Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 7D (Groups)

Identify the projective line with one-dimensional subspaces of ${\mathbb{F}}_p^2$: use $v_t=(t,1{)}^T$ for finite $t$, and $v_{\infty }=(1,0{)}^T$. This convention turns the matrix action into the stipulated Möbius transformation.

Since $x$ and $z$ are distinct projective points, $v_z,v_x$ are a basis. Write $v_y=A{v}_z+B{v}_x$. Both $A$ and $B$ are nonzero, because $y$ is different from $x,z$. The matrix with columns $A{v}_z$ and $B{v}_x$ is invertible and sends the lines of $(1,0{)}^T,(0,1{)}^T,(1,1{)}^T$ to those of $z,x,y$, respectively. This constructs the requested map without separate exceptional formulas at infinity.

A matrix fixing $0$ and $\infty$ is diagonal, say $\mathrm{diag}(a,d)$ with $a,d\ne 0$. To fix $1$ it must also satisfy $a=d$, so the stabilizer subgroup of $(0,1,\infty )$ is precisely $\{\lambda I:\lambda \in {\mathbb{F}}_p^{\times }\}$. Any two solutions differ on the right by one of these matrices. ThusEquivalently, after dividing out scalar matrices the projective general linear group action on the projective line is sharply three-transitive on a projective line.

The Möbius transformations are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the projective line has exactly three points. There are therefore five orbits of a group action.

We first count $|G{L}_2({\mathbb{F}}_p)|$. Its first column can be any nonzero vector, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. HenceFor the all-equal pattern take $(\infty ,\infty ,\infty )$. Its stabilizer subgroup consists of upper triangular matricesof order $p(p-1{)}^2$. Its group orbit has order $p+1$.

The three exactly-two-equal patterns have representatives $(\infty ,\infty ,0)$, $(\infty ,0,\infty )$ and $(0,\infty ,\infty )$. Each stabilizer subgroup fixes $0$ and $\infty$ individually, so it is the diagonal subgroup of order $(p-1{)}^2$. Each group orbit has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.

For the all-distinct pattern use $(0,1,\infty )$. Its stabilizer subgroup is the scalar subgroup of order $p-1$, and the group orbit has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the equality-pattern orbits of projective triples haveThe sizes add to $(p+1{)}^3$, and each orbit size times its stabilizer order is $|G|$, as in the orbit-stabilizer theorem.