Congruence condition on number of abelian subgroups of prime index

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This article is about a congruence condition.
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Definition

Suppose p is a prime number and P is a finite p-group that has an abelian maximal subgroup, i.e., an abelian subgroup of index p. Then, the number of abelian subgroups of index p is congruent to 1 modulo p.

More strongly, if P is non-abelian, the number is either 1 or p + 1.

Related facts

Similar facts for groups

Similar facts for rings

Facts used

  1. Jonah-Konvisser line lemma
  2. Cyclic over central implies abelian

Proof

Proof using the line lemma (weaker version)

Given: A prime p, a finite p-group P. An abelian maximal subgroup M of P. In particular, the index of M in P equals p.

To prove: The number of abelian maximal subgroups of P is congruent to 1 modulo p.

Proof:

  1. If N is another abelian maximal subgroup of P, and maximal subgroup containing M \cap N is abelian: Since both M and N are abelian and maximal, MN = P, so M \cap N is in the center of P. In particular, any subgroup generated by M \cap N and one element is abelian. Since M \cap N has index p^2, we obtain that any maximal subgroup containing it is generated by it and one element, hence is abelian. (see fact (2)).
  2. If N is also an abelian maximal subgroup of P, the number of abelian maximal subgroups containing M \cap N is congruent to 1 modulo p: From the previous step, the number of such subgroups is equal to the number of subgroups of index p containing M \cap N, which in turn equals the number of subgroups of P/(M \cap N) of order p, which is p + 1.
  3. The result now follows from fact (1), since we have essentially shown that M is an origin.

Proof of stronger version for non-abelian groups

Given: A prime p, a finite non-abelian p-group P. An abelian maximal subgroup M of P. In particular, the index of M in P equals p.

To prove: The number of abelian maximal subgroups of P is 1 or p + 1.

Proof: We consider three cases:

  1. There is exactly one abelian maximal subgroup M: In this case, we are done.
  2. There are two abelian maximal subgroups M,N, and all abelian maximal subgroups contain M \cap N: First, we note that any maximal subgroup containing M \cap N is generated by M \cap N and a single element. Since MN = P, M \cap N is central, so each such maximal subgroup containing it is abelian. Thus, the abelian maximal subgroups are in bijection with subgroups of order p in P/(M \cap N).
  3. There are three distinct abelian maximal subgroups M,N,Q such that none contains the intersection of the other two: Since MN = P, M \cap N is in the center of P. Similarly, M \cap Q is in the center of P and N \cap Q is in the center of P. Thus, we have three distinct subgroups of index p^2 in the center of P. The join is either the whole group or has index p. If it is the whole group, we are done. If it has index p, we apply fact (2) and we are again done.

References

Journal references