Exercise 14.4.2

Proof. (a) Write $\gamma ={\gamma }_{a,b},{\gamma }^{\prime }={\gamma }_{a^{\prime },b^{\prime }}$, where $a,a^{\prime }\in {𝔽}_p^{\text{∗}},b,b\in {𝔽}_p$.

Then

Since

$\delta ={\gamma }_{A,v}\in \mathrm{AGL}(2,{𝔽}_p)$, where $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a^{\prime }\hfill \end{array}\right)\in \mathrm{GL}(2,{𝔽}_p)$ (since $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=a{a}^{\prime }\ne 0$), and $v=(b,b^{\prime })\in {𝔽}_p^2$. (b) $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)=I_2$, thus $B=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ has order $2$, and is identified with the element ${\gamma }_{B,0}\in \mathrm{AGL}(2,{𝔽}_p)$.

Let $\delta ={\gamma }_{A,v}$ be any element of $\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$. With the same notations as in part (a), using (1),

where

is diagonal, therefore ${\gamma }_{B,0}\circ \delta \circ {\gamma }_{B,0}^{-1}\in \mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$. This proves that $B$ normalizes $\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$:

(c) Since $M_2$ is the subgroup of $\mathrm{AGL}(2,{𝔽}_p)$ generated by $\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$ and $B$, part (b) shows that $H=\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$ is a normal subgroup of $M_2$.

Note that, if $A\in H$, $\mathit{AB}=B(\mathit{BAB})=B{A}^{\prime }$, where $A^{\prime }=\mathit{BAB}=\mathit{BA}{B}^{-1}\in H$. Therefore every element of $M_2$ is of the form $A$ or $\mathit{BA}$, where $A\in H$. This proves that $M_2=H\cup (B\cdot H)$, so that the group homomorphism

is surjective. Since $B\notin H,B\cdot H\ne H$, thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\pi )=\{I_2\}$. This proves that $\pi$ is an isomorphism, and

This gives $|M_2|=2|\mathrm{AGL}(1,{𝔽}_p){\text{|}}^2$, so

Since $(\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p))∕\mathrm{AGL}(1,{𝔽}_p)\simeq \mathrm{AGL}(1,{𝔽}_p)$, where $\mathrm{AGL}(1,{𝔽}_p)$ is solvable, the group $\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$ is solvable.

The group $\{I_2,B\}$ is cyclic, therefore solvable. By Theorem 8.1.4, the isomoprphism $M_2∕(\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p))\simeq \{I_2,B\}$ shows that $M_2$ is solvable. (d) Note that $\delta \in H$ defined by $\delta (\alpha ,\beta )=(\mathit{\lambda \alpha }+b,\mathit{\mu \beta }+b^{\prime })$ fixes $0$ if and only if $b=b^{\prime }=0$. Therefore $H\cap {(M_2)}_0$ is generated by the matrices $\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mu \hfill \end{array}\right),\lambda ,\mu \in {𝔽}_p^{\text{∗}}$.

By part (c), $M_2=H\cup (B\cdot H)$, and $B\in {(M_2)}_0$, therefore ${(M_2)}_0\subset \mathrm{GL}(2,{𝔽}_p)$ is generated by the matrices

(e) If $H$ is the smaller subgroup $\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$, then the isotropy group $H_0$ is generated by the matrices $\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mu \hfill \end{array}\right),\lambda ,\mu \in {𝔽}_p^{\text{∗}}$, therefore is the subgroup of diagonal matrices in $\mathrm{GL}(2,{𝔽}_p)$.

We prove that $H_0$ is not irreducible. The nontrivial subspace $V={𝔽}_p\times \{0\}\subset {𝔽}_p^2$ is such that $h(V)\subset V$ for all $h=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mu \hfill \end{array}\right)\in H_0$: For all $(\gamma ,0)\in V$,

□