= Solution
<Iwasawa's simplicity lemma> states the following. Suppose $G$ acts faithfully and primitively on a set, and a <stabilizer subgroup> $H$ has an abelian <normal subgroup> $A$ whose $G$-conjugates generate $G$. Then every nontrivial <normal subgroup> $N$ contains $G'$. In particular, \b[if $G$ is nontrivial and perfect, then $G$ is simple].
Indeed a nontrivial <normal subgroup> in a faithful primitive action is transitive, so $G=NH$. Since $H$ normalizes $A$, all conjugates of $A$ have the same image in $G/N$. Those images generate the quotient, which is therefore abelian. This gives $G'\leq N$ and proves the stated conclusion.
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