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Keywords:
Thompson's conjecture; conjugacy class size; symmetric groups; prime graph
Thompson's conjecture; conjugacy class size; symmetric groups; prime graph
Summary:
Let $G$ be a finite group and let $N(G)$ denote the set of conjugacy class sizes of $G$. Thompson's conjecture states that if $G$ is a centerless group and $S$ is a non-abelian simple group satisfying $N(G)=N(S)$, then $G\cong S$. In this paper, we investigate a variation of this conjecture for some symmetric groups under a weaker assumption. In particular, it is shown that $G\cong {\rm Sym}(p+1)$ if and only if $|G|=(p+1)!$ and $G$ has a special conjugacy class of size $(p + 1)!/p$, where $p>5$ is a prime number. Consequently, if $G$ is a centerless group with $N(G)=N({\rm Sym}(p+1))$, then $G \cong {\rm Sym}(p+1)$.
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