For a finite group $G$ and a fixed Sylow $p$-subgroup $P$ of $G$, Ballester-Bolinches and Guo proved in 2000 that $G$ is $p$-nilpotent if every element of $P\cap G^{\text{'}}$ with order $p$ lies in the center of $N_G\left(P\right)$ and when $p=2$, either every element of $P\cap G^{\text{'}}$ with order $4$ lies in the center of $N_G\left(P\right)$ or $P$ is quaternion-free and $N_G\left(P\right)$ is $2$-nilpotent. Asaad introduced weakly pronormal subgroup of $G$ in 2014 and proved that $G$ is $p$-nilpotent if every element of $P$ with order $p$ is weakly pronormal in $G$ and when $p=2$, every element of $P$ with...

A theorem of Burnside asserts that a finite group $G$ is $p$-nilpotent if for some prime $p$ a Sylow $p$-subgroup of $G$ lies in the center of its normalizer. In this paper, let $G$ be a finite group and $p$ the smallest prime divisor of $\left|G\right|$, the order of $G$. Let $P\in {\mathrm{Syl}}_p\left(G\right)$. As a generalization of Burnside’s theorem, it is shown that if every non-cyclic $p$-subgroup of $G$ is self-normalizing or normal in $G$ then $G$ is solvable. In particular, if $P\ncong \langle a,b|a^{p^{n-1}}=1,b^2=1,b^{-1}ab=a^{1+p^{n-2}}\rangle$, where $n\ge 3$ for $p>2$ and $n\ge 4$ for $p=2$, then $G$ is $p$-nilpotent or $p$-closed. ...

Let $\sigma =\{{\sigma }_i:i\in I\}$ be some partition of the set of all primes $\mathbb{P}$, $G$ be a finite group and $\sigma \left(G\right)=\{{\sigma }_i:{\sigma }_i\cap \pi \left(G\right)\ne \varnothing \}$. A set $\mathscr{H}$ of subgroups of $G$ is said to be a complete Hall $\sigma$-set of $G$ if every non-identity member of $\mathscr{H}$ is a Hall ${\sigma }_i$-subgroup of $G$ and $\mathscr{H}$ contains exactly one Hall ${\sigma }_i$-subgroup of $G$ for every ${\sigma }_i\in \sigma \left(G\right)$. $G$ is said to be $\sigma$-full if $G$ possesses a complete Hall $\sigma$-set. A subgroup $H$ of $G$ is $\sigma$-permutable in $G$ if $G$ possesses a complete Hall $\sigma$-set $\mathscr{H}$ such that $H{A}^x$= $A^{x}H$ for all $A\in \mathscr{H}$ and all $x\in G$. A subgroup $H$ of $G$ is $\sigma$-permutably embedded in...

Let $N$ be a normal subgroup of a group $G$. The structure of $N$ is given when the $G$-conjugacy class sizes of $N$ is a set of a special kind. In fact, we give the structure of a normal subgroup $N$ under the assumption that the set of $G$-conjugacy class sizes of $N$ is $(p_{1n_1}^{a_{1n_1}},\cdots ,p_{11}^{a_{11}},1)\times \cdots \times (p_{r{n}_r}^{a_{r{n}_r}},\cdots ,p_{r1}^{a_{r1}},1)$, where $r>1$, $n_i>1$ and $p_{ij}$ are distinct primes for $i\in \{1,2,\cdots ,r\}$, $j\in \{1,2,\cdots ,n_i\}$.

Let $G$ be a group and $\omega \left(G\right)$ be the set of element orders of $G$. Let $k\in \omega \left(G\right)$ and $m_k\left(G\right)$ be the number of elements of order $k$ in $G$. Let nse$\left(G\right)=\{m_k\left(G\right):k\in \omega \left(G\right)\}$. Assume $r$ is a prime number and let $G$ be a group such that nse$\left(G\right)=$ nse$\left(S_r\right)$, where $S_r$ is the symmetric group of degree $r$. In this paper we prove that $G\cong S_r$, if $r$ divides the order of $G$ and $r^2$ does not divide it. To get the conclusion we make use of some well-known results on the prime graphs of finite simple groups and their components.