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 solvable (in particular, is a finite solvable group).

In fact, something stronger is true: the group is either itself a nilpotent group or has a nilpotent maximal normal subgroup such that the quotient group is a group of prime order.

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 nilpotency

• 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: $G$ has a nilpotent maximal normal subgroup such that the quotient group is cyclic of prime order. In particular, it is metanilpotent and hence 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 maximal normal subgroup, say $N$, that is nilpotent.		Every proper subgroup of $G$ is nilpotent	Step (1)	direct from Step (1).
3	$G/N$ is cyclic of prime order.	Facts (2), (3)	inductive hypothesis	Step (2)	[SHOW MORE] Since $N$ is maximal normal, $G/N$ must be simple. However, by Fact (3), every proper subgroup of $G/N$ is nilpotent, hence it satisfies the conditions for applying Fact (2). Therefore, if $G/N$ is not nilpotent, it cannot be simple. This forces $G/N$ to be nilpotent and simple, and hence, to be a simple abelian group, and therefore a cyclic group of prime order.
4	$G$ is solvable.	Fact (4)		Steps (2), (3)	Step-fact combination direct.