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

## Statement

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

## Proof

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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).