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

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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

References

Journal references

• Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)More info}, Theorem 6.1