Abelian-to-normal replacement theorem for prime-cube order

Statement in terms of weak normal replacement condition
Let $$p$$ be a prime number. Then, the collection of abelian groups of order $$p^3$$ is a fact about::collection of groups satisfying a weak normal replacement condition.

Hands-on statement
Suppose $$p$$ is a prime number and $$P$$ is a finite $$p$$-group. If $$A$$ is an abelian subgroup of $$P$$ of order $$p^3$$, there is an abelian normal subgroup $$B$$ of $$P$$ of order $$p^3$$.

Related facts

 * Congruence condition on number of abelian subgroups of prime-cube order
 * Congruence condition on number of abelian subgroups of prime-fourth 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 $$p$$ odd and $$0 \le k \le 5$$, the number of abelian subgroups of order $$p^k$$ is either zero or $$1$$ mod $$p$$.
 * Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime
 * Congruence condition on number of abelian subgroups of small prime power order and bounded exponent 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 finite $$p$$-group of order $$p^n$$ has an abelian normal subgroup of order $$p^k$$.

Proof
Given: A finite $$p$$-group $$P$$ of order $$p^n, n \ge 3$$, containing an abelian subgroup $$A$$ of order $$p^3$$.

To prove: $$P$$ contains an abelian normal subgroup $$B$$ of order $$p^3$$.

Proof: If $$A = P$$, then it is normal and we can set $$B = A$$. Thus, we assume that $$A$$ is a proper subgroup of $$P$$.

In this case, since $$n \ge 4$$, fact (1) tells us that $$P$$ has an abelian normal subgroup of order $$p^3$$, and we are done.