Groups of Prime-Power Orders' Subgroups (4.3.29)

Dummit and Foote Abstract Algebra, section 4.3, exercise 29:

Let G be a group of order p^a for some prime p. Prove that G has a subgroup of order p^b for 0 ≤ b ≤ a (apply theorem 8 and then use induction on a).

Proof: To start off, first observe that any subgroup contained in the center of its parent group is normal within its parent group. (1)

Part 1: By theorem 8, we have that Z(G) is nontrivial, and by the nature of G it is a group of order p^c for some c > 0. By Cauchy's Theorem, take x∈Z(G) such that | x | = p. Since Z(G) is abelian, we have that < x > is normal in Z(G) and thus Z(G) / < x > is an abelian group of order p^c-1 (and what's important is that there is now a group of such order in G contained in Z(G)). Inductively apply this reasoning on Z(G) / < x > until the basis is the identity subgroup, and thus every group of order p^d for 0 ≤ d ≤ c has been demonstrated to exist in G.

Part 2: By (1), we have that G / Z(G) is a group of order p^a-c, and refer its equivalent within G as H. Since Z(G) ≤ Z(H) for Z(G) ≤ H (2) and by (1), we have that J = H / Z(G) is a subgroup of order p^a-2c. Iterate this process to obtain the set of subgroups of order p^a-nc for integers n such that | Z(G) | = c ≤ a-nc. Now, by (2) we have that the subgroups constructed in part 1 are normal within the subgroups just now constructed, such that, taking the subgroup F of order p^a-mc and a subgroup I ≤ Z(G) ≤ Z(F) of order pⁱ for i ≤ c (there exists one for every i ≤ c by part 1), we have that F / I is a subgroup of order p^a-mc-i. This "bridges the gaps" for the integers between each a-mc and a-(m+1)c, such that every subgroup of order p^z for 0 ≤ z ≤ c has been constructed in part 1, and for c ≤ z ≤ a in part 2.

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Alternatively, I suppose one could do what Dummit and Foote actually suggested, and perform a similar induction on G where you take the group of prime-power order (G in the basis case) and, using theorem 8 to prove that the center is of nontrivial prime-power order, construct and use for the inductive step the prime-power-order quotient group G / A where A = < x > for x∈Z(G) and | x | = p by Cauchy's Theorem. Honestly, that would've been a bit less complicated...