Let $k > 2$ be an integer. Is there a finite non-abelian group with exactly $k$ conjugacy classes?
The only group with $1$ conjugacy class is the trivial group, and the only group with exactly $2$ conjugacy classes is the group of order $2$, so the restriction $k>2$ is necessary. (The symmetric group $S_3$ has $3$ conjugacy classes, and is the only such non-abelian group.)
For small values of $k$, the number of non-abelian groups with exactly $k$ conjugacy classes are known.
I had originally come to this question for square-free values of $k$, but upon thinking about it, I realised that there was an even more basic form of the question, whose answer I did not know and could not find, as stated above.
I thought that it might be possible to construct a group of the form $G = A\rtimes B$, with $A$ abelian of order $k$, and $B\leq\operatorname{Aut}(A)$ so that $G$ has exactly $k$ conjugacy classes, but I was unable to come up with a general argument. It does not always work for cyclic $A\cong C_k$.
