Characteristically metacyclic and commutator-realizable implies abelian

Statement with symbols
The statement has two forms:


 * 1) Suppose $$G$$ is a fact about::characteristically metacylic group: in other words, there exists a cyclic characteristic subgroup $$K$$ of $$G$$ such that $$G/K$$ is also cyclic. Further, suppose that $$G$$ is a commutator-realizable group: there exists a group $$H$$ such that $$[H,H] = G$$, i.e., $$G$$ occurs as the commutator subgroup of some group. Then, $$G$$ is an abelian group.
 * 2) Suppose $$G$$ is commutator-realizable, and has a characteristic subgroup such that the quotient is a nontrivial characteristically metacyclic quotient group. Then, that quotient group must be abelian.

A particular case of this is as follows: if $$G/G'$$ and $$G'/G$$ are both cyclic, and $$G$$ occurs as the commutator subgroup of some group, then $$G' = G$$.

Applications

 * Metacyclic derived series and commutator-realizable implies cyclic

Facts used

 * 1) uses::Commutator subgroup is normal
 * 2) uses::Characteristic of normal implies normal
 * 3) uses::Commutator subgroup centralizes cyclic normal subgroup
 * 4) uses::Cyclic over central implies abelian

Proof in the first form
Given: A group $$H$$, $$G = [H,H]$$, and $$K$$ is a characteristic subgroup of $$G$$ such that both $$K$$ and $$G/K$$ are cyclic.

To prove: $$G$$ is abelian.

Proof:


 * 1) $$G$$ is normal in $$H$$: This is fact (1).
 * 2) $$K$$ is normal in $$H$$: This follows from facts (2), the fact that $$K$$ is characteristic in $$G$$, and the fact that $$G$$ is normal in $$H$$.
 * 3) $$G$$ centralizes $$K$$: This follows from fact (3): the commutator subgroup of $$H$$ (which is $$G$$) must centralize the cyclic normal subgroup $$K$$.
 * 4) $$G$$ is abelian: From the previous step, we obtain that $$K$$ is central in $$G$$. Further, we are given that $$G/K$$ is cyclic, so by fact (4), we obtain that $$G$$ is abelian.

Proof in the second form
Given: A group $$H$$. $$G = [H,H]$$. $$G$$ has a characteristic subgroup $$L$$ such that $$G/L$$ is characteristically metacyclic.

To prove: $$G/L$$ is abelian.

Proof:


 * 1) $$G$$ is normal in $$H$$: This is fact (1).
 * 2) $$L$$ is normal in $$H$$: This is fact (2), along with the given data that $$L$$ is characteristic in $$G$$ and $$G$$ is normal in $$H$$.
 * 3) $$G/L = [H/L,H/L]$$, and hence, $$G/L$$ is realizable as a commutator subgroup: Since both $$G$$ and $$H$$ contain $$L$$, this follows by the definition of commutator subgroup.

The result now follows by applying the result in its first form to the group $$G/L$$.

Textbook references

 * , Page 198, Exercise 18, Section 6.1