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The supersolvable residual of a finite group factorized by pairwise permutable seminormal subgroups
A. A. Trofimuk A. S. Pushkin Brest State University
Abstract:
A subgroup $A$ is seminormal in a finite group $G$ if there exists a subgroup $B$ such that $G=AB$ and $AX$ is a subgroup for each subgroup $X$ from $B$. We study a group $G=G_1G_2\ldots G_n$ with pairwise permutable supersolvable groups $G_1,\ldots,G_n$ such that $G_i$ and $G_j$ are seminormal in $G_iG_j$ for any $i,j\in\{1,\ldots,n\}$, $i\neq j$. It is stated that $G^\mathfrak U=(G^\prime)^\mathfrak N$. Here $\mathfrak N$ and $\mathfrak U$ are the formations of all nilpotent and supersolvable groups, and $H^\mathfrak X$ and $H^{\prime}$ are the $\mathfrak X$-residual and the derived subgroup, respectively, of a group $H$. It is proved that a group $G=G_1G_2\ldots G_n$ with pairwise permutable subgroups $G_1,\ldots,G_n$ is supersolvable provided that all Sylow subgroups of $G_i$ and $G_j$ are seminormal in $G_iG_j$ for any $i,j\in\{1,\ldots,n\}$, $i\neq j$.
Keywords:
supersolvable group, nilpotent group, seminormal subgroup, derived subgroup, $\mathfrak X$-residual, Sylow subgroup.
Received: 29.11.2020 Revised: 18.10.2021
Citation:
A. A. Trofimuk, “The supersolvable residual of a finite group factorized by pairwise permutable seminormal subgroups”, Algebra Logika, 60:3 (2021), 313–326; Algebra and Logic, 60:3 (2021), 207–216
Linking options:
https://www.mathnet.ru/eng/al2666 https://www.mathnet.ru/eng/al/v60/i3/p313
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Abstract page: | 200 | Full-text PDF : | 37 | References: | 40 | First page: | 2 |
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