Past exam of the options II course of the University of Oxford / 2007 / A / A2

Every nonzero vector can be extended to a basis, and the invertible map between two such bases carries the first vector to the second. The action is therefore transitive. Counting ordered bases givesThe stabilizer $H$ of one nonzero vector consequently has order $168/7=24$. There are three planes containing that vector, corresponding to the three lines of the two-dimensional quotient space. Adapted bases show $H$ acts transitively on these planes, so the stabilizer of one such plane has order $24/3=8$.

Let $Q$ be any subgroup of order eight. Its orbit lengths on the seven nonzero vectors are powers of two. Since their sum is odd, at least one orbit has length one. Fix such a vector $v$. The action on the three planes through $v$ likewise has a fixed plane $P$. Thus $Q$ is contained in the stabilizer of the flag $⟨v⟩\subset P$, and equality follows because that stabilizer already has order eight. The full group acts transitively on these flags, so all the subgroups are conjugate.

There are $7\cdot 3=21$ flags. To see they yield distinct subgroups, use the standard upper unitriangular matrices relative to an adapted basis. Their only common nonzero fixed vector is the first basis vector: varying entries in positions $(1,2)$ and $(1,3)$ forces the other two coordinates to vanish. On the quotient by that line, varying position $(2,3)$ leaves only the line of the second basis vector fixed. Hence this subgroup fixes exactly one full flag. Conjugation transfers uniqueness to each flag stabilizer, giving