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

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Suppose G is a non-abelian finite group such that every proper subgroup of G is an abelian group. Then, G is not a simple group.

Related facts

Further facts about every proper subgroup being abelian


  • 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.


Facts similar to ideas used in the proof techniques

Facts used

  1. Center is normal
  2. Subgroup of index two is normal
  3. Finite and any two maximal subgroups intersect trivially implies not simple non-abelian


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

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 G is centerless, i.e., its center is trivial Fact (1) G is simple non-abelian -- [SHOW MORE]
2 Any two maximal subgroups of G intersect trivially every proper subgroup of G is abelian Step (1) [SHOW MORE]
3 We have the desired contradiction. Fact (3) G is finite simple non-abelian Step (2) Follows from Fact (3).


Textbook references

  • 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