# Article

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Keywords:
set of the numbers of elements of the same order; prime graph
Summary:
Let $G$ be a group and $\omega (G)$ be the set of element orders of $G$. Let $k\in \omega (G)$ and $m_k(G)$ be the number of elements of order $k$ in $G$. Let nse$(G) = \{m_k(G) \colon k \in \omega (G)\}$. Assume $r$ is a prime number and let $G$ be a group such that nse$(G)=$ nse$(S_r)$, 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.
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