Exercise 10, § 6

Let G be a solvable finite group of order $mn$, where m and n have no common prime factor. Show that there exists a subgroup H of G of order m and that every subgroup of G whose order divides m is contained in a conjugate of H ("Hall's Theorem")!

(Argue by induction on the order of G. If $mn \neq 1$, choose a commutative subgroup A of G which is normal and not equal to $\{e\}$ and whose order a is a power of a prime number. Apply this induction hypothesis to $G/A$ and to $(m/a, n)$ or $(m, n/a)$ according to whether a divides m or n. In the second case, use Exercise 9 to pass from $G/A$ to G.)

† Here again it can be shown that two such sections are conjugate, cf. Feit and Thompson, Proc. Nat. Acad. Sci., U.S.A., 48, 1962, 968–970.