Finite non-abelian and every proper subgroup is abelian implies not simple
Further facts about every proper subgroup being abelian
- Finite non-abelian and every proper subgroup is abelian implies metabelian
- Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center
- Classification of finite non-abelian groups in which every proper subgroup is abelian
- Schmidt-Iwasawa theorem
- The analogous statement for infinite groups is not true. In fact, Tarski has constructed infinite groups in which every proper nontrivial subgroup is cyclic of prime order.
- Classification of cyclicity-forcing numbers: This is a classification of all natural numbers such that every group of order is cyclic.
Facts similar to ideas used in the proof techniques
- Center is normal
- Subgroup of index two is normal
- Finite and any two maximal subgroups intersect trivially implies not simple non-abelian
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Given: A finite non-abelian group such that every proper subgroup of is abelian.
To prove: is not simple.
Proof: We assume that is simple, and derive a contradiction. Let be the number of elements of .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||is centerless, i.e., its center is trivial||Fact (1)||is simple non-abelian||--||[SHOW MORE]|
|2||Any two maximal subgroups of intersect trivially||every proper subgroup of is abelian||Step (1)||[SHOW MORE]|
|3||We have the desired contradiction.||Fact (3)||is finite simple non-abelian||Step (2)||Follows from Fact (3).|
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 149, Exercise 53 of Section 4.5 (Sylow's theorem), (Hint for solution given)More info