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

## Statement

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

### Analogues

• 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.

### Applications

• Classification of cyclicity-forcing numbers: This is a classification of all natural numbers $n$ such that every group of order $n$ is cyclic.

## Proof

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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).

## 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 53 of Section 4.5 (Sylow's theorem), (Hint for solution given)More info