Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 4 / a / 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 $G=N{G}_{\alpha }$.

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 $G^{\prime }\le N$. This proves the Iwasawa simplicity lemma in its possibly nonfaithful form: