Abstract
Let R be a cyclic group of prime order which acts on the extraspecial group F in such a way that F=[F,R]. Suppose RF acts on a group G so that CG(F)=1 and (|R|,|G|)=1. It is proved that F(CG(R))⊆F(G). As corollaries to this, it is shown that the Fitting series of CG(R) coincides with the intersections of CG(R) with the Fitting series of G, and that when |R| is not a Fermat prime, the Fitting heights of CG(R) and G are equal.
Highlights
IntroductionThey consider a soluble finite group G admitting a ‘Frobenius-like’ group of automorphisms F R of odd order such that |F | is of prime order, CG(F ) = 1, and (|G|, |R|) = 1
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If a group A acts on a group G in such a way that CG(A) = 1, one can often say something about the structure of G given properties of A
Summary
They consider a soluble finite group G admitting a ‘Frobenius-like’ group of automorphisms F R of odd order such that |F | is of prime order, CG(F ) = 1, and (|G|, |R|) = 1. (‘Frobenius-like’ means that F is a nilpotent normal subgroup and F R/F is a Frobenius group with Frobenius kernel F/F and complement R.) Theorem A of that paper asserts that Fi(CG(R)) = Fi(G) ∩ CG(R) for all i and f (CG(R)) = f (G) These results are very similar to Corollaries 1.3 and 1.4 of the present paper. We will make it clear later how Proposition 1.5 can be used to shorten the proof of Theorem 1.2
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