In:
Matematicheskie Zametki, Steklov Mathematical Institute, Vol. 113, No. 1 ( 2023), p. 109-117
Abstract:
Let $G$ be a finite group. For every element $x\in G$, the set $\{x^g=g^{-1}xg: g\in G\}$ is called the conjugacy class of $x$ in $G$ and is denoted by $x^G$. The conjugacy class size of $x$ in $G$ is denoted by $|x^G|$ and is equal to $|G:C_G(x)|$. An element $y$ of $G$ is said to be primary or biprimary if the order of $y$ is divisible by exactly one or two distinct primes. For a positive integer $n$ and a prime $p$, if $e 〉 0$ is an integer such that $p^e$ divides $n$ and $p^{e+1}$ does not divide $n$, then $p^e$ is called the $p$-part of $n$. Let $p$ be a prime divisor of $p$ such that $(p-1,|G|)=1$. We prove that $G$ is solvable and $p$-nilpotent if the conjugacy sizes of all noncentral primary and biprimary elements in $G$ have the same $p$-part. On the other hand, suppose that $N$ is a normal subgroup of $G$; we write $\operatorname{cs}_G(N)=\{|x^G|:x\in N\}$. Suppose that $\operatorname{cs}_G(N)=\{1,n_1,n_2,…,n_t\}$, where $1 〈 n_1 〈 n_2 〈 \cdots 〈 n_t$. Denote by $$ M_N(G)=\langle x^G:x\in N, x^G=1 or n_1\rangle $$ a subgroup of $G$. We prove that if $C_G(F(G))\le F(G)$ and $[x,F(G)]$ is a normal subset of $F(G)$ for every $x\in N$ with $|x^G|=1$ or $n_1$, then $M_N(G)$ is a nilpotent group with nilpotency class at most 2.
Type of Medium:
Online Resource
ISSN:
0025-567X
,
2305-2880
Language:
Russian
Publisher:
Steklov Mathematical Institute
Publication Date:
2023
detail.hit.zdb_id:
2550622-5
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