Finite non-abelian and every proper subgroup is abelian implies metabelian

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Statement

Suppose G is a non-abelian finite group such that every proper subgroup of G is abelian. Then, G is a metabelian group: it has derived length two. In fact, G has an abelian maximal normal subgroup that has prime index.

Related facts

Facts used

  1. Finite non-abelian and every proper subgroup is abelian implies not simple
  2. Fourth isomorphism theorem
  3. Abelianness is quotient-closed
  4. Simple abelian implies cyclic of prime order

Proof

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Given: A finite non-Abelian group G in which every proper subgroup is abelian.

To prove: G is a metabelian group: it has derived length two.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 G has at least one maximal normal subgroup. Suppose N is a maximal normal subgroup of G. Then, N is abelian and G/N is simple. G is finite and non-abelian; every proper subgroup of G is abelian [SHOW MORE]
2 Every proper subgroup of G/N is abelian. Every proper subgroup of G is abelian Step (1) [SHOW MORE]
3 G/N is a simple abelian group. Steps (1), (2) [SHOW MORE]

Thus, we have found an abelian normal subgroup N such that G/N is also abelian, yielding that G is metabelian, i.e., has derived length two.

References

Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 149, Exercise 56 of Section 4.5 (Sylow's theorem), (states a somewhat weaker version: only requires a proof of solvability)More info