Abelian-to-normal replacement theorem for prime-cube index for odd prime
This article defines a replacement theorem
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Suppose is an odd prime. Then, if a finite -group contains an abelian subgroup of index , it contains an abelian normal subgroup (hence, an Abelian normal subgroup of group of prime power order (?)) of index .
- Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
- Prime power order implies nilpotent, nilpotent implies every maximal subgroup is normal
The Jonah-Konvisser proof
Given: An odd prime , a finite -group , an abelian subgroup of of index .
To prove: has an abelian normal subgroup of index .
- Let be a maximal subgroup of containing : Such a exists and has index in .
- By fact (1), the number of abelian subgroups of of index is either equal to or congruent to modulo .
- If has exactly two abelian subgroups of index , then both of them are normal in : Since is maximal in , it is normal (See fact (2)). Thus, any inner automorphism of sends subgroups of to subgroups of . Since the sizes of orbits are either or powers of , and there are only two abelian subgroups of index , they must both be normal in .
- If the number of abelian subgroups of of index is congruent to modulo , then there exists a subgroup of index in that is normal in : The inner automorphisms of permute abelian subgroups of index in , since is normal in . All orbits have size either or a nontrivial power (and hence, multiple) of . Since the total number is modulo , there must be at least one orbit of size one.
Alperin's proofPLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- Large abelian subgroups of p-groups by Jonathan Lazare Alperin, Transactions of the American Mathematical Society, Volume 117, Page 10 - 20(Year 1965): Official copyMore info
- 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