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Given an integer $n\geq 1$, it is known that there are only finitely many finite groups with exactly $n$ conjugacy classes.

Question: For $n\geq 3$, does there exists a finite non-abelian group, with exactly $n$ conjugacy classes?

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    $\begingroup$ Dihedral group of order 4n-6. $\endgroup$ – user641 Sep 27 '12 at 6:35
  • $\begingroup$ Steve shouldn't that be 2n-6? $\endgroup$ – Nicky Hekster Sep 27 '12 at 8:42
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    $\begingroup$ No 4n-6 works for all n>1 $\endgroup$ – user641 Sep 27 '12 at 14:26
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For $k$ even the number of conjugacy classes of the dihedral group $D_k$ of order $2k$ is $(k+6)/2$. Hence take $k=2n-6$. If $n=3$, then take $S_3$,

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