Exercise 14.4.1

Proof. Recall that $\mathrm{AGL}(1,{𝔽}_{p^2})∕{𝔽}_p\simeq {𝔽}_p^{\text{∗}}$, where $({𝔽}_p,+)$ and $({𝔽}_p^{\text{∗}},\times )$ are cyclic, thus solvable, hence $\mathrm{AGL}(1,{𝔽}_{p^2})$ is solvable by Theorem 8.1.4.

By Exercise 14.3.3(a), $\mathrm{AGL}(1,{𝔽}_{p^2})$ has index 2 in $\mathrm{A\Gamma L}(1,{𝔽}_{p^2})$, therefore $\mathrm{AGL}(1,{𝔽}_{p^2})$ is a normal subgroup of $\mathrm{AGL}(1,{𝔽}_{p^2})$, and

The group $\{-1,1\}$ is cyclic, of prime order $2$, therefore is solvable, and $\mathrm{AGL}(1,{𝔽}_{p^2})$ is solvable. The same Theorem 8.1.4 shows that $M_1=\mathrm{A\Gamma L}(1,{𝔽}_{p^2})$ is solvable.

By Exercise 14.3.26,

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