12
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ In a first step you can ask about estimates on the order of such a group (possibly with $\le k$ conjugacy classes rather than exactly $k$), and ask about groups with $\le k$ conjugacy classes of cardinal "asymptotically" maximal. $\endgroup$
    – YCor
    Apr 27, 2016 at 23:04
  • $\begingroup$ I didn't know the first statement of your question. Do you know a reference? $\endgroup$
    – LSpice
    Apr 27, 2016 at 23:21
  • 7
    $\begingroup$ E.Landau proved in around 1895 that for a fixed $k$, there are only finitely many solutions to $\sum_{j=1}^{k} \frac{1}{n_{j}} = 1$ in positive integers. Apply this to the class equation of a finite group. $\endgroup$ Apr 28, 2016 at 0:02
  • 9
    $\begingroup$ The groups are known up to $k=14$ at least, see Vera-López, A., Sangroniz, Josu, The finite groups with thirteen and fourteen conjugacy classes. Math. Nachr. 280 (2007), no. 5-6, 676–694. $\endgroup$
    – verret
    Apr 28, 2016 at 3:08
  • 2
    $\begingroup$ See also mathoverflow.net/questions/58794/… and math.stackexchange.com/questions/46981/… $\endgroup$ Apr 28, 2016 at 3:43

1 Answer 1

7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.