A well-known theorem of O. Schmidt

Question

Prove that if all the proper subgroups of a finite group $G$ are nilpotent, then $G$ is soluble.

How to I prove it?

Thanks in advance.

Answer 1

A rough outline of the proof is given here: http://groupprops.subwiki.org/wiki/Schmidt-Iwasawa_theorem.

Answer 2

This theorem was proved in the paper (unfortunately in Russian):

Otto Schmidt, Über Gruppen, deren sämtliche Teiler spezielle Gruppen sind. Mat. Sb., 31:3-4 (1924), 366–372

See http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=6900&option_lang=eng

Answer 3

Outline for the usual proof:

Proposition 1: Suppose that $G$ is a nonabelian finite group and that intersections of distinct maximal subgroups is trivial. Then $G$ is not simple.

Proof idea: Suppose that $G$ is simple. Now there exists a maximal subgroup $K$ in $G$. Show that $K$ along with its conjugates contains $[G:K](|K| - 1) = |G| - [G:K]$ nonidentity elements. Conclude that there exists a maximal subgroup $L$ that is not conjugate to $K$. As before, the subgroup $L$ along with its conjugates contains $|G| - [G:L]$ nonidentity elements. Since $[G:K] \leq |G|/2$ and $[G:L] \leq |G|/2$, it follows that are $\geq |G|$ nonidentity elements in $G$, a contradiction.

Proposition 2: Suppose that $G$ is a nonabelian finite group and that every proper subgroup of $G$ is nilpotent. Then $G$ is not simple.

Proof idea: Suppose that $G$ is simple. Since $G$ is nonabelian, we can choose two distinct maximal subgroups $K$ and $L$ such that the order of $D = K \cap L$ is as large as possible. Try to prove that the only maximal subgroup containing $N_K(D)$ is $K$ and that the only maximal subgroup containing $N_L(D)$ is $L$. Then $N_G(D)$ is not contained in a maximal subgroup, so $N_G(D) = G$. Since $G$ is simple, we have $D = 1$, but this is a contradiction by proposition 1.

With proposition 2, the theorem should not be too difficult to prove.