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:

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:

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:

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.