Exercise 28, § 2

Let $G$ be a group, $M$ a maximal subgroup of $G$ and $U$ a normal subgroup of $M$, satisfying condition (R) of no. 7. Assume that $G$ is equal to its commutator subgroup, that it is generated by the union of the conjugates of $U$ and that the intersection of the conjugates of $M$ reduces to the unit element. Show that $G$ is simple. (Consider a normal subgroup $N$ of $G$ distinct from $\{1\}$; show that $G = N.M$ and then that $G = N.U.$)