Group/Examples/Linear Functions

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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):


G0: Closure

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 $\circ$.

$\Box$


G1: Associativity

We have that Composition of Mappings is Associative.

$\Box$


G2: Identity

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

\(\displaystyle \theta_{a, b} \circ \theta_{1, 0}\) \(=\) \(\displaystyle \theta_{1 \times a, \, 0 \times a + b}\) $\quad$ Composition of Linear Real Functions $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \theta_{a, b}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \theta_{a \times 1, \, b \times 1 + 0}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \theta_{1, 0} \circ \theta_{a, b}\) $\quad$ Composition of Linear Real Functions $\quad$

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

$\Box$


G3: Inverses

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:

\(\displaystyle \theta_{1 / a, \, -b / a} \circ \theta_{a, b}\) \(=\) \(\displaystyle \theta_{a \times 1 / a, \, b \times 1 / a + \paren {-b / a} }\) $\quad$ Composition of Linear Real Functions $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \theta_{1, 0}\) $\quad$ $\quad$


and:

\(\displaystyle \theta_{a, b} \circ \theta_{1 / a, \, -b / a}\) \(=\) \(\displaystyle \theta_{\paren {1 / a} \times a, \, \paren {-b / a} \times a + b}\) $\quad$ Composition of Linear Real Functions $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \theta_{1, 0}\) $\quad$ $\quad$


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

$\Box$


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

$\Box$


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.

$\blacksquare$


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