About relation between simple section and section in induced representation on orbits

Question

The following is from Dixon & Mortimer, Permutation Groups, page 74.

A group $S$ is called a section of a group $G$ if for some subgroups $H$ and $K$ of $G$ we have $K \unlhd H$ and $H/K \cong S$. This means that $S$ is the homomorphic image of some subgroup of $G$. Further if $G \le Sym(\Omega)$ and for some subset $\Gamma \subseteq \Omega$ we have $\Gamma^x = \Gamma$ for each $x \in G$, then each element $x \in G$ could be seen as a function from $\Gamma$ to $\Gamma$, denoted by $x^{\Gamma} : \Gamma \to \Gamma$, in that case set $G^{\Gamma} := \{ x^{\Gamma} : x \in G \}$.

Theorem: Suppose that $G$ is a finite primitive subgroup of $Sym(\Omega)$. Let $\alpha \in \Omega$ and let $\Gamma$ be a nontrivial orbit of $G_{\alpha}$. Then every simple section of $G_{\alpha}$ is isomorphic to a section of the group $G_{\alpha}^{\Gamma}$ which $G_{\alpha}$ induces on $\Gamma$. In particular, each composition factor of $G_{\alpha}$ is isomorphic to a section of $G_{\alpha}^{\Gamma}$.

In the book the following corollary of noted:

Corollary: If $G$ is a finite, nonregular, primitive group and $\Gamma \ne \{\alpha\}$ is an orbit of a point stabilizer $G_{\alpha}$, then:

(i) each prime dividing $|G_{\alpha}|$ divides $|G_{\alpha}^{\Gamma}|$,

(ii) $G_{\alpha}$ is solvable whenever $G_{\alpha}^{\Gamma}$ is solvable.

For (i) if $p$ divides $|G_{\alpha}|$, then there exists an element $x$ of order $p$, set $H = \langle x \rangle$. This is a section of itself by the trivial group $K = 1$, and it is simple. Hence it is isomorphic to a section of $G_{\alpha}^{\Gamma}$, let $H', K' \le G_{\alpha}^{\Gamma}$ be such that $K' \unlhd H'$ and $H'/K' \cong H$. Then as $p$ equals the index of $K'$ in $H'$ it must divide $|H'|$, hence it must also divide the order $G_{\alpha}^{\Gamma}$.

But why does (ii) holds? It might be simple but I do not see it...

Answer 1

1. If $G_{\alpha}^{\Gamma}$ is solvable, then all it's subgroup or quotient group are solvable.

2. Since there is a lemma showing that every simple solvable group must be abelian, so every simple section of $G_{\alpha}^{\Gamma}$ is an abelian group.

3. Moreover, every simple sections of $G_{\alpha}$ must be abelian, since it's isomorphic to a simple section of $G_{\alpha}^{\Gamma}$.

4. In particular, every composition factor of $G_{\alpha}$ is an abelian group, which implies that every composition sequence of $G_{\alpha}$ is a sub-normal sequence of $G_{\alpha}$ with the property that every factor group is abelian. By the equivalent condition of solvable group, we know that $G_{\alpha}$ is solvable.