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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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Dihedral group of order 4n-6. –  user641 Sep 27 '12 at 6:35
    
Steve shouldn't that be 2n-6? –  Nicky Hekster Sep 27 '12 at 8:42
1  
No 4n-6 works for all n>1 –  user641 Sep 27 '12 at 14:26

1 Answer 1

up vote 2 down vote accepted

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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