PGL(V) Sharply 3-transitive iff n=2

Proposition

Let $G={\mathrm{PGL}}_n(q)$, and let $V={\mathbb{F}}_q^n$. Then $G$ is sharply $3$-transitive on projective line iff $n=2$.

\begin{proof} Note that $G$ must send $3$ collinear points to $3$ other collinear points. So when $n⩾3$ $G$ is not $3$-transitive.

When $n=2$, any two elements $(v_1,v_2,v_3)$ and $(w_1,w_2,w_3)$ can be mapped to $(e_1,e_2,f_1)$ and $(e_1,e_2,f_2)$ by $g_1$ and $g_2$, respectively. Since $f_i=a_i{e}_1+b_i{e}_2$ for $a_i,b_i\in {\mathbb{F}}_q$, there is a $g=\left(\begin{array}{cc}{a}_2/a_1& \\ & b_2/b_1\end{array}\right)$ such that $\overline{g}\in G$ satisfying $(e_1,e_2,f_1{)}^{\overline{g}}=(e_1,e_2,f_2)$. And $g_{1}g{g}_2^{-1}$ is the unique such map that does so. \end{proof}