Finite non-nilpotent and every proper subgroup is nilpotent implies not simple
- Nilpotent implies normalizer condition: In a nilpotent group, every proper subgroup is properly contained in its normalizer.
- Finite and any two maximal subgroups intersect trivially implies not simple non-abelian
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Given: A finite non-nilpotent group such that every proper subgroup of is nilpotent.
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||The intersection of any two maximal subgroups of is normal in .||Fact (1)|| Every proper subgroup of is nilpotent.
|2||Any two maximal subgroups of intersect trivially.||is simple||Step (1)||[SHOW MORE]|
|3||We have the desired contradiction.||Fact (2)||is simple non-nilpotent, hence simple non-abelian||Step (2)||Follows directly from Fact (2).|