So I'm trying to prove a mobius function forms a group under composition. I know this has been asked (3 years ago) here: Proving the set $\mathcal H$ of Möbius transformations is a group under composition and finding a transformation that satisfies certain conditions
But the part that I'm stuck on is a part that was skipped. I am having trouble proving $M\,o\,M^{-1}(z)=z$. Here is what I have so far, skipping the introduction to the proof and such because I am already aware of how to prove identity, associativity and closure.
Let $M={az+b\over cz+d}$ and $M^{-1}={dz-b\over -cz+a}$ due to finding inverses via matrices. Now, from here I believe I need to show that $M\,o\,M^{-1}(z)=z$ and vice versa.
$M\,o\,M^{-1}(z)={a{dz-b\over -cz+a}+b\over c{dz-b\over -cz+a}+d}$ = ${{adz-ab\over -cz+a}+b}\over{{-cdz+cb\over -cz+a}+a}$ and then it becomes a bit of an issue. I simplified it down to $adz-bcz\over -czd+cb-acz+a^2$ and I'm pretty sure I did that right, but maybe I didn't because I can't seem to simplify this anymore.
Multiply the fraction that by $-cz + a$.
$MM^{-1}(z)=\frac{(adz - ab) + b(-cz + a)}{(-cdz+cb) + a(-cz + a)} = \frac{(ad - bc)z}{ad -bc}$