This question arises when looking at a certain constant associated to (a certain Banach algebra built out of) a given compact group, and specializing to the case of finite groups, in order to try and do calculations for toy examples. It feels like the answer should be (more) obvious to those who play around with finite groups more than I do, or are at least know some more of the literature.


To be more precise: let $G$ be a finite group; let $d(G)$ be the maximum degree of an irreducible complex representation of $G$; and (with apologies to Banach-space theorists reading this) let $K\_G$ denote the order of $G$ divided by the number of conjugacy classes.


Some easy but atypical examples:


* if $G$ is abelian, then $K\_G=1=d(G)$;
* if $G=Aff(p)$ is the affine group of the finite field $F\_p$, $p$ a prime, then $$K\_{Aff(p)}=\frac{p(p-1)}{p}=p-1=d(Aff(p))$$


**Question.**
Does there exist a sequence $(G\_n)$ of finite groups such that $d(G\_n)\to\infty$ while
$$\sup\_n K\_{G\_n} <\infty ? $$


To give some additional motivation: when $d(G)$ is small compared to the order of $G$, we might regard this as saying that $G$ is not too far from being abelian. (In fact, we can be more precise, and say that $G$ has an abelian subgroup of small index, although I can't remember the precise dependency at time of writing.) Naively, then, is it the case that having $K\_G$ small compared to the order of $G$ will also imply that $G$ is not too far from being abelian?


**Other thoughts.**
Since the number of conjugacy classes in $G$ is equal to the number of mutually inequivalent complex irreps of $G$, and since $|G|=\sum\_\pi d\_\pi^2$, we see that $K\_G$ is also equal to the mean square of the degress of complex irreps of $G$. Now it is very easy, given any large positive $N$, to find a sequence $a\_1,\dots, a\_m$ of strictly positive integers such that
$$ \frac{1}{m}\sum\_{i=1}^m a\_i^2 \hbox{is small while} \max\_i a\_i > N $$
so the question is whether we can do so in the context of degrees of complex irreps -- and if not, why not? The example of $Aff(p)$ shows that we can find examples with only one large irrep, but as seen above such groups won't give us a counterexample.



Can we bound degrees of complex irreps in terms of the average conjugacy class size?
Yes, take 2-extraspecial group $2^{2n+1}$, plus or minus should not matter. It has $2^{2n}$ irreducible representations of degree 1 and one of degree $2^n$. So your $K\_G$ is about 2 while $d(G)=2^n$.


