in \R: \map \psi y = \dfrac 1 p y + \paren {-\dfrac q p}$
where $\dfrac 1 p \in \R_{\ne 0}$ and $-\dfrac q p \in \R$.
It is seen that $\psi \in S$.
For all $x \in \R$, we have:
\(\ds \map {\phi \circ \psi} x\) | \(=\) | \(\ds \map \phi {\map \psi x}\) | Definition of Composition of Mappings | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\dfrac 1 p x - \dfrac q p}\) | Definition of $\psi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds p {\dfrac 1 p x -\dfrac q p} + q\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac p p x - \dfrac {p q} p + q\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
Hence:
- $\phi \circ \psi = I$
and so $\psi$ is the inverse of $\phi$.
We have then that $\struct {S, \circ}$ is closed under inversion.
The result follows by the Two-Step Subgroup Test.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{B iv}$