Congruence condition on number of elementary abelian subgroups of prime-cube and prime-fourth order for odd prime

From Groupprops
Jump to: navigation, search
This article defines a replacement theorem
View a complete list of replacement theorems| View a complete list of failures of replacement
This article is about a congruence condition.
View other congruence conditions

Statement

In terms of a universal congruence condition

Suppose p is an odd prime and k = 3 or k = 4. Then, the singleton set comprising the elementary abelian group of order p^k is a Collection of groups satisfying a universal congruence condition (?) for the prime p.

Hands-on statement

Let p be an odd prime and k = 3 or k = 4. Then, if P is a finite p-group, the number of elementary abelian subgroups of P of order p^k is either equal to zero or congruent to 1 modulo p.

Related facts

Facts used

  1. Jonah-Konvisser lemma for elementary abelian-to-normal replacement for prime-cube and prime-fourth order: This states that if E_1, E_2 are two distinct elementary abelian normal subgroups of P, and N = E_1E_2 satisfies |N'| \le p^2 or |N/Z(N)| \le p^2, then any maximal subgroup of N containing E_1 \cap E_2 contains an elementary abelian subgroup of order p^k.
  2. Support of good lines corollary to line lemma, Jonah-Konvisser line lemma

Proof

Preliminary lemma

Given: p odd, k = 3 or k = 4, and E_1,E_2 are normal subgroups of a finite p-group both of order p^k. N = E_1E_2.

To prove: [N,N] has order at most p^2 or N/Z(N) has order at most p^2.

Proof: Since N = E_1E_2 and both E_1 and E_2 are abelian and normal, [N,N] = [E_1,E_2] \le E_1 \cap E_2 \le Z(N). We consider the cases:

  1. E_1 \cap E_2 has order at most p^2: In this case, [N,N] has order at most p^2.
  2. E_1 \cap E_2 has order p^3: In this case, N = E_1E_2 has order p^5, so N/Z(N) has order p^2.

Main proof

We prove the statement by induction on the order of the finite p-group. Here k = 3 or k = 4.

Given: A finite p-group P containing an elementary abelian subgroup of order p^k.

To prove: The number of elementary abelian subgroups of P of order p^k is congruent to 1 modulo p.

Proof: If P itself has order p^k, the number of such subgroups is 1. Thus, we assume that the order of P is greater than p^k.

  1. There is a maximal subgroup M of P containing the elementary abelian subgroup: This follows because the elementary abelian subgroup is proper.
  2. There is an elementary abelian subgroup E of G of order p^k (in fact, E \le M): This follows from the various equivalent formulations of universal congruence condition for M. Explicitly, the number of elementary abelian subgroups of order p^k in M is congruent to 1 modulo p. The action of G on these subgroups by conjugation has orbits of size 1 or multiples of p, so there is at least one orbit of size 1, and this gives a normal subgroup.
  3. Any pair of distinct elementary abelian normal subgroups supports good lines for the collection of elementary abelian subgroups of order p^k. In other words, if E_1,E_2 are distinct elementary abelian normal subgroups of order p^k, any maximal subgroup containing E_1 \cap E_2 contains an elementary abelian subgroup of order p^k:
    1. If N = E_1E_2, then either |N'| \le p^2 or |N/Z(N)| \le p^2: This is the lemma proved above.
    2. Any maximal subgroup of N containing E_1 \cap E_2 contains an elementary abelian subgroup of order p^k: This follows from the previous step and fact (1).
    3. Any maximal subgroup of P containing E_1 \cap E_2 contains an elementary abelian subgroup of order p^k. In other words, the pair E_1,E_2 supports good lines: This follows from the previous step and the fact that any maximal subgroup of P contains a maximal subgroup of E_1E_2.
  4. The result now follows from fact (2).

References