The Fitting length of solvable \(H_{p^ n}\)-groups
Let p be a prime and \(n\geq 1\) an integer. The generalized Hughes subgroup \(H_{p^ n}(G)\) of a finite group G is defined by \(H_{p^ n}(G)=<x\in G:\) \(x^{p^ n}\neq 1>\). It is known that if \(H_ p(G)<G\), then \(H_ p(G)\) is nilpotent. The present paper is concerned with bounding the nilpotent or Fitting length of \(H_{p^ n}(G)\), under the assumption that it is soluble and a proper subgroup of G. Actually this gives the same bound as the Fitting length h(G) of G. Under the assumptions just stated, it is shown that h(G)\(\leq 2n\) if \(p\neq 2\) and h(G)\(\leq 4n\) if \(p=2\). This improves a quadratic bound obtained by \textit{A. Rae} and the reviewer [Bull. Lond. Math. Soc. 5, 197-198 (1973; Zbl 0273.20015)]. Problems of this type are analogous to fixed point free automorphism questions, since each element \(x\in G\setminus H_{p^ n}(G)\) has order dividing \(p^ n\) and acts fixed point freely on every invariant p'-section of \(H_{p^ n}(G)\), and so the method of proof involves studying actions of sections of the upper Fitting series on one another by representation theoretic methods. It is suggested that the correct bound is perhaps h(G)\(\leq n\) if \(H_{p^ n}(G)\) is a soluble proper subgroup of G, and there is some discussion of this. \textit{A. Espuelas} [J. Algebra (to appear)] has since verified that this is the case if \(p\neq 2\). Also Turau has shown that with the same hypothesis, the 2-length of \(H_{2^ n}(G)\) is at most 2n.
scientific article; zbMATH DE number 3935291
scientific article