# 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.

## 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