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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Contents
Statement
Statement with symbols
The statement has two forms:
- Suppose is a Characteristically metacylic group (?): in other words, there exists a cyclic characteristic subgroup of such that is also cyclic. Further, suppose that is a commutator-realizable group: there exists a group such that , i.e., occurs as the commutator subgroup of some group. Then, is an abelian group.
- Suppose 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 and are both cyclic, and occurs as the commutator subgroup of some group, then .
Applications
Facts used
- Commutator subgroup is normal
- Characteristic of normal implies normal
- Commutator subgroup centralizes cyclic normal subgroup
- Cyclic over central implies abelian
Proof
Proof in the first form
Given: A group , , and is a characteristic subgroup of such that both and are cyclic.
To prove: is abelian.
Proof:
- is normal in : This is fact (1).
- is normal in : This follows from facts (2), the fact that is characteristic in , and the fact that is normal in .
- centralizes : This follows from fact (3): the commutator subgroup of (which is ) must centralize the cyclic normal subgroup .
- is abelian: From the previous step, we obtain that is central in . Further, we are given that is cyclic, so by fact (4), we obtain that is abelian.
Proof in the second form
Given: A group . . has a characteristic subgroup such that is characteristically metacyclic.
To prove: is abelian.
Proof:
- is normal in : This is fact (1).
- is normal in : This is fact (2), along with the given data that is characteristic in and is normal in .
- , and hence, is realizable as a commutator subgroup: Since both and contain , this follows by the definition of commutator subgroup.
The result now follows by applying the result in its first form to the group .
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