Exercise 10, § 6

Let $G$ be a finite solvable group of order $mn$, where $m$ and $n$ have no common prime factor. Now prove 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 theorem”).

(Reason by induction on the order of G. If $mn \neq 1$, choose a commutative subgroup A of G, distinguished, not reduced to $e$, and whose order $a$ is a power of a prime number. Apply the induction hypothesis to G/A and to $\left( \frac{m}{a}, n \right)$ or $\left( m, \frac{n}{a} \right)$ according as $a$ divides $m$ or divides $n$. In the second case, use Exercise 9 of I, p. 136 to pass from G/A to G.)