Dtyjlui

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

If $Win X$ 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 $u,v$?

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 $u,vin V$ 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.

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 $hcdot w' in w'$ 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

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