Iwasawa's lemma

Lemma (Iwasawa) If $(G,\Omega)$ is a primitive permutation group with $G$ perfect and for some $\alpha \in \Omega$, $G_\alpha$ has a normal abelian subgroup $A$ whose conjugates generate $G$, then $G$ is simple.
proof: Suppose $1 \not = N \unlhd G$, we will gradually show that $N$ must be the whole group. By primitivity $N$ is transitive on $\Omega$ and $G_\alpha$ is a maximal subgroup, so $N \not \le G_\alpha$ and $N G_\alpha = G$. Any $g$ may be written $n x$ for some $n \in N$, $x \in G_\alpha$ so $A^g = A^{nx} = A^x \le AN$ and these conjugates cover the whole group so $AN = G$. Now $G/N \simeq A/(A \cap N)$ is abelian, but $G$ is perfect so $N = G$.