$PGL_2(q)$ acts on $\Omega$ $3-$transitively? Anyone who studies Permutation Groups will be encountering the following definition:

A group $G$ acting on a set $\Omega$ is said to be “Sharply m-Transitive” iff  $$\forall (a_1,a_2…,a_m) , (b_1,b_2…,b_m) \in \Omega^{m};\   ∃! g \in G  , a_i^g=b_i, 1\leq i\leq m$$

While reviewing my written notes in the class about group $PGL_2(q)$, I have faced to this matter that $PGL_2(q)$ acting on set $\Omega=GF(q)\cup\{\infty\}$ is sharply $3-$transitive. The idea for judging that is as follows:

Since
$PGL_2(q)=\{f|f:\Omega\longrightarrow\Omega, f(z)=\frac{az+b}{cz+d},ad-bc\neq 0; a,b,c,d\in GF(q)\}$ so we have
$PGL_2(q)_{\infty}=\{f|f:\Omega\longrightarrow\Omega, f(z)= az+b ,a\neq 0; a,b\in GF(q)\}$,
$PGL_2(q)_{\infty,0}=\{f|f:\Omega\longrightarrow\Omega, f(z)= az ,a\neq 0; a\in GF(q)\}$ and
$PGL_2(q)_{\infty,0,1}=\{f|f:\Omega\longrightarrow\Omega, f(z)= z\}=\{id\}$.
Knowing that $|PGL_2(q)|=q(q^2-1)$, we get $| PGL_2(q)_{\infty}|=q(q-1)$ and then
$| PGL_2(q)_{\infty,0}|=q-1$.Now, since $|PGL_2(q):PGL_2(q)_{\infty}|=q+1=|\Omega|$ then $PGL_2(q)$ is
acting transitively on $\Omega$; and because of having $|PGL_2(q))_{\infty,0}:PGL_2(q)_{\infty}|=q=|\Omega-\{\infty\}|$
therefore $PGL_2(q)$ is acting $2-$transitively on $\Omega$. Finally, it was concluded in the class that
since $|PGL_2(q)_{\infty,0,1}|=1$; $|PGL_2(q)|$ is acting 3-transitively on $\Omega$. I should confess, I can't
reach to the last result and above definition is hopeless for me. My question is "Why the group is
sharply 3-transitive on $\Omega$. Thanks and sorry for my long question here.

 A: To show that a group is acting sharply $3$-transitively, you need to show that it acts $3$-transitively and that the stabilizer of one (and hence of any) triplet of distinct points is trivial. Since you know that $|PGL_2(q)_{\infty,0,1}|=1$, you need to show just that the action is $3$-transitive. For that, it is sufficient to show that the action of the stabilizer of two points, i.e. $PGL_2(q)_{\infty,0}$, is transitive on $\Omega\setminus\{\infty,0\}$. Indeed, if $x,y\in GF(q)\setminus\{0\}$, then take $a=yx^{-1}$ and you have $f(z)=az$ such that $f(x)=y$. 
A: Just to point out, this is just basic algebraic manipulation.  You solve a few systems of linear equations.  The inclusion of infinity requires a little projective fiddling, but it is just a few extra cases to check.
If $f(z) = \frac{az+b}{cz+d}$ then $f(\infty) = \frac{a}{c}$, $f(0) = \frac{b}{d}$ and $f(1) = \frac{a+b}{c+d}$.  Hence you are solving a system of three linear equations in four variables.
If $x,y,z \neq \infty$, set $a=x$, $b=\frac{xy-yz}{z-y}$, $c=1$, and $d=\frac{x-z}{z-y}$.
If $x=\infty$, set $c=0$ and solve for the rest.  If $y=\infty$, set $d=0$ and solve for the rest. If $z=\infty$, set $c=-d$ and solve for the rest.
