Exercise 14.4.13

Proof. The subgroup $N={𝔽}_p^{\text{∗}}{I}_2$ is a normal subgroup of $G=\mathrm{GL}(2,{𝔽}_p)$:

If $A\in G$, and $H=\lambda I_2$, then $\mathit{AH}{A}^{-1}=\mathit{A\lambda }{I}_p{A}^{-1}=\mathit{\lambda A}{A}^{-1}=\lambda I_2\in H$.

Therefore $G_{0}N$ is a subgroup of $G$. This implies that $G_{0}N$ is the smallest subgroup containing $G_0$ and $N$, so that $G_{0}N$ is the subgroup generated by $G_0$ and ${𝔽}_p^{\text{∗}}{I}_2$.

By the Second Isomorphism Theorem (see the beginning of Exercise 12),

By hypothesis, $G_0$ is solvable, thus its subgroup $N\cap G_0$ is solvable, and the quotient group $G_0∕N\cap G_0$ is solvable (Proposition 8.2.4).

$N$ is cyclic, thus $N$ is solvable. Using Proposition 8.2.4 anew, $G_{0}N$ is solvable.

We have proved that if $G_0\subset \mathrm{GL}(2,{𝔽}_p)$ is solvable, the subgroup generated by $G_0$ and ${𝔽}_p^{\text{∗}}{I}_2$ is also solvable. □