Automorphism group of the complex plane

The automorphism group of the complex plane is $$\text{Aut}(\Bbb C)=\{\text{analytic bijections}\;f:\Bbb C\to \Bbb C\}.$$ Such automorphisms have the form $$f(z)=az+b,\quad a,b\in \Bbb C, a\neq 0.$$ (To be proved later) If $f(z)=az+b,g(z)=a'z+b'$, then $$f\circ g(z)=aa'z+(ab'+b)\\f^{-1}(z)=a^{-1}z-a^{-1}b.$$ Expressed in matrix form, we have $$\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}a'&b'\\0&1\end{pmatrix}=\begin{pmatrix}aa'&ab'+b\\0&1\end{pmatrix}\\ \begin{pmatrix}a&b\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}a^{-1}&-a^{-1}b\\0&1\end{pmatrix},$$ which encode the formulas for composing and inverting automophisms of the plane.

The correspondence $$f(z)=az+b \longleftrightarrow \begin{pmatrix}a&b\\0&1\end{pmatrix}$$ is a natural group isomorphism. We now define the parabolic group as $$P=\bigg\{\begin{pmatrix}a&b\\0&1\end{pmatrix}\bigg|\; a,b\in \Bbb C, a\neq 0\bigg\}.$$ Thus $$\text{Aut}(C)\cong P.$$ Two subgroups of $P$ are its Levi component $$M=\bigg\{\begin{pmatrix}a&0\\0&1\end{pmatrix}\bigg|\; a\in \Bbb C, a\neq 0\bigg\},$$ describing the dilations $f(z)=az$, and its unipotent radical $$N=\bigg\{\begin{pmatrix}1&b\\0&1\end{pmatrix}\bigg|\; b\in \Bbb C\bigg\},$$ describing the translations $f(z)=z+b$. The parabolic group takes the form $$P=MN=NM$$ since $$\begin{pmatrix}a&b\\0&1\end{pmatrix}=\begin{pmatrix}a&0\\0&1\end{pmatrix}\begin{pmatrix}1&\color{blue}{a^{-1}b}\\0&1\end{pmatrix}=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}.$$ Geometrically, any affine map is the composition of a translation and a dilation, in which the order of translation and dilation does not matter. Also, $M$ normalizes $N$, meaning that $$m^{-1}nm\in N\quad \forall m\in M, n\in N$$ since $$\begin{pmatrix}a^{-1}&0\\0&1\end{pmatrix}\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}=\begin{pmatrix}1&a^{-1}b\\0&1\end{pmatrix}.$$ Geometrically, it says that a dilation followed by a translation followed by the reciprocal dilation is again a translation. On the other hand, $N$ does not normalize $M$.