## A finite group whose every proper subgroup is nilpotent is solvable

Prove that if $G$ is a finite group in which every proper subgroup is nilpotent, then $G$ is solvable.

Let $\mathcal{C}$ denote the class of all finite groups $G$ such that every proper subgroup of $G$ is nilpotent and $G$ is not solvable. Choose $G \in \mathcal{C}$ of minimal order.

Suppose $G$ is not simple; say $N \leq G$ is nontrivial and normal. Note that every proper subgroup of $N$ and of $G/N$ is nilpotent, and that $N$ and $G/N$ have cardinality strictly less than that of $G$. Since $G$ is minimal in $\mathcal{C}$, $N$ and $G/N$ are solvable, hence $G$ is solvable, a contradiction.

Thus we may assume that $G$ is simple; that is, that $G$ has no normal subgroups. Suppose 1 is maximal in $G$; then $G$ has no proper subgroups and we have $G \cong Z_p$ for some prime $p$. But then $G$ is abelian, hence solvable, a contradiction. Thus no maximal subgroup of $G$ is trivial.

Let $M$ and $N$ be distinct maximal subgroups and suppose $M \cap N \neq 1$; then $N_G(M \cap N) \neq G$. Note that $M$ and $N$ are each nilpotent, so that $M \cap N < N_M(M \cap N)$ and $M \cap N < N_N(M \cap N)$ are strict. Thus $M \cap N \leq N_M(M \cap N) \cap N_N(M \cap N)$. Recall that if $H \leq G$, then $N_H(A) = N_G(A) \cap H$ for all subsets $A \subseteq G$, so that we have $M \cap N \leq N_G(M \cap N) \cap M \cap N$, hence $M \cap N = N_G(M\cap N)$, and thus $M \cap N = N_M(M \cap N)$. Since $M$ is nilpotent, by Theorem 3, $M \cap N \leq M$ is not proper and in fact $M = N$, a contradiction. Thus for all distinct maximal subgroups $M$ and $N$ of $G$, $M \cap N = 1$.

Let $M \leq G$ be maximal. Now $M$ is not normal in $G$, so that by §4.3 #23, the number of elements contained in conjugates of $M$ is at most $(|M|-1)[G:M] = |G| - |G|/|M| < |G|$. Some element of $G$ is not contained in a conjugate of $M$; since $G$ is finite, every element is contained in some maximal subgroup. Thus $G$ has at least two nonconjugate maximal subgroups. Call these $M$ and $N$.

Note that $MN = G$ and $M \cap N = 1$; moreover, no two maximal subgroups intersect nontrivially. Thus, via the proof of this previous exercise, counting the number of nonidentity elements in conjugates of $M$ and $N$, respectively, we see that $G$ contains at least $(|M|-1)[G:M] + (|N|-1)[G:N] = |G| + [|M| \cdot |N| - (|M| + |N|)]$ elements. Since $M$ and $N$ are nontrivial, $|M| \cdot |N| > |M| + |N|$, and $G$ has too many elements.

Thus no such group $G$ exists.