Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
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Statement
Suppose is an odd prime and is a finite -group having an abelian subgroup of index . Then, one of these two cases holds:
- The number of abelian subgroups of index is congruent to modulo .
- There are exactly two abelian subgroups of index . (In particular, since , 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 .
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