Characteristically metacyclic and commutator-realizable implies abelian

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This fact is related to the problem of realization related to the following subgroup-defining function: commutator subgroup
Realization problems are usually about which groups can be realized as subgroups/quotients related to a subgroup-defining function.
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Statement

Statement with symbols

The statement has two forms:

1. Suppose $G$ is a 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 a cyclic 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 cyclic.

A particular case of this is as follows: if $G/G^{\prime }$ and $G^{\prime }/G^{″}$ are both cyclic, and $G$ occurs as the commutator subgroup of some group, then $G^{\prime }=G^{″}$.

Applications

• Metacyclic derived series and commutator-realizable implies cyclic

Facts used

1. Commutator subgroup is normal
2. Characteristic of normal implies normal
3. Commutator subgroup centralizes cyclic normal subgroup
4. Cyclic over central implies abelian

Proof

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.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 198, Exercise 18, Section 6.1