Past exam of the options II course of the University of Oxford / 2008 / A (Groups in Action)

An invertible matrix is an ordered basis of $K^n$. After choosing $j$ independent columns, their span has $q^j$ elements, so the next column has $q^n-q^j$ choices. Thus

For each flag choose an adapted basis: its first $n_i$ vectors span $V_i$, obtained by extending a basis successively through the nested subspaces and finally through $V$. For two flags, the invertible linear transformation taking one adapted basis to the other takes each member of the first flag to the corresponding member of the second. Hence the action is transitive.

Yes. Suppose $g$ maps the first flag to the second and has determinant $\delta \ne 0$. Relative to an adapted basis of the first flag, scaling its first basis vector by ${\delta }^{-1}$ and fixing all others defines a transformation $h$ preserving that flag, with determinant ${\delta }^{-1}$. Then $gh$ still maps the first flag to the second and has determinant one. Thus the special linear subgroup acts transitively too.

The stabilizer of the indicated flag consists of block upper triangular matrices with diagonal block sizes $1,2,2$. The diagonal blocks can be any elements of $\mathrm{GL}(1,q)$, $\mathrm{GL}(2,q)$ and $\mathrm{GL}(2,q)$. The upper off-diagonal blocks have $2+2+4=8$ arbitrary entries. Thus the stabilizer order is $(q-1)|\mathrm{GL}(2,q){|}^2{q}^8$. The orbit-stabilizer theorem gives