In Guralnick and Moreto (Conjugacy classes, characters and products of elements, arXiv:1807.03550v1, Theorem 4.2) it has been shown that if p≠q\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$p\ne q$$\end{document} are two odd primes, π={2,p,q}\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$\pi =\{2,p,q\}$$\end{document} and G is a finite group such that for every π\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$\pi $$\end{document}-elements x,y∈G\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$x,y \in G$$\end{document} with (O(x),O(y))=1\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$(O(x),O(y))=1$$\end{document}, (xy)G=xGyG\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$(xy)^G=x^Gy^G$$\end{document}, then G does not have any composition factors of order divisible by pq. In this note, inspired by the above result, we show that if p and q are two primes (not necessarily odd) and G is a finite group such that for every p-element x and q-element y∈G\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$y \in G$$\end{document}, (xy)G=xGyG\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$(xy)^G=x^Gy^G$$\end{document}, then G does not have any composition factors of order divisible by pq. In particular, we show that if p is an odd prime and G is a finite group such that for every p-element x and 2-element y∈G\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$y \in G$$\end{document}, (xy)G=xGyG\documentclass[12pt]{minimal}
\usepackage{amsmath}
\usepackage{wasysym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{mathrsfs}
\usepackage{upgreek}
\setlength{\oddsidemargin}{-69pt}
\begin{document}$$(xy)^G=x^Gy^G$$\end{document}, then G is p-solvable.