Let be a prime and let be a group of order . Prove that has a subgroup of order for all . [Hint: Use Theorem 8 and induction.]
For the base case, suppose . Then and are subgroups of order and , respectively. Hence has a subgroup of order for all .
For the inductive step, suppose that every group of order , where , has a subgroup of order for all . Let be a group of order . Now by Theorem 8 in D&F, is nontrivial. If , then is abelian; let be a subgroup of order by Cauchy’s Theorem. If , then is a proper nontrivial normal subgroup; let . In either case, note that (by Lagrange’s Theorem) both and are groups of order , with . Thus has a subgroup of order for all dividing , as does . By the Lattice Isomorphism Theorem, has a subgroup of order for all .