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.

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  • Lennon  On February 27, 2011 at 2:06 pm

    in the paragraph starting with “Let M and N be distinct maximal subgroups..” you say that:

    (MnN) is contained in N_G(MnN)n(MnN),which is always true.why should this imply N_G(MnN)=MnN?

    • nbloomf  On February 27, 2011 at 10:41 pm

      Good question. I can’t see how to fix this at the moment, so I’ll mark this one “Incomplete” for now.

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