Abelian-to-normal replacement theorem for prime-square index
This article defines a replacement theorem
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Contents
Statement
Let be any prime number (possibly ). Then, if is a finite -group (i.e., a group of prime power order where the underlying prime is ) and is an abelian subgroup of index in , then there is an abelian normal subgroup of of index . Moreover, we can choose this abelian normal subgroup so that it is contained in the normal closure of in .
Related facts
Stronger facts
Facts used
Proof
Given: A finite -group for some prime , an abelian subgroup of of index .
To prove: There exists an abelian normal subgroup of of index , contained inside the normal closure of in .
Proof:
- If is normal in , we are done. Otherwise, is a 2-subnormal subgroup and its normal closure is a maximal subgroup of containing . is normal and has index in , and is normal and has index in .
- By fact (1), the number of abelian subgroups of is congruent to modulo .
- Since is normal in , acts on by conjugation and the orbits of abelian maximal subgroups are therefore of size equal to a power of . Since the total number of abelian maximal subgroups is modulo , there exists an orbit of size , and the member of this orbit is thus an abelian normal subgroup of index .
References
Journal references
- Large abelian subgroups of p-groups by Jonathan Lazare Alperin, Transactions of the American Mathematical Society, Volume 117, Page 10 - 20(Year 1965): ^{Official copy}^{More info}: This seems to be the first paper where a proof appeared, though the result may have been known or observed by others earlier.
- 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}: Here, they prove stronger results including abelian-to-normal replacement theorem for prime-cube index for odd prime and congruence condition on number of abelian subgroups of prime-square index for odd prime.