Maximal among abelian normal implies self-centralizing in supersolvable

From Groupprops
Jump to: navigation, search
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.
View all subgroup property implications in supersolvable groups | View all subgroup property non-implications in supersolvable groups | View all subgroup property implications | View all subgroup property non-implications
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
View other similar statements

Statement

Verbal statement

Suppose G is a supersolvable group and H is maximal among abelian normal subgroups of G: in other words, H is an Abelian normal subgroup of G, and there is no Abelian normal subgroup of G strictly containing H. Then, H is a self-centralizing subgroup; in other words:

C_G(H) = H

where C_G(H) denotes the centralizer of H.

Definitions used

Supersolvable group

Further information: Supersolvable group

A group G is termed supersolvable if there exists a normal series of G such that all the factor groups are cyclic groups.

Self-centralizing subgroup

Further information: Self-centralizing subgroup

A subgroup H of a group G is termed self-centralizing if C_G(H) \le H. When H is Abelian, this is equivalent to saying C_G(H) = H.

Related facts

Facts used

  1. Normality is centralizer-closed: The centralizer of a normal subgroup is normal.
  2. Normality satisfies image condition: The image of any normal subgroup under a surjective homomorphism, is a normal subgroup of the image.
  3. Supersolvability is quotient-closed: A quotient of a supersolvable group by any normal subgroup is still supersolvable.
  4. Supersolvable implies every nontrivial normal subgroup contains a cyclic normal subgroup
  5. Normality satisfies inverse image condition: The inverse image of a normal subgroup, under any homomorphism, is a normal subgroup.
  6. Cyclic over central implies abelian

Proof

Given: A supersolvable group G, a subgroup H that is maximal among Abelian normal subgroups.

To prove: C_G(H) = H

Proof:

  1. Suppose C = C_G(H) is the centralizer of H. Then, C is normal and contains H: Since H is normal, so is C (by fact (1)). Since H is abelian, H \le C.
  2. Let \overline{C} = C/H and \overline{G} = G/H. Then, \overline{C} is normal in \overline{G}: This follows from fact (2).
  3. \overline{G} is a supersolvabe group: This follows from fact (3).
  4. \overline{C} contains a cyclic normal subgroup of \overline{G}, say, generated by \overline{x}, x \in G: This follows from fact (4).
  5. The subgroup \langle H,x \rangle is an abelian normal subgroup of G properly containing H:
    • This is abelian, since, by definition x \in C, so x commutes with all the elements of H, which is itself abelian.
    • This is normal in G, since it is the inverse image of a normal subgroup of \overline{G} under a quotient map (and by fact (5)).

Hence we have found an abelian normal subgroup of G properly containing H, a contradiction to the assumption of maximality.

References

Textbook references