Finite non-abelian and every proper subgroup is abelian implies metabelian
From Groupprops
Statement
Suppose is a non-abelian finite group such that every proper subgroup of is abelian. Then, is a metabelian group: it has derived length two. In fact, has an abelian maximal normal subgroup that has prime index.
Related facts
- Finite non-abelian and every proper subgroup is abelian implies not simple
- Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center
Facts used
- Finite non-abelian and every proper subgroup is abelian implies not simple
- Fourth isomorphism theorem
- Abelianness is quotient-closed
- Simple abelian implies cyclic of prime order
Proof
This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format
Given: A finite non-Abelian group in which every proper subgroup is abelian.
To prove: is a metabelian group: it has derived length two.
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | has at least one maximal normal subgroup. Suppose is a maximal normal subgroup of . Then, is abelian and is simple. | is finite and non-abelian; every proper subgroup of is abelian | [SHOW MORE] | ||
2 | Every proper subgroup of is abelian. | Every proper subgroup of is abelian | Step (1) | [SHOW MORE] | |
3 | is a simple abelian group. | Steps (1), (2) | [SHOW MORE] |
Thus, we have found an abelian normal subgroup such that is also abelian, yielding that is metabelian, i.e., has derived length two.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 149, Exercise 56 of Section 4.5 (Sylow's theorem), (states a somewhat weaker version: only requires a proof of solvability)^{More info}