For $f, g \in PGL(2, \mathbb{C})$, $fg=gf$ if and only if one of three conditions holds. 
For $f, g \in PGL(2, \mathbb{C})$, $fg=gf$ if and only if at least one
of the following three conditions holds.
i) $f$ or $g$ is identity.
ii) $f$ and $g$ have the same fixed point sets.
iii) $f^2=g^2=1$ with fixed points $x_1,x_2;y_1,y_2$ that satisfies
$(x_1,x_2;y_1,y_2)=-1$.

i) is obvious.
ii) I know that $f,g \in PGL(2,\mathbb{C})$ having fixed points is equivalent to $\widetilde{f} , \widetilde{g}$ (the representatives in $GL(2, \mathbb{C})$ of $f,g$) having some eigenvalues.
But, I am stuck showing for the subcase when $\widetilde{f} , \widetilde{g}$ having only one eigenvector, which would make them nondiagonalizable. How do I show that if ii), then $fg=gf$?
iii) I think the cross-ratio means a lot for fixed points, but I can't clearly see how they are related. What does being harmonic of collinear pairs of points tell us?
Lastly, to wrap up the only if direction, how do we claim that $fg=gf$ is only possible when i), ii) or iii)? Can it be done by contrapositive statement?
 A: Commuting transformations must at least permute each other's fixed point sets. If $x$ is a fixed point of $f$ for example, then $f(gx)=g(fx)=gx$ so $gx$ is also a fixed point, so $g$ permutes $f$'s fixed points.
Conjugation affects fixed point sets in a predictable way. In particular, if $\{x_1,\cdots\}$ is the fixed point set of $f$ then $\{ hx_1,\cdots\}$ is the fixed point set of the conjugate $hfh^{-1}$. Conjugation also preserves commutaativity: $f$ and $g$ commute iff $hfh^{-1}$ and $hgh^{-1}$ commute.

All nontrivial elements of $PGL_2\Bbb C$ have exactly one or two fixed points. If you set $\frac{az+b}{cz+d}=z$, either the equation is an identity (e.g. $z=z$) so the transformation is the identity, or the equation is a contradiction (e.g. $z+b=z$ with $b\ne0$) and the only solution is $\infty$, or the equation is linear (e.g. $az+b=z$ with $a\ne1$) so there is one finite fixed point and $\infty$ is the other fixed point, or the equation is quadratic (after clearing denominators) so there are two finite fixed points (use the quadratic formula).
$PGL_2\Bbb C$ acts sharply $3$-transitively: given any three distinct points $(a_1,a_2,a_3)$ on the Riemann sphere / extended complex plane / complex projective line and any other three distinct points $(b_1,b_2,b_3)$, there is exactly one Mobius transformation $f$ with $b_i=f(a_i)$ for $i=1,2,3$. To show existence, it suffices to show it is possible for an $f$ to turn any three distinct points $a_1,a_2,a_3$ into $0,\infty,1$, because if you could do that then you could also find a $g$ to turn $b_1,b_2,b_3$ into $0,\infty,1$ too and then $g^{-1}f$ would turn $a_1,a_2,a_3$ into $b_1,b_2,b_3$.
This can be achieved with the cross-ratio: $\displaystyle\frac{z-a_1}{z-a_2}\frac{a_3-a_2}{a_3-a_1}$ sends $a_1,a_2,a_3$ to $0,\infty,1$. If two transformations $f$ and $g$ moved three distinct points to the same three other distinct points (in the correct order), then $g^{-1}f$ would have at least three fixed points, forcing it to be the identity transformation, i.e. $f=g$.
So $(z,a_1,a_2,a_3)$ (as I've written it; there are other conventions) is the unique transformation of $z$ sending $a_1,a_2,a_3$ to $0,\infty,1$. That would mean $(fz,fa_1,fa_2,fa_3)$ does the same, so it must be the same transformation. Thus, the cross-ratio is invariant.

Suppose $f$ and $g$ commute. They share either two, one, or no fixed points.
If they share two fixed points, they have the same fixed point set.
Suppose they share one fixed point $x$, and one of them (say $g$) has a second fixed point $y$. Since $f$ permutes $g$'s fixed point set $\{x,y\}$ and fixes $x$, it must also fix $y$. Thus, if the two functions share exactly one fixed point, neither can have a second fixed point. Thus, they have the same fixed point set.
Suppose they share no fixed points in common. If either of them had only one fixed point, it would also be a fixed point of the other, therefore they must each have two fixed points. Note $f^2$ fixes both $f$'s fixed points and $g$'s fixed points (because $f$ swaps $g$'s fixed points) so $f^2=\mathrm{id}$, and similarly for $g$. By conjugating as appropriate, wlog $f$ has fixed point set $\{0,\infty\}$ so it is of the form $f(w)=aw$ for some $a\in\mathbb{C}^\times$. The condition $f^2=\mathrm{id}$ implies $a=-1$ (note $a=1$ is impossible since $\mathrm{id}$ does in fact share fixed points in common with $g$). Thus, the fixed point set of $g$ is $\{y,-y\}$ for some $y\ne0$, and then the cross ratio is $-1$.

Conversely, suppose $f$ and $g$ have the same fixed point set. Either it is one fixed point, wlog $\infty$ so they're both translations which commute. Or it is two fixed points, wlog $\{0,\infty\}$ so they're both of the form $f(w)=aw$ so they commute.
Or suppose $f^2=g^2=1$ and their fixed points (each have two, none in common) have cross ratio $-1$. By conjugating as appropriate we can say $f$'s fixed point set $\{0,\infty\}$, and the cross-ratio tells us $g$'s fixed point set is $\{y,-y\}$. By conjugating by a function of the form $h(w)=aw$ we can preserve $f$'s fixed point set and make $g$'s fixed point set $\{+1,-1\}$. Under stereographic projection to the Riemann sphere, we see $f$ and $g$ are $180^\circ$ rotations around perpendicular axes in 3D, so they commute.
