Set of Affine Mappings on Real Line under Composition forms Group

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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

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