Finite non-nilpotent and every proper subgroup is nilpotent implies not simple

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Suppose G is a finite group that is not a nilpotent group, but every proper subgroup of G is a nilpotent group (and in particular, a finite nilpotent group). Then, G cannot be a simple group.

Facts used

  1. Nilpotent implies normalizer condition: In a nilpotent group, every proper subgroup is properly contained in its normalizer.
  2. Finite and any two maximal subgroups intersect trivially implies not simple non-abelian


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Given: A finite non-nilpotent group G such that every proper subgroup of G is nilpotent.

To prove: G is not simple.

Proof: We assume that G is simple, and derive a contradiction. Let n be the number of elements of G.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The intersection of any two maximal subgroups of G is contained in an intersection of two maximal subgroups that is a normal subgroup of G. Fact (1) Every proper subgroup of G is nilpotent.
G is finite
2 Any two maximal subgroups of G intersect trivially. G is simple Step (1) [SHOW MORE]
3 We have the desired contradiction. Fact (2) G is simple non-nilpotent, hence simple non-abelian Step (2) Follows directly from Fact (2).