Abelian-to-normal replacement theorem for prime-cube index for odd prime

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This article defines a replacement theorem
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History

The result appears to have been first proved in a paper by Alperin in 1965 and later proved as part of a larger array of results in a paper by David Jonah and Marc Konvisser in 1975.

Statement

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 .

Facts used

  1. Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
  2. Prime power order implies nilpotent, nilpotent implies every maximal subgroup is normal

Proof

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 .

Proof:

  1. Let be a maximal subgroup of containing : Such a exists and has index in .
  2. By fact (1), the number of abelian subgroups of of index is either equal to or congruent to modulo .
  3. 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 .
  4. 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 proof

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References

Journal references