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

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 fact about::collection of groups satisfying a universal congruence condition. Thus, it is also a fact about::collection of groups satisfying a strong normal replacement condition and a fact about::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.


 * Congruence condition on number of abelian subgroups of prime-fourth order
 * Congruence condition on number of abelian subgroups of order eight and exponent dividing four
 * Congruence condition on number of abelian subgroups of order sixteen and exponent dividing eight
 * Abelian-to-normal replacement theorem for prime-cube order
 * Abelian-to-normal replacement theorem for prime-fourth order
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: For odd $$p$$, the result extends to $$p^4$$ and $$p^5$$.
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
 * Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime

Facts used

 * 1) uses::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) uses::Congruence condition on number of abelian subgroups of prime index
 * 3) uses::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) uses::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 $$p$$.