Exercise 14.4.11

Proof. By definition, $M_3={\pi }^{-1}(N(H))$, where

Consider the restriction (and co-restriction) of $\pi$, given by

Since $M_3={\pi }^{-1}(N(H))$, $\tilde{\pi }$ is surjective. Moreover, ${\gamma }_{A,v}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\tilde{\pi })$ if and only if $[A]=[I_2]$, that is $A\in {𝔽}_p^{\text{∗}}{I}_2$, thus

We know that ${\gamma }_{A,v}={\gamma }_{B,w}\Rightarrow A=B,v=w$, therefore the map

is bijective. This gives

The First Isomorphism Theorem gives

therefore

Moreover, if we give the structure of semidirect product to ${𝔽}_p^2\times {𝔽}_p^{\text{∗}}$, we see that

is a group isomorphism: By formula (1) in Ex. 14.3.2, we obtain

We have seen that $\psi$ is bijective, and, following the formula (6.9) for the semidirect product,

Therefore

This proves that $\psi$ is a group isomorphism, and $\ker <!--\; FUNCTION\; APPLICATION\; -->(\tilde{\pi })\simeq {𝔽}_p^2⋊{𝔽}_p^{\text{∗}}$.

By exercise 6.4.8, we know that ${𝔽}_p^2⋊{𝔽}_p^{\text{∗}}∕{𝔽}_p^2\times \{1\}\simeq {𝔽}_p^{\text{∗}}$, where ${𝔽}_p^2\times \{1\}\simeq {𝔽}_p^2$. Since ${𝔽}_p^{\text{∗}}$ is cyclic, ${𝔽}_p^{\text{∗}}$ is solvable, and ${𝔽}_p^2$ is also cyclic and solvable, Theorem 8.1.4 shows that the semidirect product ${𝔽}_p^2⋊{𝔽}_p^{\text{∗}}$ is solvable, thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\tilde{\pi })$ is solvable.

We know that $S_4$ is solvable (Exercise 8.1.1), thus $N(H)\simeq S_4$ is solvable.

Using again Theorem 8.1.4 with $N(H)\simeq M_3∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\tilde{\pi })$, we obtain that $M_3$ is solvable. □