Maximal among abelian normal implies self-centralizing in supersolvable
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a supersolvable group. That is, it states that in a Supersolvable group (?), every subgroup satisfying the first subgroup property (i.e., Maximal among abelian normal subgroups (?)) must also satisfy the second subgroup property (i.e., Self-centralizing subgroup (?)). In other words, every maximal among abelian normal subgroups of supersolvable group is a self-centralizing subgroup of supersolvable group.
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This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a group being self-centralizing. In other words, the centralizer of the subgroup in the group is contained in the subgroup
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Contents
Statement
Verbal statement
Suppose is a supersolvable group and is maximal among abelian normal subgroups of : in other words, is an Abelian normal subgroup of , and there is no Abelian normal subgroup of strictly containing . Then, is a self-centralizing subgroup; in other words:
where denotes the centralizer of .
Definitions used
Supersolvable group
Further information: Supersolvable group
A group is termed supersolvable if there exists a normal series of such that all the factor groups are cyclic groups.
Self-centralizing subgroup
Further information: Self-centralizing subgroup
A subgroup of a group is termed self-centralizing if . When is Abelian, this is equivalent to saying .
Related facts
- Maximal among abelian normal implies self-centralizing in nilpotent
- Maximal among abelian normal not implies self-centralizing in solvable
Facts used
- Normality is centralizer-closed: The centralizer of a normal subgroup is normal.
- Normality satisfies image condition: The image of any normal subgroup under a surjective homomorphism, is a normal subgroup of the image.
- Supersolvability is quotient-closed: A quotient of a supersolvable group by any normal subgroup is still supersolvable.
- Supersolvable implies every nontrivial normal subgroup contains a cyclic normal subgroup
- Normality satisfies inverse image condition: The inverse image of a normal subgroup, under any homomorphism, is a normal subgroup.
- Cyclic over central implies abelian
Proof
Given: A supersolvable group , a subgroup that is maximal among Abelian normal subgroups.
To prove:
Proof:
- Suppose is the centralizer of . Then, is normal and contains : Since is normal, so is (by fact (1)). Since is abelian, .
- Let and . Then, is normal in : This follows from fact (2).
- is a supersolvabe group: This follows from fact (3).
- contains a cyclic normal subgroup of , say, generated by : This follows from fact (4).
- The subgroup is an abelian normal subgroup of properly containing :
- This is abelian, since, by definition , so commutes with all the elements of , which is itself abelian.
- This is normal in , since it is the inverse image of a normal subgroup of under a quotient map (and by fact (5)).
Hence we have found an abelian normal subgroup of properly containing , a contradiction to the assumption of maximality.
References
Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 185, Lemma 3.12 (Section 5.3): proves it in the case that is a group of prime power order