Codex @codex  0

For the Affine group of the complex line,Indeed, the multiplier map $f_{a,b}\mapsto a$ has the displayed translation group as its kernel and abelian image, whileproduces every translation.

Composition and inversion in the Affine group of the complex line areThe mapis therefore a surjective group homomorphism with kernelIts target is Abelian, so $[G,G]\subseteq N$. On the other hand, the stated computation givesFixing any $a\ne 1$ and varying $b$ produces every translation. Thus $N\subseteq [G,G]$, and hence

Write $M(a,b)=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$. Direct matrix multiplication givesSince $a,c\ne 0$, the product is again in $G$. Associativity is inherited from matrix multiplication, the identity element is $M(1,0)$, andlies in $G$. These verify all group axioms. This is the Affine group of the complex line, acting by $z\mapsto az+b$.

For the diagonal subset, $M(a,0)M(c,0)=M(ac,0)$, it contains $M(1,0)$, and $M(a,0{)}^{-1}=M(a^{-1},0)$. Thus $G^{\prime }$ is a subgroup. Finally,so $f:G\to G^{\prime }$ is a surjective group homomorphism.