Abstract
Let G be a group and ω(G) be the set of element orders of G. Let k ∈ ω(G) and mk(G) be the number of elements of order k in G. Let nse(G) = {mk(G): k ∈ ω(G)}. Assume r is a prime number and let G be a group such that nse(G) = nse(Sr), where Sr is the symmetric group of degree r. In this paper we prove that G ≅ Sr, if r divides the order of G and r2 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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