Congruence condition on number of abelian subgroups of prime index
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Contents
Definition
Suppose is a prime number and is a finite -group that has an abelian maximal subgroup, i.e., an abelian subgroup of index . Then, the number of abelian subgroups of index is congruent to modulo .
More strongly, if is non-abelian, the number is either or .
Related facts
Similar facts for groups
- Elementary abelian-to-normal replacement theorem for prime-square order
- Elementary abelian-to-normal replacement theorem for prime-cube and prime-fourth order
- Jonah-Konvisser elementary abelian-to-normal replacement theorem
- Jonah-Konvisser abelian-to-normal replacement theorem
- Jonah-Konvisser congruence condition for abelian subgroups of prime-square index
Similar facts for rings
Facts used
Proof
Proof using the line lemma (weaker version)
Given: A prime , a finite -group . An abelian maximal subgroup of . In particular, the index of in equals .
To prove: The number of abelian maximal subgroups of is congruent to modulo .
Proof:
- If is another abelian maximal subgroup of , and maximal subgroup containing is abelian: Since both and are abelian and maximal, , so is in the center of . In particular, any subgroup generated by and one element is abelian. Since has index , we obtain that any maximal subgroup containing it is generated by it and one element, hence is abelian. (see fact (2)).
- If is also an abelian maximal subgroup of , the number of abelian maximal subgroups containing is congruent to modulo : From the previous step, the number of such subgroups is equal to the number of subgroups of index containing , which in turn equals the number of subgroups of of order , which is .
- The result now follows from fact (1), since we have essentially shown that is an origin.
Proof of stronger version for non-abelian groups
Given: A prime , a finite non-abelian -group . An abelian maximal subgroup of . In particular, the index of in equals .
To prove: The number of abelian maximal subgroups of is or .
Proof: We consider three cases:
- There is exactly one abelian maximal subgroup : In this case, we are done.
- There are two abelian maximal subgroups , and all abelian maximal subgroups contain : First, we note that any maximal subgroup containing is generated by and a single element. Since , is central, so each such maximal subgroup containing it is abelian. Thus, the abelian maximal subgroups are in bijection with subgroups of order in .
- There are three distinct abelian maximal subgroups such that none contains the intersection of the other two: Since , is in the center of . Similarly, is in the center of and is in the center of . Thus, we have three distinct subgroups of index in the center of . The join is either the whole group or has index . If it is the whole group, we are done. If it has index , we apply fact (2) and we are again done.
References
Journal 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}