Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / A / 2

An invertible matrix is exactly an ordered basis of its column space. The first column has $q^n-1$ choices. After $j$ independent columns have been chosen, their span has $q^j$ elements, so the next column has $q^n-q^j$ choices. Therefore

Choose a basis for the first subspace of a flag, extend it to a basis for the second, and continue until obtaining a basis of the whole vector space. Do the same for the target flag. The unique linear map sending the first adapted basis to the second is invertible and sends every flagged subspace to its counterpart. Hence the action is transitive.

Yes. Let $g$ send the first flag to the second and write $d=\det g\ne 0$. In an adapted basis of the target flag, scale its first vector by $d^{-1}$ and fix the others. This defines a flag-preserving map $h$ with determinant $d^{-1}$. Then $hg$ still sends the first flag to the second and has determinant one. Thus $\mathrm{SL}(n,K)$ is transitive as well.

Relative to the three successive blocks of sizes $1,2,2$, the stabilizer consists of the block upper triangular matricesThe off-diagonal blocks contain $2+2+4=8$ arbitrary field entries. Thus its order is $(q-1)|\mathrm{GL}(2,q){|}^2{q}^8$. By orbit-stabilizer theorem,Equivalently, choose the line first and then a two-dimensional subspace of the four-dimensional quotient:This also checks the stabilizer count independently.