Result: On Composition Factors of Finite Groups Having the Same Set of Element Orders as the Group U3(q): On composition factors of finite groups having the same set of element orders as the group \(U_3(q)\)

For a finite group \(G\) denote by \(\omega(G)\) the set of its element orders. The article continues the study of finite groups \(G\) such that \(\omega(G)=\omega(U_3(q))\). \textit{V.~D.~Mazurov, M.~C.~Xu}, and \textit{H.~P.~Cao} [Algebra Logika 39, No. 5, 567-585 (2000; Zbl 0979.20019)] proved that, if \(q\) is even, the above condition forces \(G\) to be isomorphic to \(U_3(q)\). The author of the article under review considers the case in which the group \(G\) is nonsoluble and \(q\) is odd. The main part of the article is devoted to a proof of the following statement: If \(q>5\) then \(U_3(q)\leq G/N\leq\Aut(U_3(q))\) for some normal subgroup \(N\) in \(G\). The author also establishes some properties of \(N\) and determines the factor \(G/O_2(G)\) for \(q=3,5,7,9,11\). Note that \textit{V.~D.~Mazurov} [Algebra Logika 36, No. 1, 37-53 (1997; Zbl 0880.20007)] has completely studied finite groups \(G\) with the property \(\omega(G)=\omega(U_3(q))\), \(q=3,5,7\).