Constructing the largest finite group with a fixed number of conjugacy classes

Question

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.

Answer 1

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, MR880291, MR2308490. See also OEIS/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; B): 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.