A p-group contains subgroups of every order allowed by Lagrange’s Theorem

Let p be a prime and let G be a group of order p^a. Prove that G has a subgroup of order p^b for all 0 \leq b \leq a. [Hint: Use Theorem 8 and induction.]


For the base case, suppose |G| = p. Then 1 and G are subgroups of order 1 and p, respectively. Hence G has a subgroup of order p^b for all 0 \leq b \leq 1.

For the inductive step, suppose that every group of order p^a, where 1 \leq a \leq n, has a subgroup of order p^b for all 0 \leq b \leq a. Let G be a group of order p^{n+1}. Now by Theorem 8 in D&F, Z(G) is nontrivial. If Z(G) = G, then G is abelian; let H \leq G be a subgroup of order p by Cauchy’s Theorem. If Z(G) \neq G, then Z(G) is a proper nontrivial normal subgroup; let H = Z(G). In either case, note that (by Lagrange’s Theorem) both H and G/H are groups of order p^a, with 1 \leq a \leq n. Thus H has a subgroup of order p^b for all p^b dividing |H|, as does G/H. By the Lattice Isomorphism Theorem, G has a subgroup of order p^b for all 0 \leq b \leq n+1.

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