Let G be a finite group and ω(G) be the set of element orders of G. Let k ∈ ω(G) and m k be the number of elements of order k in G. Let nse(G) = {m k |k ∈ ω(G)}. In this paper, we prove that if G is a finite group such that nse(G) = nse(H), where H = PSU(3, 3) or PSL(3, 3), then G ∼ = H.Keywords: element order; number of elements of the same order; projective special linear group; projective special unitary group; simple K n -group

‎Let $G$ be a finite group and $\omega(G)$ be the set of element orders of $G$‎. ‎Let $k\in\omega(G)$ and $m_k$ be the number of elements of order $k$ in $G$‎. ‎Let $ nse(G)=\{m_k|k\in \omega(G)\}$‎. ‎The aim of this paper is to prove that‎, ‎if $G$ is a finite group such that nse($G$)=nse($U_4(2)$)‎, ‎then $G\cong U_4(2)$.