TITLE: Groups of order $8n$ have at least five distinct conjugacy classes
QUESTION [7 upvotes]: It was brought to my attention by Kevin Dong that every finite group whose order is a multiple of 8 must have at least five distinct conjugacy classes. This can be seen as follows:
If $|G| = 8n$, then $$8n = \sum_{\text{irreps of G}} (\text{dim}(V))^2 = 1 + \sum_{\text{nontrivial irreps of G}} (\text{dim}(V))^2$$
Using that perfect squares are always $0$, $1$, or $4$ $\pmod{8}$, a little casework shows there must be at least four nontrivial irreducible representations of $G$. Hence, there are at least five conjugacy classes.
How could one prove this without using any representation theory?

REPLY [2 votes]: In fact, it turns out that there are only finitely many finite groups with a given number of conjugacy classes. This is not hard to prove and it's not too hard to find all of them with up to four conjugacy classes, say:
http://groupprops.subwiki.org/wiki/There_are_finitely_many_finite_groups_with_bounded_number_of_conjugacy_classes