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

From Groupprops
Revision as of 22:14, 20 December 2011 by Vipul (talk | contribs) (Created page with "==Statement== Suppose <math>G</math> is a finite group that is ''not'' a nilpotent group, but every proper subgroup of <math>G</math> is a [[fact about::nilpotent gr...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Statement

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

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-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 normal in G. Fact (1) Every proper subgroup of G is nilpotent.
G is finite
[SHOW MORE]
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).