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).

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) uses::Nilpotent implies solvable
 * 2) uses::Finite non-nilpotent and every proper subgroup is nilpotent implies not simple
 * 3) uses::Nilpotency is quotient-closed
 * 4) uses::Solvability is extension-closed

Proof
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: