# Affine Group of One Dimension is Group

## Contents

## Theorem

The $1$-dimensional affine group on $\R$ $\operatorname{Af}_1 \left({\R}\right)$ is a group.

## Proof 1

Taking the group axioms in turn:

### G0: Closure

Let $f_{ab}, f_{cd} \in \operatorname{Af}_1 \left({\R}\right)$.

Then:

\(\displaystyle f_{ab} \circ f_{cd} \left({x}\right)\) | \(=\) | \(\displaystyle a \left({c x + d}\right) + b\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a c x + a d + b\) | $\quad$ | $\quad$ |

By the field axioms, $a c \in \R_{\ne 0}$ and $a d + b \in \R$.

Thus $f_{ab} \circ f_{cd} \in \operatorname{Af}_1 \left({\R}\right)$ and so $\operatorname{Af}_1 \left({\R}\right)$ is closed.

$\Box$

### G1: Associativity

From Composition of Mappings is Associative, it follows directly that $\circ$ is associative on $\operatorname{Af}_1 \left({\R}\right)$.

$\Box$

### G2: Identity

By Identity of Affine Group of One Dimension, $\operatorname{Af}_1 \left({\R}\right)$ has $f_{1, 0}$ as an identity element.

$\Box$

### G3: Inverses

By Inverse in Affine Group of One Dimension, every element $f_{a b}$ of $\operatorname{Af}_1 \left({\R}\right)$ has an inverse $f_{c d}$ where $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.

$\Box$

All the group axioms are thus seen to be fulfilled, and so $\operatorname{Af}_1 \left({\R}\right)$ is a group.

$\blacksquare$

## Proof 2

It follows from Affine Group of One Dimension as Semidirect Product and Semidirect Product of Groups is Group.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Exercise $(2)$