Groups of order $8n$ have at least five distinct conjugacy classes 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? 
 A: Proof in the case $n$ is odd
By Sylow's 1st theorem there is a subgroup $H \vartriangleleft G$ of order (at least) 2 × 2  × 2 = 8.  This could be one of five possibilities: 


*

*$\mathbb{Z}_8, \mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ and additionally

*$ D_8 \simeq \mathbb{Z}_4 \rtimes \mathbb{Z}_2, Q = (\mathbb{Z}_2 \times \mathbb{Z}_2) \rtimes \mathbb{Z}_2 $.  The  dihedral group and quaternion group.


In the Abelian case, it is clear there are eight conjugacy classes - 1 for each element.  Since $H$ is normal, these extend to conjugacy classes of $G$.
In the Dihedral case, the rotations form a normal subgroup: $\mathbb{Z}_4 \vartriangleleft D_8 $.  This leads to 4 conjugacy classes + 2 for the reflections.  These are the symmetries of a square (See List all subgroups of the symmetry group of $n$-gon).
The quaternion group $\langle i,j,k | i^2 = j^2 = k^2 = ijk=-1\rangle$ has 6 conjugacy classes.

Representation theory makes it as simple as the sumset identity ($7$ is missing):
$$ \{ 0^2,1^2,2^2\} + \{ 0^2,1^2,2^2\} + \{ 0^2,1^2,2^2\} \subset \{0, 1,2,3,4,5,6,7\}$$
Sylow theorem guaranteed us a large enough normal subgroup; this is clearly the way to go.  
The problem is reduced to the detailed study of the partitions and permutations involves the number 8 and related coincidences.
A: 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
