# 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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## 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 $C_G(H) = H$.

## 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.