Maximal among abelian normal implies self-centralizing in supersolvable

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 $$H$$.

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
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

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

Facts used

 * 1) uses::Normality is centralizer-closed: The centralizer of a normal subgroup is normal.
 * 2) uses::Normality satisfies image condition: The image of any normal subgroup under a surjective homomorphism, is a normal subgroup of the image.
 * 3) uses::Supersolvability is quotient-closed: A quotient of a supersolvable group by any normal subgroup is still supersolvable.
 * 4) uses::Supersolvable implies every nontrivial normal subgroup contains a cyclic normal subgroup
 * 5) uses::Normality satisfies inverse image condition: The inverse image of a normal subgroup, under any homomorphism, is a normal subgroup.
 * 6) uses::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 $$H$$:
 * 6) * This is abelian, since, by definition $$x \in C$$, so $$x$$ commutes with all the elements of $$H$$, which is itself abelian.
 * 7) * 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.

Textbook references

 * , Page 185, Lemma 3.12 (Section 5.3): proves it in the case that $$G$$ is a group of prime power order