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.