Given proof:
Consider the action of $\operatorname{PGL}_2(\mathbb{F}_q)$ on the projective line $\mathbb{P}^1(\mathbb{F}_q)$. This action is faithful and 3-transitive, so $$ \phi:\operatorname{PGL}_2(\mathbb{F}_q) \to S_{q+1} $$ is a monomorphism (note: $q+1 = |\mathbb{P}^1(\mathbb{F}_q)|$, $S_n$ is the symmetric group). Therefore, $\operatorname{im} \phi$ has to be a 3-transitive subgroup of $S_{q+1}$. So for $q=2$ or $q=3$, this is only possible if $\operatorname{im}\phi = S_{q+1}$ ...
Question: I don't see why the last sentence is true. For $q=2$, $\operatorname{im}\phi$ should map every 3-tuple of elements in $S_3$ to every other 3-tuple of elements in $S_3$; so we have to consider all possible permutations of 3 elements, which would explain $\operatorname{im}\phi = S_3$. But why does it hold for $q=3$ as well?
Thanks.
The number of $3$-tuples of distinct elements of $\{1,2,3,4\}$ is ${}_4P_3=24$, so the image of $\mathrm{PGL}_2\mathbb{F}_3$ must have size divisible by $24$, forcing it to be all of $S_4$. Thus $\mathrm{PSL}_2\mathbb{F}_3$ must be an index $2$ subgroup, which means it must be $A_4$. You can either use a theorem about index $2$ subgroups of symmetric groups or you can verify a couple generators of $\mathrm{PSL}_2\mathbb{F}_3$ act as even permutations.