Affine Group of One Dimension as Semidirect Product

Theorem
Let $\operatorname{Af}_1 \left({\R}\right)$ be the $1$-dimensional affine group on $\R$.

Let $\R^+$ be the additive group of real numbers.

Let $\R^\times$ be the multiplicative group of real numbers.

Let $\phi: \R^\times \to \operatorname{Aut} \left({\R^+}\right)$ be defined as:
 * $\forall b \in \R^\times: \phi \left({b}\right) = \left({a \mapsto a b}\right)$

Let $\R^+ \rtimes_\phi \R^\times$ be the corresponding semidirect product.

Then:
 * $\operatorname{Af}_1 \left({\R}\right) \cong \R^+ \rtimes_\phi \R^\times$

where $\cong$ denotes (group) isomorphism.

Proof
By definition, a (group) isomorphism is a (group) homomorphism which is a bijection.

Recall the definition of underlying set of $1$-dimensional affine group on $\R$:


 * $S = \left\{ {f_{a b}: x \mapsto a x + b : a \in \R_{\ne 0}, b \in \R}\right\}$

So the bijection $\psi: \operatorname {Af}_1 \left({\R}\right) \to \R^+ \rtimes_\phi \R^\times$ defined by $\psi \left({f_{a b} }\right) = \left({b, a}\right)$ arises naturally.

It remains to show that $\psi$ is a (group) homomorphism:

Let $f_{a b}, f_{c d} \in \operatorname{Af}_1 \left({\R}\right)$.

Then:

Let $\left({b, a}\right), \left({d, c}\right) \in \R^+ \rtimes_\phi \R^\times$.

Then:

So $\psi \left({f_{a b} }\right) \psi \left({f_{c d} }\right) = \psi \left({f_{a b} \circ f_{c d} }\right)$.

So the bijection $\psi$ is a (group) homomorphism, and thus a (group) isomorphism.

Also see

 * Affine Group as Semidirect Product