Action of $GL(\Bbb F_2^3)$ on the sub-spaces of $\Bbb F_2^3$ of dimension $2$

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 $$u,v$$?

• The statement that you say it is enough to verify is false, for example with $u=e_1$ and $v=e_3$. – Derek Holt Jan 15 at 14:09
• Also, what does "for some arbitrary" mean? The words "some" and "arbitrary" contradict each other. – Derek Holt Jan 15 at 14:12
• So how should we verify that the action is transitive? – John Cataldo Jan 15 at 14:25
• 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. – Adam Higgins Jan 15 at 14:34
• 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. – Derek Holt Jan 15 at 14:34

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.

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

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