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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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.


Suppose p is an odd prime. Then, if a finite p-group contains an abelian subgroup of index p^3, it contains an abelian normal subgroup (hence, an Abelian normal subgroup of group of prime power order (?)) of index p^3.

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


The Jonah-Konvisser proof

Given: An odd prime p, a finite p-group P, an abelian subgroup A of P of index p^3.

To prove: P has an abelian normal subgroup of index p^3.


  1. Let M be a maximal subgroup of P containing A: Such a M exists and has index p in P.
  2. By fact (1), the number of abelian subgroups of M of index p^2 is either equal to 2 or congruent to 1 modulo p.
  3. If M has exactly two abelian subgroups of index p^2, then both of them are normal in P: Since M is maximal in P, it is normal (See fact (2)). Thus, any inner automorphism of P sends subgroups of M to subgroups of M. Since the sizes of orbits are either 1 or powers of p, and there are only two abelian subgroups of index p^2, they must both be normal in P.
  4. If the number of abelian subgroups of M of index p^2 is congruent to 1 modulo p, then there exists a subgroup of index p^2 in M that is normal in P: The inner automorphisms of P permute abelian subgroups of index p^2 in M, since M is normal in P. All orbits have size either 1 or a nontrivial power (and hence, multiple) of p. Since the total number is 1 modulo p, there must be at least one orbit of size one.

Alperin's proof



Journal references