Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime

In terms of universal congruence condition
Suppose $$p$$ is an odd prime, and $$0 \le d \le k \le 5$$. Then, the collection of abelian groups of order $$p^k$$ and exponent dividing $$p^d$$ is a fact about::collection of groups satisfying a universal congruence condition.

Hands-on statement
Suppose $$p$$ is an odd prime, and $$P$$ is a finite $$p$$-group. Suppose $$0 \le d \le k \le 5$$. Suppose $$P$$ has an abelian subgroup of order $$p^k$$ and exponent dividing $$p^d$$. Then, the following equivalent statements hold:


 * 1) The number of abelian subgroups of $$P$$ of order $$p^k$$ and exponent dividing $$p^d$$ is either equal to zero or congruent to $$1$$ modulo $$p$$.
 * 2) The number of abelian normal subgroups of $$P$$ of order $$p^k$$ and exponent dividing $$p^d$$ is congruent to $$1$$ modulo $$p$$.
 * 3) For any finite $$p$$-group $$L$$ containing $$P$$, the number of abelian subgroups of $$P$$ of order $$p^k$$ and exponent dividing $$p^d$$ that are normal in $$L$$ is congruent to $$1$$ modulo $$p$$.

Related facts

 * 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