 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 $G^{\prime }$. In particular, 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 $G^{\prime }\le N$ and proves the stated conclusion.