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

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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.

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