Maximal elementary abelian subgroup of prime-square order implies rank at most the prime for odd prime

Statement
Let $$p$$ be an odd prime.

Suppose $$P$$ is a group of prime power order (i.e., a finite $$p$$-group for some prime $$p$$) and $$A$$ is an elementary abelian subgroup of $$P$$ of order $$p^2$$ that is not contained in any bigger elementary abelian subgroup. In other words, $$P$$ is a fact about::group of prime power order having a maximal elementary abelian subgroup of prime-square order.

Then, the rank of $$P$$ is at most $$p$$. In other words, every elementary abelian subgroup of $$P$$ is of order at most $$p^p$$.

Related facts

 * Maximal elementary abelian subgroup of prime-square order implies normal rank at most the prime
 * Maximal elementary abelian subgroup of order four implies subgroup rank at most four
 * There exists a 2-group with a maximal elementary abelian subgroup of order four and rank four