Congruence condition on number of elementary abelian subgroups of prime-cube and prime-fourth order for odd prime
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Contents
Statement
In terms of a universal congruence condition
Suppose is an odd prime and or . Then, the singleton set comprising the elementary abelian group of order is a Collection of groups satisfying a universal congruence condition (?) for the prime .
Hands-on statement
Let be an odd prime and or . Then, if is a finite -group, the number of elementary abelian subgroups of of order is either equal to zero or congruent to modulo .
Related facts
- Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime: This is a stronger version that includes all .
- Congruence condition on number of elementary abelian subgroups of prime-square order for odd prime: An easier version covering the case .
- Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: This is a stronger version that includes all .
Facts used
- Jonah-Konvisser lemma for elementary abelian-to-normal replacement for prime-cube and prime-fourth order: This states that if are two distinct elementary abelian normal subgroups of , and satisfies or , then any maximal subgroup of containing contains an elementary abelian subgroup of order .
- Support of good lines corollary to line lemma, Jonah-Konvisser line lemma
Proof
Preliminary lemma
Given: odd, or , and are normal subgroups of a finite -group both of order . .
To prove: has order at most or has order at most .
Proof: Since and both and are abelian and normal, . We consider the cases:
- has order at most : In this case, has order at most .
- has order : In this case, has order , so has order .
Main proof
We prove the statement by induction on the order of the finite -group. Here or .
Given: A finite -group containing an elementary abelian subgroup of order .
To prove: The number of elementary abelian subgroups of of order is congruent to modulo .
Proof: If itself has order , the number of such subgroups is . Thus, we assume that the order of is greater than .
- There is a maximal subgroup of containing the elementary abelian subgroup: This follows because the elementary abelian subgroup is proper.
- There is an elementary abelian subgroup of of order (in fact, ): This follows from the various equivalent formulations of universal congruence condition for . Explicitly, the number of elementary abelian subgroups of order in is congruent to modulo . The action of on these subgroups by conjugation has orbits of size or multiples of , so there is at least one orbit of size , and this gives a normal subgroup.
- Any pair of distinct elementary abelian normal subgroups supports good lines for the collection of elementary abelian subgroups of order . In other words, if are distinct elementary abelian normal subgroups of order , any maximal subgroup containing contains an elementary abelian subgroup of order :
- If , then either or : This is the lemma proved above.
- Any maximal subgroup of containing contains an elementary abelian subgroup of order : This follows from the previous step and fact (1).
- Any maximal subgroup of containing contains an elementary abelian subgroup of order . In other words, the pair supports good lines: This follows from the previous step and the fact that any maximal subgroup of contains a maximal subgroup of .
- The result now follows from fact (2).
References
- 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}, Theorem 2.3