= Solution
The orbits of a <normal subgroup> $N$ are blocks for the <primitive group action>. If every orbit is a singleton then $N\le G_{[\Omega]}$, giving the second alternative. Otherwise $N$ is transitive, and $G=NG_\alpha$.
Work in $G/N$. Every $g\in G$ can be written $nh$ with $n\in N$, $h\in G_\alpha$. <Conjugation> by $n$ has no effect on the <image> of a <subgroup> in this quotient, while $h$ normalizes $A$. Thus every conjugate of $A$ has the same <image> $AN/N$. The generation hypothesis says this <image> generates all of $G/N$. But it is already an abelian <subgroup>, so $G/N$ is abelian and $G'\le N$. This proves the <Iwasawa simplicity lemma> in its possibly nonfaithful form:
$$
\boxed{G'\le N\quad\text{or}\quad N\le G_{[\Omega]}.}
$$
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