Exercise 15, § 4

Let $\mathscr{F}$ be a set of stable subgroups of a group with operators $G$; $\mathscr{F}$ is said to satisfy the maximal condition (resp. minimal condition) if every subset of $\mathscr{F}$, ordered by inclusion, has a maximal (resp. minimal) element.

Suppose that the set of all stable subgroups of a group G satisfies the minimal condition.

(a) Prove that there exists no stable subgroup of G isomorphic to G and distinct from G (argue by reductio ad absurdum by showing that the hypothesis would imply the existence of a strictly decreasing infinite sequence of stable subgroups of G).

(b) The minimal elements of the set of normal stable subgroups of G not consisting only of e are called minimal normal stable subgroups. Let $\mathfrak{M}$ be a set of minimal normal stable subgroups of G and S the smallest stable subgroup of G containing all the subgroups belonging to $\mathfrak{M}$; show that S is the direct product of a finite number of minimal normal stable subgroups of G (let $(M_n)$ be a sequence of minimal normal stable subgroups of G belonging to $\mathfrak{M}$ and such that $M_{n+1}$ is not contained in the stable subgroup generated by the union of $M_1, M_2, \ldots, M_n$; let $S_k$ be the stable subgroup generated by the union of the $M_n$ of index $n \geq k$; show that $S_{k+1} = S_k$ after a certain rank and therefore that $(M_n)$ is a finite sequence; finally use no. 9, Proposition 15).

(c) If G is a group without operators, show that every minimal normal subgroup M of G is the direct product of a finite number of simple groups isomorphic to one another (let N be a minimal normal subgroup of M; show that M is the smallest subgroup of G containing all the subgroups $aNa^{-1}$ where a runs through G, and then apply (b) to the group M).