Does this question make sense about $P\Gamma L_2(q)$ ?

Question

We know that since $PGL_2(q)$ is a subgroup of $P\Gamma L_2(q)$ and $PGL_2(q)$ is 2-transitive on the projective line, so $P\Gamma L_2(q)$ is 2-transitive also. Now, does it make sense to find the conditions in which $P\Gamma L_2(q)$ is 4-transitive group? We know that the only 4-transitive groups are the symmetric groups $S_n$ for $n>4$, the alternating groups $A_n$ for $n>6$, and the Mathieu groups $M_{24}$, $M_{23}$, $M_{12}$ and the $M_{11}$.

Answer 1

While the classification of $4$-transitive groups does answer this question, it is an extravagantly difficult result to quote for the present purpose. We prove that $q=3$ and $q=4$ are the only cases where ${\text P}\Gamma{\text L}_2(q)$ acts 4-transitively on the projective line over ${\bf F}_q$, simply because once $q>4$ the group is too small to act $4$-transitively on a set of size $q+1$. Indeed a $4$-transitive group of permutations of $q+1$ objects must have at least $(q+1)q(q-1)(q-2)$ elements. But if $q=p^e$ for some prime $p$ then ${\text P}\Gamma{\text L}_2(q)$ has order $(q+1)q(q-1)e$. Therefore $q-2 \leq e$. This clearly cannot happen for $q$ at all large, and in fact we soon see that $q$ can only be $3$ or $4$. Both $q=3$ and $q=4$ yield $4$-transitive actions because in each case ${\text P}\Gamma{\text L}_2(q)$ is the full permutation group on $q+1$ letters, so we're done.