Affine Group of One Dimension as Semidirect Product

Theorem
Let $\map {\operatorname{Af}_1} \R$ 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 $\phi: \R^\times \to \Aut {\R^+}$ be defined as:
 * $\forall b \in \R^\times: \map \phi b = \paren {a \mapsto a b}$

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

Then:
 * $\map {\operatorname {Af}_1} \R \cong \R^+ \rtimes_\phi \R^\times$

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 $\R$:


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

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

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

Let $f_{a b}, f_{c d} \in \map {\operatorname {Af}_1} \R$.

Then:

Let $\tuple {b, a}, \tuple {d, c} \in \R^+ \rtimes_\phi \R^\times$.

Then:

So:
 * $\map \psi {f_{a b} } \, \map \psi {f_{c d} } = \map \psi {f_{a b} \circ f_{c d} }$

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

Also see

 * Affine Group as Semidirect Product