Exercise 9, § 6

Let $F \to E \to G$ be an extension of finite groups. Assume that no prime number divides both the order of $F$ and the order of $G$. (a) Assume $F$ is solvable. Show that there exists a section $s : G \to E$ and that any two such sections are conjugate by an element of $F$. (One shall reason by induction on the solvability class of $F$; when $F$ is commutative, one shall use the preceding exercise.) (b) Show the existence of a section$^1 s : G \to E$ without assuming $F$ is solvable. (Reason by induction on the order of $G$. If $p$ divides the order of $F$, choose a $p$-Sylow subgroup $P$ of $F$ and consider its normalizer $N$ in $E$. The image of $N$ in $G$ is equal to $G$, cf. I, p. 139, exerc. 25. If $N \neq E$, conclude by means of the induction hypothesis; if $N = E$, use (a) as well as the induction hypothesis applied to $F/P \to E/P \to G.$)

1 Here again, one can prove that any two such sections are conjugate, cf. Feit-Thompson, Proc. Nat. Acad. Sci. U.S.A., XLVIII, 1962, p. 968–970.

EXERCISES