= Solution
The orbits of a <normal subgroup> form a <block system>: $g(N\omega)=N(g\omega)$, so every <group> element permutes these orbits. By primitivity the orbits are either all singletons or one whole orbit. In the singleton case $N$ fixes every point, giving $N\leq G_{[\Omega]}$.
In the other case $N$ is transitive. For each $g\in G$ choose $n\in N$ with $n\alpha=g\alpha$. Then $n^{-1}g\in G_\alpha$, so $G=NG_\alpha$. Work in the quotient $G/N$. If $g=nh$ with $h\in G_\alpha$, the image of $A^g=g^{-1}Ag$ is the image of $A^h$, which equals the image $\overline A$ of $A$ because $A\triangleleft G_\alpha$. All conjugates of $A$ thus have the same image in $G/N$.
By hypothesis those conjugates generate $G$, so the quotient is generated by the single <subgroup> $\overline A$, and equals it. Since $A$ is abelian, $G/N$ is abelian. The <kernel of a group homomorphism> of a <group homomorphism> to an <abelian group> contains the <commutator subgroup>, yielding
$$
\boxed{G'\leq N\quad\text{or}\quad N\leq G_{[\Omega]}.}
$$
This is the normal-subgroup argument behind the <Iwasawa simplicity lemma>. Faithfulness is not needed for this version: retaining the pointwise <kernel of a group homomorphism> as the second alternative is essential. In a faithful <primitive group action> it says every nontrivial <normal subgroup> contains $G'$.
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