# Why do sporadic simple groups have so few conjugacy classes?

In finite group theory, there's a general intuition that the further away a group is from abelian, the fewer conjugacy classes it will have. So it is to be expected that non-abelian finite simple groups have a smallish number of conjugacy classes relative to size. Even with this intuition though, some of the numbers appearing in the list of sporadic simple groups are a bit surprising. For instance, the Monster group, with more than $10^{53}$ elements and $15$ prime divisors, has fewer than $200$ conjugacy classes. (An alternating group of comparable order has more than $30 000$ conjugacy classes.) $M_{22}$ has $443520$ elements and only $12$ conjugacy classes.

What is going on here? Is there something about the special combinatorial structures that allow these groups to exist that also makes the centralisers exceptionally small?

• An equivalent question is, why do sporadic simple groups lack small representations? – S. Carnahan Dec 14 '14 at 23:56
• @S.Carnahan : Well, it's maybe not quite equivalent. Having few conjugacy classes is equivalent to the mean irreducible character degree being large. For example, $M_{12}$ has an irreducible complex representation of degree $11$, Suzuki's sporadic group of order 448,345,497,600 has a $12$-dimensional irreducible complex representation. – Geoff Robinson Dec 15 '14 at 0:15
• The fact that the Monster needs nearly 200000 dimensions to be represented linearly over any field is striking in its own right, but only a few of the sporadics are like this. – Colin Reid Dec 15 '14 at 0:20
• For which sporadic simple groups have you actually checked that there is no group (whether simple or not) of the same or larger order with fewer conjugacy classes? – Stefan Kohl Dec 15 '14 at 15:00

This is also rather an expanded comment. -- Since for purely arithmetical reasons, $\ln(\ln(|G|))$ is a lower bound for the number $k(G)$ of conjugacy classes of a finite group $G$, maybe $$f(G) := \ln(k(G))/\ln(\ln(\ln(|G|)))$$ is a better measure than $k(G)/|G|$ for how many or how few conjugacy classes a group $G$ has in comparison with its order. For example we have (examples ordered by group order):

• $f({\rm A}_5) \approx 4.68799$,

• $f({\rm PSL}(2,7)) \approx 3.64930$,

• $f({\rm A}_6) \approx 3.39930$,

• $f({\rm PSL}(2,8)) \approx 3.64187$,

• $f({\rm PSL}(2,11)) \approx 3.3204$,

• $f({\rm A}_7) \approx 3.04392$,

• $f({\rm PSL}(3,3)) \approx 3.23520$,

• $f({\rm M}_{11}) \approx 2.92936$,

• $f({\rm PSL}(2,31)) \approx 3.53994$,

• $f({\rm A}_8) \approx 3.17897$,

• $f({\rm PSL}(3,4)) \approx 2.77366$,

• $f({\rm Sz}(8)) \approx 2.83466$,

• $f({\rm M}_{12}) \approx 3.03727$,

• $f({\rm J}_1) \approx 2.96687$,

• $f({\rm A}_9) \approx 3.16283$,

• $f({\rm M}_{22}) \approx 2.63787$,

• $f({\rm J}_2) \approx 3.20085$,

• $f({\rm A}_{10}) \approx 3.23851$,

• $f({\rm M}_{23}) \approx 2.76986$,

• $f({\rm A}_{11}) \approx 3.31013$,

• $f({\rm Sz}(32)) \approx 3.39405$,

• $f({\rm HS}) \approx 3.01600$,

• $f({\rm J}_3) \approx 2.88256$,

• $f({\rm M}_{24}) \approx 3.00146$,

• $f({\rm PSL}(6,2)) \approx 3.55208$,

• $f({\rm O'N}) \approx 2.85566$,

• $f({\rm Fi}_{22}) \approx 3.36345$,

• $f({\rm HN}) \approx 3.18141$,

• $f({\rm B}) \approx 3.54764$,

• $f({\rm A}_{43}) \approx 6.61233$,

• $f({\rm M}) \approx 3.34883$,

• $f({\rm A}_{44}) \approx 6.69491$.

