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Let $V=\Bbb F_2^3$ and let $G=GL(V)$ act naturally on the set $X=\{W\subset V:\text{sub-vector space,}\dim =2\}$

If $W\in X$ how do you determine the $Stab_G(W)$? and why shoud the cardinality of $Stab_G(W)$ be the same for different $W$'s?

If we want to prove that the action is transitive why is it enough to verify that there exists a matrix that sends $e_1$ to $u$, $e_2$ to $v$ and $e_3$ to $e_3$ for arbitrary linearly independent vectors $u,v$?

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    $\begingroup$ The statement that you say it is enough to verify is false, for example with $u=e_1$ and $v=e_3$. $\endgroup$ – Derek Holt Jan 15 at 14:09
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    $\begingroup$ Also, what does "for some arbitrary" mean? The words "some" and "arbitrary" contradict each other. $\endgroup$ – Derek Holt Jan 15 at 14:12
  • $\begingroup$ So how should we verify that the action is transitive? $\endgroup$ – John Cataldo Jan 15 at 14:25
  • $\begingroup$ Your statement about transitivity is incorrect, but a slight modification is. To prove that the action is transitive, it is enough to prove that given any linearly independent vectors $u,v \in V$, there exists matrix $M$ such that $Me_1 = u, Me_2 = v$, and $\textbf{importantly}$ $Me_3$ is not contained within the two dimensional subspace of $V$ spanned by $u,v$. Can see you see why this is true? Note that the requirement on the image of $e_3$ under $M$ is equivalent to the requirement that $M$ is invertible. $\endgroup$ – Adam Higgins Jan 15 at 14:34
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    $\begingroup$ It is enough to verify that for all pairs $u,v$ of linearly independent vectors, there is an element of $G$ that maps $e_1$ to $u$ and $e_2$ to $v$. That follows immediately from the definition of transitivity. $\endgroup$ – Derek Holt Jan 15 at 14:34
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To prove transitivity it would be enough to verify that there exists a matrix that sends $e_1$ to $u$, $e_2$ to $v$ and $e_3$ to $e_3$ for arbitrary linearly independent vectors $u,v$, but unfortunately such a matrix does not always exist. After all, the vectors $u$, $v$ and $e_3$ may be linearly dependent.

Fortunately, it is already enough to show that for every pair of linearly independent vectors $u,v\in V$ there exists a matrix in $G$ that maps $e_1$ to $u$ and $e_2$ to $v$. This would imply that the hyperplanes $\operatorname{span}(e_1,e_2)$ and $\operatorname{span}(u,v)$ are in the same orbit under the action of $G$, and hence that all hyperplanes are in the same orbit, i.e. that the action is transitive.

It is a basic fact (or simple exercise) on group actions that elements in the same $G$-orbit have conjugate stabilizers; in particular their stabilizers have the same cardinality.

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Your question on transitivity has been answered in the comments, but with regards to your question on $\operatorname{Stab}_{G}(W)$ for $W$ a two dimensional subspace of $V$. Note that because we have transitivity, $\operatorname{Stab}_{G}(W) \cong \operatorname{Stab}_{G}(W')$ for any pairs $W,W' \in X$. Indeed, suppose that $W, W' \in X$. Then by transitivity, there exists $g \in G$ such that $gW = W'$. Now suppose that $h \in \operatorname{Stab}_{G}(W')$, then $h\cdot w' \in w'$ for every $w' \in W'$, and so $(hg) \cdot w \in W'$ for every $w \in W$, and so finally $(g^{-1}hg)\cdot w \in W$ for every $w \in W$. This should make it clear that the map $\operatorname{Stab}_{G}(W') \rightarrow \operatorname{Stab}_{G}(W)$ such that $h \mapsto g^{-1}hg$ is an isomorphism. Thus, the stabalisers of pairs of elements of $X$ have the same cardinality, and so the stabiliser of all elements of $X$ must have the same cardinality.

Note that the above also allows us to calculate $\operatorname{Stab}_{G}(W)$ for any $W$ by calculating $\operatorname{Stab}_{G}(W)$ for our favourite subspace $W = \left< e_1, e_2 \right> \subseteq V$. It should be easy to see that for this $W$ we have

$$ \operatorname{Stab}_{G}(W) = \left\{ \begin{pmatrix} a & b & p \\ c & d & q \\ 0 & 0 & r \end{pmatrix} : ad - bc \neq 0, r \neq 0 \right\} $$

and so

$$ \left|\operatorname{Stab}_{G}(W) \right| = \left| \operatorname{GL}_{2}\left(\mathbb{F}_2 \right)\right|\left|\mathbb{F}_2^{\times} \right|\left| \mathbb{F}_2 \right|^{2} = 4\left| \operatorname{GL}_{2}\left(\mathbb{F}_2 \right)\right| = 4\cdot 6 = 24 $$

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