Can an order of a group with $n$ conjugacy classes be arbitrarily great

Question

This is a problem naturally comes from my homework which is ‘A group $G$ with 2 conjugacy clases is isomorphic to $\mathbb Z/2\mathbb Z$’. I solved it by representation theory, since 2 conjugacy classes means two irreducible representations and by involving $\text{Hom}_G(V,V)$ and Schur’s Lemma, $\dim V$ must be $1$ and therefore $|G|$ equals to $\sum \dim^2(V)$ is 2.

I try to extend my conclusion to arbitrary $p$ as a prime. However, I discovered that this was not that simple as situation $2$.For instance,$5$,there exist groups of order $5$ ,$14$ ,$20$, $21$ having $5$ conjugacy classes, at least.

I can’t calculate groups of greater order since categorizing them is hard. Is there an upper bound for the orders of groups with $p$ conjugacy classes?

Answer 1

The answer to your question appears to be no: for any $n>0$, there are only finitely many isomorphism classes of finite groups with exactly $n$ conjugacy classes.

See here for a proof (although no proof or reference is provided for "Fact (2)", which is used in the proof).