Formation of groups with \({\mathfrak F}\)-pronormal \({\mathfrak F}\)-normalizers and \(\pi\) (\({\mathfrak F})\)-solvable \({\mathfrak F}\)-coradicals (Q1061838)

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scientific article; zbMATH DE number 3910602
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Formation of groups with \({\mathfrak F}\)-pronormal \({\mathfrak F}\)-normalizers and \(\pi\) (\({\mathfrak F})\)-solvable \({\mathfrak F}\)-coradicals
scientific article; zbMATH DE number 3910602

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    Formation of groups with \({\mathfrak F}\)-pronormal \({\mathfrak F}\)-normalizers and \(\pi\) (\({\mathfrak F})\)-solvable \({\mathfrak F}\)-coradicals (English)
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    1984
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    For the definitions and notions used see the book of \textit{L. A. Shemetkov} [''Formations of finite groups'' (1978; Zbl 0496.20014)]. Let G be a finite group and \({\mathfrak F}\) be a local formation. Theorem 1. Let G have \(\pi\) (\({\mathfrak F})\)-soluble \({\mathfrak F}\)-coradical and \({\mathfrak F}\)- pronormal \({\mathfrak F}\)-normalizers. Then every \({\mathfrak F}\)-normalizer of G is a pronormal subgroup in G. Theorem 2. Let G have nilpotent \({\mathfrak F}\)-coradical. Then every \({\mathfrak F}\)-normalizer of G is an \({\mathfrak F}\)- pronormal subgroup in G. Theorem 3. If \({\mathfrak F}\) doesn't contain the formation of all nilpotent groups then the formation from the title is not local. The last theorem is of special interest. It gives a negative answer to Shemetkov's question about the property of locality for the formation from the title.
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    local formation
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    \({\mathfrak F}\)-coradical
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    \({\mathfrak F}\)-normalizers
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    pronormal subgroup
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