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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?

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    $\begingroup$ Wouldn't the conjugacy class equation give you a similar answer? en.m.wikipedia.org/wiki/… $\endgroup$
    – Theo Douvropoulos
    Apr 15, 2015 at 2:09
  • $\begingroup$ it can be found by examining some cases, I may write the answer later. $\endgroup$
    – mesel
    Apr 15, 2015 at 11:15

2 Answers 2

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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.

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  • $\begingroup$ I am probably missing something, but how do you use Sylow's first theorem to prove $H$ is a normal subgroup? $\endgroup$
    – Jason DeVito - on hiatus
    Aug 25, 2015 at 17:35
  • $\begingroup$ @JasonDeVito Third Sylow theorem says the maximal Sylow $p$-group is normal. I stand corrected. $\endgroup$
    – cactus314
    Aug 25, 2015 at 17:47
  • $\begingroup$ I'm sorry, but I'm still not sure why the maximal Sylow $p$-group is normal. What about a simple group, like $A_n$ ($n\geq 5$)? Such a group has order $n!/2$ which is of the form $8k$ once $n \geq 6$. $\endgroup$
    – Jason DeVito - on hiatus
    Aug 25, 2015 at 18:16
  • $\begingroup$ @JasonDeVito OK. There could be any odd number of Sylow 2-groups, all conjugate to each other, which intersect each other in interesting ways. $\endgroup$
    – cactus314
    Aug 25, 2015 at 18:36
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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

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