Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime

Statement
Suppose $$p$$ is an odd prime and $$P$$ is a finite $$p$$-group having an abelian subgroup of index $$p^2$$. Then, one of these two cases holds:


 * 1) The number of abelian subgroups of index $$p^2$$ is congruent to $$1$$ modulo $$p$$.
 * 2) There are exactly two abelian subgroups of index $$p^2$$. (In particular, since $$p > 2$$, they cannot be conjugate to each other and thus they are both normal subgroups).

In either case, we obtain that there exists an abelian normal subgroup of index $$p^2$$.

Related facts

 * Congruence condition on number of abelian subgroups of prime index
 * Jonah-Konvisser abelian-to-normal replacement theorem for prime-cube index for odd prime
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime
 * 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

Journal references

 * , Theorem 6.1