Schmidt-Iwasawa theorem

Statement

If every proper subgroup of a finite group is nilpotent (in particular, is a finite nilpotent group) then the whole group is 3 (in particular, is a finite solvable group).

Related facts

Similar facts about proper subgroups being abelian and cyclic

• Classification of finite non-abelian groups in which every proper subgroup is abelian
• Finite non-abelian and every proper subgroup is abelian implies not simple
• Finite non-abelian and every proper subgroup is abelian implies metabelian
• Classification of cyclicity-forcing numbers

Related facts about weaker conditions than nilpotence

• Finite and every proper subgroup is p-nilpotent implies p-nilpotent or solvable
• Finite and every 2-submaximal subgroup is nilpotent implies solvable or A5 or SL(2,5)

Facts used

1. Nilpotent implies solvable
2. Finite non-nilpotent and every proper subgroup is nilpotent implies not simple
3. Nilpotency is quotient-closed
4. Solvability is extension-closed

Proof

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

We prove the statement using an induction on the order. In particular, we assume that the theorem has been proved for all groups of smaller orders.

Base case for induction: The base case can be considered to be order 1, which gives the trivial group that is solvable.

Inductive hypothesis: For any finite group $H$ of order $m<n$ such that every proper subgroup of $H$ is nilpotent, $H$ is solvable.

Inductive step: We need to show that for a finite group $G$ of order $n$ such that every proper subgroup of $G$ is nilpotent, $G$ is solvable.

If the group is nilpotent

In this case, the result follows from Fact (1).

If the group is not nilpotent

Given: A finite non-nilpotent group $G$ of order $n$ such that every proper subgroup of $G$ is nilpotent. The inductive hypothesis holds for all orders smaller than $n$.

To prove: $G$ is solvable.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G$ is nontrivial and not simple.	Fact (2)	$G$ is finite non-nilpotent, every proper subgroup is nilpotent.	--	Fact-direct.
2	$G$ has a proper nontrivial normal subgroup, say $N$.			Step (1)	direct from Step (1).
3	$N$ is solvable.	Fact (1)	every proper subgroup of $G$ is nilpotent	Step (2)	$N$ is proper in $G$, hence is nilpotent from the given data. Thus, by Fact (1), it is solvable.
4	$G/N$ is solvable.	Fact (3)	inductive hypothesis	Step (2)	[SHOW MORE] Any proper subgroup of $G/N$ is a quotient of a proper subgroup of $G$, namely its inverse image under the quotient map $G\to G/N$. Thus, by Fact (3), every proper subgroup of $G/N$ is nilpotent. Further, since $N$ is nontrivial, $G/N$ has strictly smaller order than $G$, so we can apply the inductive hypothesis to it and conclude that it is solvable.
5	$G$ is solvable.	Fact (4)		Steps (2), (3)	Step-fact combination direct.