On $A$-Groups with the Same Index Set as a Nilpotent Group

Fuente: arXiv
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Main Authors: Zhou, Wei, Gorshkov, Ilya
Format: Preprint
Published: 2025
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author Zhou, Wei
Gorshkov, Ilya
author_facet Zhou, Wei
Gorshkov, Ilya
contents Let $G$ be a finite group and $N(G)$ be the set of conjugacy class sizes of $G$. For a prime $p$, let $|G||_p$ be the highest $p$-power dividing some element of $N(G)$. and define $|G|| = Π_{p\in π(G)}|G||_p$. $G$ is said to be an $A$-group if all its Sylow subgroups are abelian. We prove that if $G$ is an $A$-group such that $N(G)$ contains $|G||_p$ for every $p\in π(G)$ as well as $|G||$, then $G$ must be abelian. This result gives a positive answer to a question posed by Camina and Camina in 2006.
format Preprint
id arxiv_https___arxiv_org_abs_2506_15250
institution arXiv
publishDate 2025
record_format arxiv
spellingShingle On $A$-Groups with the Same Index Set as a Nilpotent Group
Zhou, Wei
Gorshkov, Ilya
Group Theory
20D15, 20D60
Let $G$ be a finite group and $N(G)$ be the set of conjugacy class sizes of $G$. For a prime $p$, let $|G||_p$ be the highest $p$-power dividing some element of $N(G)$. and define $|G|| = Π_{p\in π(G)}|G||_p$. $G$ is said to be an $A$-group if all its Sylow subgroups are abelian. We prove that if $G$ is an $A$-group such that $N(G)$ contains $|G||_p$ for every $p\in π(G)$ as well as $|G||$, then $G$ must be abelian. This result gives a positive answer to a question posed by Camina and Camina in 2006.
title On $A$-Groups with the Same Index Set as a Nilpotent Group
topic Group Theory
20D15, 20D60
url https://arxiv.org/abs/2506.15250