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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:)

Henry T.
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    It sounds like you are expected to handle the field of four elements as well (you end up with a copy of the alternating group $A_5$). – Jyrki Lahtonen Oct 22 '22 at 04:35
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    Your problem with the last example is likely related to the fact you did not specify the linear transformation. I switch to row vector notation, because the comments look ugly otherwise. The linear transformation that maps $(0,1)\mapsto (1,0)$ and $(1,0)\mapsto (1,1)$ won't work, because, by linearity, it would also map $(1,1)\mapsto (1,2)$. Instead you could use the transformation $T$ specified by $(0,1)\mapsto (2,0)$, $(1,0)\mapsto (1,1)$. In that case you also get $$T(1,1)=T(0,1)+T(1,0)=(2,0)+(1,1)=(0,1).$$ – Jyrki Lahtonen Oct 22 '22 at 04:45
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    (cont'd) Observe that $\langle (1,0)\rangle=\langle (2,0)\rangle$, so the 1-dimensional subspaces are still permuted as prescribed. Anyway, then $$T(1,2)=T(1,0)+2T(0,1)=(1,1)+2(0,2)=(1,1)+(0,1)=(1,2),$$ so $(1,2)$ is an eigenvector of $T$ and hence the subspace $\langle(1,2)\rangle$ is a fixed point. – Jyrki Lahtonen Oct 22 '22 at 04:47
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    But +1 for adding a lot of detail to the question. Made it easy to figure out what may be troubling you. If you think that the procedure I outlined will clear the fog for you, I will happily promote the comment to an answer (when I find the time, so others may beat me to it :-). I do need to search for duplicate before I do that, but it will have to wait :-( – Jyrki Lahtonen Oct 22 '22 at 04:50
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    See also this or this for elementary reasons the field $\Bbb{F}_4$ is the way it is. – Jyrki Lahtonen Oct 22 '22 at 13:40
  • Oh yes, you‘re right! My example was wrong. But do you think it is always possible to design the linear transformation in such a way that the fourth (or any other) vector is an eigenvector? (Then the image in $S_4$ would be all permutations that leave at least one element fixed, I assume?) I will look into it. And thank you also for pointing out the field with four elements, we didn’t really cover field theory yet, so I didn’t know about it. – Henry T. Oct 22 '22 at 14:49
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    There are quadratic polynomials without any roots in $\Bbb{F}_3$. If the matrix is, for example, the companion matrix of such a polynomial, then it won't have any eigenvalues in $\Bbb{F}_3$. Just as well because $S_4$ has permutations without fixed points. – Jyrki Lahtonen Oct 22 '22 at 15:02
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    You can also count elements. $GL_2(\Bbb{F}_3)$ has order $(3^2-1)(3^2-3)=48$. Only the two scalar matrices have all the 1-dimensional subspaces as eigenspaces, so you get $48/2=24$ permutations from them (the first isomorphism theorem, the kernel has size two). – Jyrki Lahtonen Oct 22 '22 at 15:05

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