Let $\mathbb{K}$ be a field of order $q$ and let $\mathbb{P}^1(\mathbb{K})$ denote all one dimensional $\mathbb{K}$ sub-spaces of $\mathbb{K}^2$. The natural operation of $\operatorname{GL}(\mathbb{K}^2)$ on $\mathbb{P}^1(\mathbb{K})$ factorizes through the operation $\operatorname{PGL}(\mathbb{K}^2):= \operatorname{GL}(\mathbb{K}^2)/(\mathbb{K}^{\times}\cdot \mathbb{1}_2)$
By numbering the elements in $\mathbb{P}^1(\mathbb{K})$ we get a homomorphism $\operatorname{PLG}_2(\mathbb{K}) \rightarrow S_{q+1}$, I have to determine the image of this homomorphism for $q \leq 4$. (It suffices to look at the cases $q=2, q=3$, since only these $q$ are prime).
For $q=2$, this is easy, since the operation of $\operatorname{PLG}_2(\mathbb{K})$ is 3-transitive. Because if we consider:
$$ (\left<v_1 \right>, \left<v_2 \right>, \left<v_3 \right>), (\left<w_1 \right>, \left<w_2 \right>, \left<w_3 \right>) \in (\mathbb{K}^2)^3 $$
With the three $v_i$ and three $w_i$ each in different sub-spaces, we know:
$$ v_3=rv_1+sv_2, \quad r,s \in \mathbb{K} \\ w_3=pw_1+qw_2, \quad p,w \in \mathbb{K} $$
And therefore $A\in \operatorname{GL}(\mathbb{K}^2)$, with $Av_1 = \frac{p}{r}w_1, Av_2 = \frac{q}{s}w_2$, is a map such that:
$$ A (\left<v_1 \right>, \left<v_2 \right>, \left<v_3 \right>)= (\left<w_1 \right>, \left<w_2 \right>, \left<w_3 \right>) $$.
So we can permutate any three sub-spaces in $\mathbb{K}^2$ as we like and the image is the whole of $S_3$.
I am, however, struggling with the case $q=3$, here not all permutations of the four sub-spaces
$$ \left< \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right>, \left< \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right>, \left< \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right>, \left< \begin{pmatrix} 1 \\ 2 \end{pmatrix} \right> $$
can be reached. I dont even get why an operation on this set would correspond to an element of $S_4$ because if we decide to cyclically permutate the first three vectors (which we can do by a linear operator as shown above) i.e.:
$$ \left< \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right> \mapsto \left< \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right>, \left< \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right> \mapsto \left< \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right>, \left< \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right> \mapsto \left< \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right> $$.
This leads to:
$$ \left< \begin{smallmatrix} 1 \\ 2 \end{smallmatrix} \right> = \left< \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right> + \left< \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right> \mapsto \left< \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right> + \left< \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right> = \left< \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right> $$
And thus our operation isn’t injective and can’t correspond to something in $S_4$.
I must therefore be understanding the homomorphism wrong, happy if someone could help:)