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Suppose $G$ is a finite group such that the set of all the conjugacy class size is $\{1,2,\dots,n\}$, where $n$ is a natural number. Is it true that $n\leq 3$? Thanks in advance.

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So are you saying that there is one conjugacy class of size 1, one conjugacy class of size 2, and so forth? If so, then because the size of conjugacy class must divide the order of the group, we have $n-1 | n(n+1)/2$. But $(n,n-1) = 1$, and $(n+1,n-1)$ is 1 or 2, this is impossible unless $n-1 = 1$ or 2. –  user27126 Nov 25 '13 at 6:44
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I don't mean that. There may be so many conjugacy classes of size $m$ for any $m\leq n$ and atleast one conjugacy class of size $m$ exist. –  D. N. Nov 25 '13 at 6:51
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Is there anyone out there who could produce an example or better counterexample with GAP? I have tried to get my finger around this very intriguing question. Noluck sofar. Perhaps its warrants a bounty? –  Nicky Hekster Nov 26 '13 at 16:05
    
I have check in GAP and i have found that no group of order less than or equal to 100 is a counter example to the question. –  D. N. Nov 27 '13 at 7:38
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up vote 3 down vote accepted

Yes, $n$ should be $\leq 3$. It is proved by Bianchi, Gillio and Casolo (see here) that in a finite group $G$ such that the two largest (non-central) conjugacy class sizes $n < m$ are coprimes, then any conjugacy class in $G$ has size $1$, $n$ or $m$.

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Yassine, thanks for the reference, a very nice paper indeed!!! –  Nicky Hekster Dec 3 '13 at 7:36
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