It is known that there are finitely many finite groups with a given number of conjugacy classes. How can one construct (or get a character table for) the groups $G$ that realize the maximum possible order among groups with $k$ conjugacy classes? Help even with $k=4$ or 5 would be very useful.



Constructing the largest finite group with a fixed number of conjugacy classes
Following @verret comment, here are the list of the largest finite groups $G\_k$ with a fixed class number $k\le 14$ (i.e. the number of conjugacy classes of elements, or the number of irreducible complex representations up to equiv.), coming from the papers of Vera-López and Sangroniz [MR804489](https://mathscinet.ams.org/mathscinet-getitem?mr=804489), [MR880291](https://mathscinet.ams.org/mathscinet-getitem?mr=880291), [MR2308490](https://mathscinet.ams.org/mathscinet-getitem?mr=2308490).
See also [OEIS/A002319](https://oeis.org/A002319). We will consider $c\_k:=|G\_k|^{1/k}$.


$\begin{array}{c|c|c}
k & G\_k & |G\_k| & c\_k \newline 
\hline 
1 & C\_1 & 1 & 1 \newline 
\hline 
2 & C\_2 & 2 & 1.4142\dots \newline 
\hline 
3 & S\_3 & 6 & 1.8171\dots \newline 
\hline 
4 & A\_4 & 12 & 1.8612\dots \newline 
\hline 
5 & A\_5 & 60 & 2.2679\dots \newline 
\hline 
6 & PSL(2,7) & 168 & 2.3490\dots \newline 
\hline 
7 & A\_6 & 360 & 2.3183\dots \newline 
\hline 
8 & M\_{10} & 720 & 2.2759\dots \newline 
\hline 
9 & A\_7 & 2520 & 2.3874\dots \newline 
\hline 
10 & PSL(3,4) & 20160 & 2.6943\dots \newline 
\hline 
11 & Sz(8) & 29120 & 2.5458\dots \newline 
\hline 
12 & M\_{22} & 443520 & 2.9551\dots \newline 
\hline 
13 & C\_9^2:SL(2,3) & 1944 & 1.7905\dots \newline 
\hline 
14 & PSU(3,5) & 126000 & 2.3137\dots \newline 
\end{array}$


*Monstrous question*: Is it true that $G\_{194} = M$?  

If so, $c\_{194} = 1.8960\dots$


**Bertram's problem** ([19.11](https://kourovkanotebookorg.files.wordpress.com/2018/12/19tkt-1.pdf); [B](https://mathscinet.ams.org/mathscinet-getitem?mr=2258655)): Is there $\alpha$ such that for any finite group $G$ of class number $k(G)$ then $$|G| \le \alpha^{k(G)}$$


*Bonus question 1*: Is $c\_k \le c\_{12}$ for all $k$?


If so, it would solve Bertram's problem explicitly with $\alpha = c\_{12}$.  

In particular, we would have $|G|< 3^{k(G)}$ for any $G$ (checked by GAP for $G$ simple with $|G|<10^7$).


*Bonus question 2*: Is it true that $c\_k \to 1$ when $k \to \infty$?


If so, it would solve Bertram's problem implicitly.  

I guess it is known (from CFSG) that for $G$ simple, $|G|^{1/k(G)} \to 1$ when $|G| \to \infty$. We can reduce to the infinite families, ok for $C\_p$ and $A\_n$, now if $X(q)$ is a generic finite simple group of Lie type then $\exists n,m \ge 1$ such that $|X(q)| = O(q^n)$ and $k(X(q)) = O(q^m)$, but $q^{n/{q^m}} \to 1$ when $q \to \infty$.  

Then, an eventual reduction of Bertram's problem to the simple group case should solve it.


