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Let $N(G)$ be the set of conjugacy class sizes of finite group $G$. Let $G$ be a finite group for which $N(G)=N(A_n)$, where $A_n$ is the alternating group of degree $n$.

Let $p$ and $q$ be distinct prime numbers satisfying $n/2<p,q<n$. Moreover, assume that the Sylow $p$-subgroups and Sylow $q$-subgroups of $G$ are cyclic. Is it true to say if $x\in G$ is of order $p$, then $q\mid|x^G|$?

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  • $\begingroup$ Why do you say every non-identity conjugacy class of $A_n$ has size divisible by $p$ or by $q$? $\endgroup$
    – Anna
    Jun 23, 2015 at 12:59
  • $\begingroup$ I was mistaken. $\endgroup$ Jun 23, 2015 at 13:03
  • $\begingroup$ Professor Holt, I cannot follow it would you please explain more? $\endgroup$
    – Anna
    Jun 23, 2015 at 13:33
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    $\begingroup$ @GeoffRobinson I think I have misunderstood the question. So $G$ is any finite group in which the set of conjugacy class sizes is equal to the set of conjugacy class sizes in $A_n$. I will retract my vote to close. $\endgroup$
    – Derek Holt
    Jun 23, 2015 at 14:37
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    $\begingroup$ What is Thompson's conjecture? Also do we just know the set of conjugacy class sizes, or do we know it as a multiset? In the latter case we would know the order of $G$. $\endgroup$
    – Derek Holt
    Jun 23, 2015 at 14:54

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