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

Further facts about every proper subgroup being abelian

• Finite non-abelian and every proper subgroup is abelian implies metabelian
• Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center
• Classification of finite non-abelian groups in which every proper subgroup is abelian
• Schmidt-Iwasawa theorem

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.

Facts similar to ideas used in the proof techniques

• Union of all conjugates is proper
• Product of conjugates is proper

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

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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] Since $G$ is simple, its center, which is a normal subgroup (fact (1)), must be either equal to $G$ or trivial. The center of $G$ cannot equal $G$ since $G$ is non-abelian, so the center of $G$ must be trivial.
2	Any two maximal subgroups of $G$ intersect trivially		every proper subgroup of $G$ is abelian	Step (1)	[SHOW MORE] Suppose $A,B\leq G$ are maximal, and $g\in A\cap B$. Then the centralizer of $g$ in $G$ contains both $A$ and $B$ since, by assumption, $A$ and $B$ are abelian subgroups. Thus, $g$ is centralized by $\langle A,B\rangle$. Further, since $A$ and $B$ are maximal subgroups, we obtain $\langle A,B\rangle =G$, and thus, $G$ centralizes $g$, yielding that $g$ is in the center of $G$. Since $G$ is centerless by step (1), we obtain that $g$ is the identity element, yielding that $A\cap B$ is trivial.
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}