For a finite group $G$ denote by $N\left(G\right)$ the set of conjugacy class sizes of $G$. In 1980s, J. G. Thompson posed the following conjecture: If $L$ is a finite nonabelian simple group, $G$ is a finite group with trivial center and $N\left(G\right)=N\left(L\right)$, then $G\cong L$. We prove this conjecture for an infinite class of simple groups. Let $p$ be an odd prime. We show that every finite group $G$ with the property $Z\left(G\right)=1$ and $N\left(G\right)=N\left(A_i\right)$ is necessarily isomorphic to $A_i$, where $i\in \{2p,2p+1\}$.

Let $G$ be a finite group, and let $N\left(G\right)$ be the set of conjugacy class sizes of $G$. By Thompson’s conjecture, if $L$ is a finite non-abelian simple group, $G$ is a finite group with a trivial center, and $N\left(G\right)=N\left(L\right)$, then $L$ and $G$ are isomorphic. Recently, Chen et al. contributed interestingly to Thompson’s conjecture under a weak condition. They only used the group order and one or two special conjugacy class sizes of simple groups and characterized successfully sporadic simple groups (see Li’s PhD dissertation)....

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.

Let $G$ be a finite group. An element $g\in G$ is called a vanishing element if there exists an irreducible complex character $\chi$ of $G$ such that $\chi \left(g\right)=0$. Denote by $\mathrm{Vo}\left(G\right)$ the set of orders of vanishing elements of $G$. Ghasemabadi, Iranmanesh, Mavadatpour (2015), in their paper presented the following conjecture: Let $G$ be a finite group and $M$ a finite nonabelian simple group such that $\mathrm{Vo}\left(G\right)=\mathrm{Vo}\left(M\right)$ and $\left|G\right|=\left|M\right|$. Then $G\cong M$. We answer in affirmative this conjecture for $M=Sz\left(q\right)$, where $q=2^{2n+1}$ and either $q-1$, $q-\sqrt{2q}+1$ or $q+\sqrt{2q}+1$ is a prime number, and $M=F_4\left(q\right)$, where...

Given an integer $n\ge 3$, let $u_1,...,u_n$ be pairwise coprime integers $\ge 2$, $𝒟$ a family of nonempty proper subsets of $\{1,...,n\}$ with “enough” elements, and $\epsilon$ a function $𝒟\to \{\pm 1\}$. Does there exist at least one prime $q$ such that $q$ divides ${\prod }_{i\in I}{u}_i-\epsilon \left(I\right)$ for some $I\in 𝒟$, but it does not divide $u_1\cdots u_n$? We answer this question in the positive when the $u_i$ are prime powers and $\epsilon$ and $𝒟$ are subjected to certain restrictions. We use the result to prove that, if ${\epsilon }_0\in \{\pm 1\}$ and $A$ is a set of three or more primes that contains all prime divisors of any...

It is proved that real functions on $ℝ$ which can be represented as the difference of two semiconvex functions with a general modulus (or of two lower $C^1$-functions, or of two strongly paraconvex functions) coincide with semismooth functions on $ℝ$ (i.e. those locally Lipschitz functions on $ℝ$ for which $f_{+}^{\text{'}}\left(x\right)={lim}_{t\to x+}{f}_{+}^{\text{'}}\left(t\right)$ and $f_{-}^{\text{'}}\left(x\right)={lim}_{t\to x-}{f}_{-}^{\text{'}}\left(t\right)$ for each $x$). Further, for each modulus $\omega$, we characterize the class $DS{C}_{\omega }$ of functions on $ℝ$ which can be written as $f=g-h$, where $g$ and $h$ are semiconvex with modulus $C\omega$ (for some $C>0$) using a new...

We consider the equation $$-y^{\text{'}}\left(x\right)+q\left(x\right)y(x-\varphi \left(x\right))=f\left(x\right),x\in ℝ,$$ where $\varphi$ and $q$ ($q\ge 1$) are positive continuous functions for all $x\in ℝ$ and $f\in C\left(ℝ\right)$. By a solution of the equation we mean any function $y$, continuously differentiable everywhere in $ℝ$, which satisfies the equation for all $x\in ℝ$. We show that under certain additional conditions on the functions $\varphi$ and $q$, the above equation has a unique solution $y$, satisfying the inequality $$\parallel y^{\text{'}}{\parallel }_{C\left(ℝ\right)}+{\parallel qy\parallel }_{C\left(ℝ\right)}\le c{\parallel f\parallel }_{C\left(ℝ\right)},$$ where the constant $c\in (0,\infty )$ does not depend on the choice of $f$.