Nilpotency of \(p\)-complements and \(p\)-regular conjugacy class sizes.
This is one in a series of papers where the authors consider what information can be gained about a group if we are told information about the sizes of \(p\)-regular classes of a finite \(p\)-soluble group for some prime \(p\). The reviewer [\textit{A. R. Camina}, J. Lond. Math. Soc., II. Ser. 5, 127-132 (1972; Zbl 0242.20025)] proved that if the only conjugacy class sizes were \(\{1,m,n,mn\}\) where \(\gcd(m,n)=1\) and both \(m\) and \(n\) were prime powers then the group is nilpotent. Recently in two papers the authors have shown that the assumption about \(m\) and \(n\) were unnecessary, [J. Algebra 296, No. 1, 253-266 (2006; Zbl 1091.20017), J. Group Theory 9, No. 6, 787-797 (2006; Zbl 1108.20021)]. In this paper they are considering \(p\)-soluble groups with the condition that the only sizes of \(p'\)-conjugacy classes are \(\{1,m,n,mn\}\) where \(\gcd(m,n)=1\) and ask whether this would imply that a \(p\)-complement would be nilpotent. What they show is that if \(n=p^a\) then \(m=q^b\) for some prime \(q\) and that the \(p\)-complements are nilpotent. The proof is carried out in a series of ten steps and is quite ingenious. It would be interesting to know if there are any non \(p\)-soluble counter examples.
scientific article; zbMATH DE number 5144429