Congruence condition on number of abelian subgroups of prime-cube order

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This article is about a congruence condition.
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Statement

Statement in terms of universal congruence condition

Let p be a prime number (odd or p = 2). The collection of abelian p-groups of order p^3 is a Collection of groups satisfying a universal congruence condition (?). Thus, it is also a Collection of groups satisfying a strong normal replacement condition (?) and a Collection of groups satisfying a weak normal replacement condition (?).

Hands-on statement

Let p be any prime (including the cases of odd p and p = 2). Then, if P is a finite p-group and A is an abelian subgroup of P of order p^3, the number of abelian subgroups of P of order p^3 is congruent to 1 modulo p.

Related facts

For a summary of facts about universal congruence conditions, refer collection of groups satisfying a universal congruence condition#Examples/facts.

Facts used

  1. Existence of abelian normal subgroups of small prime power order: This states that if n \ge 1 + k(k-1)/2, then any group of order p^n contains an abelian normal subgroup of order p^k.
  2. Congruence condition on number of abelian subgroups of prime index
  3. Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group: This states that if n(G) is the number of subgroups of G isomorphic to one of the groups in a given collection \mathcal{S} and G \notin \mathcal{S}, then n(G) \equiv \sum n(M) \pmod p.
  4. Congruence condition on number of subgroups of given prime power order

Proof

Given: A group G of order p^n, n \ge 3, that contains an abelian subgroup of order p^3.

To prove: The number of abelian subgroups of G of order p^3 is congruent to 1 modulo p.

Proof: We consider three cases:

  1. n = 3: In this case, there is exactly one abelian subgroup of order p^3, namely G itself.
  2. n = 4: In this case, fact (2) completes the proof.
  3. n \ge 5: In this case, we use induction on n. By fact (1), every maximal subgroup of G contains an abelian normal subgroup (normality being relative to the maximal subgroup) of order p^3, so by the inductive hypothesis, the number of abelian subgroups of order p^3 contained in each maximal subgroup is congruent to 1 modulo p. Since the number of maximal subgroups is itself congruent to 1 modulo p by fact (4), fact (3) tells us that the number of abelian subgroups of G of order p^3 is congruent to 1 modulo p.