Exercise 19, § 4

(a) Let G be a group such that every normal subgroup of G distinct from G is contained in a direct factor of G distinct from G. Show that G is the restricted sum of a family of simple subgroups. (Let K be the subgroup of G generated by the union of all the simple subgroups of G. Suppose $K \neq G$; if $x \in G - K$, consider a normal subgroup M of G which is maximal amongst those containing K and not containing x. If $G = M' \times S$, where $M'$ contains M and $S \neq \{e\}$, show that $x \notin M'$ and therefore $M' = M$; on the other hand, if N is a normal subgroup of S, show that $M \times N = G$ and conclude that S is simple, whence a contradiction. Conclude with the aid of Exercise 18.) Obtain the converse (cf. Exercise 18).

(b) In order that there exist in a group G a family $(N_\alpha)$ of normal simple subgroups such that the $G/N_\alpha$ are simple and $\bigcap_\alpha N_\alpha = \{e\}$, it is necessary and sufficient that, for every normal subgroup $N \neq \{e\}$ of G, there exist a normal subgroup $N' \neq G$ such that $NN' = G$. (To see that the condition is necessary, consider an $x \in N$ distinct from e and an $N_\alpha$ not containing x. To see that the condition is sufficient, for all $x \neq e$ in G, consider the normal subgroup N generated by x and a normal subgroup $N' \neq G$ such that $NN' = G$. Show that if a normal subgroup M of G is maximal amongst the normal subgroups of G containing $N'$ and not containing x, it is maximal in the set of all normal subgroups $\neq G$ and deduce the conclusion.)