# 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 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 $G' = G''$.

## Proof

### 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 $G/L$.

## 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