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For four distinct projective points on the complex projective line, use the conventionwhere $[p_i,p_j]$ is the determinant of chosen representative vectors. Rescaling any representative cancels, and applying an invertible matrix multiplies every bracket by its determinant, which also cancels. Thus the expression is a projective invariant. The projective general linear group action on the projective line has a unique element $T$ taking the first three points to $\infty ,0,1$. In those coordinates the displayed expression is exactly $T(P_3)$. Consequently the cross-ratio completely determines the orbit of the ordered quadruple under the projective linear group. Its value lies in $\mathbb{C}\setminus \{0,1\}$.

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.

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.