Group/Examples/Linear Functions

Theorem
Let $G$ be the set of all real functions $\theta_{a, b}: \R \to \R$ defined as:
 * $\forall x \in \R: \map {\theta_{a, b} } x = a x + b$

where $a, b \in \R$ such that $a \ne 0$.

The algebraic structure $\struct {G, \circ}$, where $\circ$ denotes composition of mappings, is a group.

$\struct {G, \circ}$ is specifically non-abelian.

Proof
We verify the group axioms, in the following order (for convenience):

Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$.

From Composition of Linear Real Functions:
 * $\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$

As $a \ne 0$ and $c \ne 0$ it follows that $a c \ne 0$.

Thus $\theta_{a c, b c + d} \in G$

This proves closure of $\circ$.

We have that Composition of Mappings is Associative.

We assert that $\theta_{1, 0}$ is the identity with respect to $\circ$.

So indeed $\theta_{1, 0}$ is the identity for $\circ$.

For any element $\theta_{a, b} \in S$, we claim that $\theta_{1 / a, \, -b / a}$ is the inverse for $\theta_{a, b}$.

From Inverse of Linear Function on Real Numbers:
 * $\theta^{-1}_{a, b} = \theta_{1 / a, \, -b / a}$

Verifying this:

and:

We conclude that $\theta_{1 / a, \, -b / a}$ is indeed the inverse for $\theta_{a, b}$.

It follows that $\struct {G, \circ}$ is indeed a group.

Non-Abelian
Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$.

From Condition for Composition of Linear Real Functions to be Commutative, it is not true in general that $\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$.

Thus $\struct {G, \circ}$ is non-abelian by definition.