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

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Statement

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 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.

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