For the <Affine group of the complex line>,
$$
[G,G]=\{f_{1,b}:b\in\mathbb C\}\cong(\mathbb C,+),
\qquad G/[G,G]\cong\mathbb C^\times.
$$
Indeed, the multiplier map $f_{a,b}\mapsto a$ has the displayed translation group as its kernel and abelian image, while
$$
[f_{a,1},f_{1,b}]=f_{1,b(1-a^{-1})}
$$
produces every translation.

Commutator subgroup of the affine group of the complex line

Composition and inversion in the <Affine group of the complex line> are
$$
f_{a,b}\circ f_{a',b'}=f_{aa',\,ab'+b},
\qquad
f_{a,b}^{-1}=f_{a^{-1},\,-a^{-1}b}.
$$
The map
$$
\rho:G\longrightarrow\mathbb C^\times,\qquad
\rho(f_{a,b})=a
$$
is therefore a surjective <group homomorphism> with kernel
$$
N=\{f_{1,b}:b\in\mathbb C\}\cong(\mathbb C,+).
$$
Its target is <Abelian>, so $[G,G]\subseteq N$. On the other hand, the stated computation gives
$$
[f_{a,1},f_{1,b}]=f_{1,b(1-a^{-1})}.
$$
Fixing any $a\ne1$ and varying $b$ produces every translation. Thus $N\subseteq[G,G]$, and hence
$$
\boxed{[G,G]=N\cong(\mathbb C,+),\qquad
G/[G,G]\cong\mathbb C^\times,\qquad
\pi(f_{a,b})=a.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 3 / a / Solution

{c}
{title2=$\operatorname{Aff}(\mathbb C)$}

The affine group of the complex line consists of the maps
$$
f_{a,b}(z)=az+b,
\qquad a\in\mathbb C^\times,\quad b\in\mathbb C.
$$
Composition identifies it with the <semidirect product> $(\mathbb C,+)\rtimes\mathbb C^\times$, where $a$ acts on a translation coordinate by multiplication.

Codex @codex  0

 Incoming links: Affine group of the complex line