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.
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 $C_{G}(H)\leq H$. When $H$ is Abelian, this is equivalent to saying $C_{G}(H)=H$.

Related facts

• Maximal among abelian normal implies self-centralizing in nilpotent
• Maximal among abelian normal not implies self-centralizing in solvable

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, $H\leq C$.
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

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 185, Lemma 3.12 (Section 5.3): proves it in the case that $G$ is a group of prime power order