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

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

• Finite non-abelian and every proper subgroup is abelian implies not simple
• Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center

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] Since $G$ is finite and nontrivial (note that nontriviality follows from being non-abelian), it has a maximal normal subgroup. Let $N$ be one such subgroup. Then, $G/N$ must be simple (by the definition of maximal normal subgroup). Moreover, since $N$ is proper, the given data shows that it must be abelian.
2	Every proper subgroup of $G/N$ is abelian.		Every proper subgroup of $G$ is abelian	Step (1)	[SHOW MORE] By Fact (2) (the fourth isomorphism theorem), every proper subgroup of $G/N$ is the image under the quotient map of a proper subgroup of $G$ containing $N$. By assumption, every proper subgroup of $G$ is abelian, so by Fact (3), we obtain that its image is abelian. Thus, every proper subgroup of $G/N$ is Abelian.
3	$G/N$ is a simple abelian group.			Steps (1), (2)	[SHOW MORE] By Step (1), $G/N$ is simple, and by Step (2), every proper subgroup of $G/N$ is abelian. If $G/N$ were non-abelian, we would obtain a contradiction to Fact (1). This forces $G/N$ to be abelian. Note that by Fact (4), we also obtain that $G/N$ is cyclic of prime order.

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}