${\mathrm{GL}}(4,{\mathbb{F}}_q)$ action on the $(2,4)$-Grassmann Let ${\mathbb{F}}_q$ be the finite field of order $q$ and ${\mathcal{G}}(2,4)$ denote the set of all $2$-dimensional subspaces of ${\mathbb{F}}^4_q$. Let $P$ denote the $(2,2)$-parabolic (which is the subgroup of all matrices of ${\mathrm{GL}}(4,{\mathbb{F}}_q)$ which has lower triangular $2 \times 2$ block zero while written in $2 \times 2$ sub-blocks).
There is a natural transitive left action of ${\mathrm{GL}}(4,{\mathbb{F}}_q)$ on ${\mathcal{G}}(2,4)$: if $W \in {\mathcal{G}}(2,4), g \in {\mathrm{GL}}(4,{\mathbb{F}}_q)$ with $W = {\mathrm{span}} \{ v_1, v_2 \}$, then $g \cdot W := {\mathrm{span}} \{ g \cdot v_1, g \cdot v_2 \}$. Setting $\{ e_1, e_2, e_3, e_4 \}$ as the canonical basis of ${\mathbb{F}}^4_q$, clearly $P$ turns out to be the stabiliser of $W_0 = {\mathrm{span}} \{ e_1, e_2 \}$.
${\mathrm{\bf Question.1:}}$ Show that ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$ can be identified with ${\mathcal{G}}(2,4)$.
Since $P$ is not a normal subgroup ${\mathrm{GL}}(4,{\mathbb{F}}_q)$, this identification doesn't seems to come from the natural action. What is the best way to see this?
${\mathrm{\bf Question.2:}}$ Under this identification, two elements of ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$ are in the same orbit of $P$ if and only if the corresponding subspaces intersect in the same dimensional subspaces of ${\mathrm{span}} \{ e_1, e_2 \}$.
What is the meaning of orbit here?
One can make an equivalence relation on ${\mathcal{G}}(2,4)$ satisfying this property to explain the orbit. However, I don't see a natural connection with ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$. What am I missing?
 A: If $G$ acts on a set $\Omega$, the action is called transitive if for every $\omega,\omega'\in\Omega$ there exists a $g\in G$ for which $\omega'=g\omega$. The stabilizer $\mathrm{Stab}(\omega)\le G$ of a point $\omega\in G$ is the set of all $g\in G$ which fix $\omega$, i.e. $g\omega=\omega$. The orbit $\mathrm{Orb}(\omega)$ of a point $\omega\in\Omega$ is the set of all "translates" $g\omega$ of $\omega$ (for all $g\in G$). Note a $G$-set $\Omega$ is transitive if and only if $\Omega=\mathrm{Orb}(\omega)$ for any (or all) $\omega\in\Omega$, i.e. it's all one orbit. Two $G$-sets $\Omega_1$ and $\Omega_2$ are called equivalent or isomorphic if there is a $G$-equivariant bijection between them. A coset space $G/H$ is itself a $G$-set, with the action by left-multiplication.
The orbit-stabilizer theorem says if $G\curvearrowright\Omega$ is a transitive action then $\Omega$ is equivalent (as a $G$-set) to a coset space $G/\mathrm{Stab}(\omega)$ (for any point $\omega\in\Omega$). Specifically, $g\mathrm{Stab}(\omega)\leftrightarrow g\omega$. Note $\mathrm{Stab}(g\omega)=g\mathrm{Stab}(\omega)g^{-1}$ so this sets up a one-to-one correspondence between isomorphism classes of transitive $G$-sets and conjugacy classes of subgroups $H$.
In particular, $\mathrm{GL}_4\mathbb{F}_2$ acts on $\mathrm{Gr}_2\mathbb{F}_2^4$ transitively, and the stabilizer of the first-two-coordinates-plane is $P$, so we automatically know that $\mathrm{GL}_4\mathbb{F}_2/P$ is equivalent to $\mathrm{Gr}_2\mathbb{F}_2^4$. This answers (1).
For (2) you'll want to explore how $P$ acts on $\mathrm{Gr}_2\mathbb{F}_2^4$...
