Exercise 7.5.11

Proof.

(a)

Write ${\gamma }_{a,b}:F\to F,\alpha \mapsto {\gamma }_{a,b}(\alpha )=\mathit{a\alpha }+b$. For all $\alpha \in F,$

$({\gamma }_{a,b}\circ {\gamma }_{c,d})(\alpha )=a(\mathit{c\alpha }+d)+b=\mathit{ac\alpha }+\mathit{ad}+b={\gamma }_{\mathit{ac},\mathit{ad}+b}(\alpha )$,

thus

$${\gamma }_{a,b}\circ {\gamma }_{c,d}={\gamma }_{\mathit{ac},\mathit{ad}+b}.$$

Let

$$\phi :\{\begin{array}{ccc}\hfill \mathrm{AGL}(1,F)\hfill & \hfill \to \hfill & \hfill \mathrm{PGL}(2,F)\hfill \\ \hfill {\gamma }_{a,b}\hfill & \hfill \mapsto \hfill & \hfill \left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\hfill \\ \hfill \hfill \end{array}.$$

$$\begin{array}{llll}\hfill \phi ({\gamma }_{a,b})\phi ({\gamma }_{c,d})& =\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill c\hfill & \hfill d\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]& \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill \mathit{ac}\hfill & \hfill \mathit{ad}+b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]& \hfill & \\ \hfill & =\phi ({\gamma }_{\mathit{ac},\mathit{ad}+b})& \hfill & \\ \hfill & =\phi ({\gamma }_{a,b}\circ {\gamma }_{c,d}).& \hfill & \end{array}$$

$\phi :\mathrm{AGL}(1,F)\to \mathrm{PGL}(2,F)$ is so a group homomorphism.

$$\begin{array}{llll}\hfill {\gamma }_{a,b}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& ⟺\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=[I_2]& \hfill & \\ \hfill & ⟺\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \lambda \hfill \end{array}\right),\lambda \in F^{\text{∗}}& \hfill & \\ \hfill & ⟺a=1,b=0& \hfill & \\ \hfill & ⟺{\gamma }_{a,b}=1_F& \hfill & \end{array}$$

$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1_F\}$, thus $\phi$ is an injective group homomorphism, which embeds $\mathrm{AGL}(1,F)$ in $\mathrm{PGL}(2,F)$.

(b)

Write $G_{\infty }$ the stabilizer of $\infty$ in $\mathrm{PGL}(2,F)$. $$\begin{array}{llll}\hfill \left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\in G_{\infty }& ⟺\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)=\lambda \left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right),\lambda \in F^{\text{∗}}& \hfill & \\ \hfill & ⟺c=0& \hfill & \end{array}$$

Let $\gamma =\left(\begin{array}{cc}\hfill r\hfill & \hfill s\hfill \\ \hfill t\hfill & \hfill u\hfill \end{array}\right)\in \mathrm{GL}(2,F)$. If $[\gamma ]\in \phi (\mathrm{AGL}(1,F))$, then $[\gamma ]=\phi ({\gamma }_{a,b})=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$, thus $[\gamma ]\in G_{\infty }$ by the preceding equivalence.

Conversely, if $[\gamma ]\in G_{\infty }$, then $t=0$, therefore $\det <!--\; FUNCTION\; APPLICATION\; -->(\gamma )=\mathit{ru}\ne 0$, so $u\ne 0$.

$\left(\begin{array}{cc}\hfill r\hfill & \hfill s\hfill \\ \hfill t\hfill & \hfill u\hfill \end{array}\right)=u\left(\begin{array}{cc}\hfill r∕u\hfill & \hfill s∕u\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),u\in F^{\text{∗}}$, thus $[\gamma ]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$, where $a=r∕u,b=s∕u$ : $[\gamma ]=\phi ({\gamma }_{a,b})\in \phi (\mathrm{AGL}(1,F))$.

$$G_{\infty }=\phi (\mathrm{AGL}(1,F)).$$

So $\mathrm{AGL}(1,F)$ is identified with the stabilizer of $\infty$ in $\mathrm{PGL}(2,F)$ and is isomorphic to a subgroup of $\mathrm{PGL}(2,F)$.