• This list is very interesting. It suggests to me almost the opposite of what the OP asked, namely that the number of conjugacy classes in a simple group is a very well-behaved statistic. I wonder whether one could prove an absolute upper bound for your function $f(G)$, as $G$ ranges over all of the (non-alternating) simple groups? This seems too much to ask, but you never know... I presume that $f(G)$ is unbounded for $A_n$? Could one also prove that $f(G)$ takes its minimal value when $G=M_{22}$? – Nick Gill Dec 15 '14 at 20:23
• @NickGill: We have indeed $\lim_{n \rightarrow \infty} f({\rm A}_n) = \infty$, as one can check using the well-known approximation formulas for $n!$ and the partition function $p(n)$. As to your other questions, I don't know. – Stefan Kohl Dec 15 '14 at 21:10
• @NickGill: At least we also have $\lim_{p \rightarrow \infty} f({\rm PSL}(2,p)) = \infty$. – Stefan Kohl Dec 15 '14 at 21:46
• Yes, that makes sense as an invariant. I suppose it is only some of the sporadics that stand out. Maybe the issue is more that $A_n$ has unusually many classes by simple group standards? – Colin Reid Dec 15 '14 at 22:01
• @ColinReid: The former yes, but the latter I doubt: I computed the approximate values $f({\rm PSL}(n,2))$ for $n = 3, 4, \dots, 32$. -- They are $3.6493, 3.17897, 3.22354, 3.55208, 3.81153, 4.1534, 4.46381, 4.79852, 5.11825, 5.45075, 5.77278, 6.10037, 6.42311, 6.74737, 7.06869, 7.39041, 7.71006, 8.02942, 8.34732, 8.66458, 8.98068, 9.29605, 9.61042, 9.92402, 10.2367, 10.5487, 10.8599, 11.1703, 11.48, 11.789$. For larger $n$, increasing $n$ by $1$ seems to add about $0.30$ to $0.33$. – Stefan Kohl Dec 15 '14 at 23:31

This isn't really an answer, but an extended comment that might shed a little light.

Let $k(G)$ be the number of conjugacy classes in a group $G$, and let me interpret your question as a query about the proportion $k(G)/|G|$. The following paper suggests that the behaviour you describe for sporadic groups is actually shared by the finite groups of Lie type too.... but that the proportion $k(G)/|G|$ "takes longer" to reach zero.

Liebeck, Martin W.; Pyber, László Upper bounds for the number of conjugacy classes of a finite group. J. Algebra 198 (1997), no. 2, 538–562.

The main result of this paper is that $k(G)\leq (6q)^{\ell}$ for a group of Lie type of untwisted rank $\ell$ over a field of $q$ elements.

In the case, for instance, where $G={\rm PSL}_{\ell+1}(q)$, we know that $|G|>q^{\frac12\ell^2}$ for $q$ big enough, so one obtains immediately that (for $\ell>2$) $\lim\limits_{|G|\to\infty} \frac{k(G)}{|G|}=0$. When $\ell=1, 2$, one can check directly that the same is true.

Indeed, for the other families, one has similar lower bounds on the size of $|G|$, except that one might need to replace the $\frac12$ by some other constant. Thus, in general, it is true that for $G$ in the family of simple groups of Lie type $$\lim\limits_{|G|\to\infty} \frac{k(G)}{|G|}=0.$$ Thus the groups of Lie type are exhibiting exactly the same behaviour asymptotically as what you observe for sporadic groups.

Of course, I have chosen just one interpretation of your question, and even from this perspective, this is only a partial answer: I cannot explain why the sporadic groups reach such small values for $\frac{k(G)}{|G|}$ so much more quickly than the groups of Lie type. In truth, though, I doubt an "answer" such as this really exists, since saying anything about the general behaviour of sporadic groups seems incredibly difficult.

Added later: Actually, I just looked at the ATLAS and I must say I don't see a great deal of evidence that the sporadic simple groups in general have much fewer conjugacy classes than groups of Lie type of similar sizes. For instance:

• $k(M_{24})=26, \, \, k(G_2(4))=32$
• $k(HN)=54, \, \, k(Fi_{22})=65, \, \, k(F_4(2))=95$
• $k(O'N)=39, \, \, k(C0_3)=42, \, \, k(O_{10}^+(2))=97$

In each row, groups have comparable size. In each case the sporadic group(s) have a few less conjugacy classes but not by a great deal. Perhaps your question should really be about the Monster?

• As a matter of interest, It is proved in a 2006-ish paper of Bob Guralnick and myself that in general, $\frac{k(G)}{|G|} \to 0$ as $[G:F(G)] \to \infty$ ( for finite $G$). – Geoff Robinson Dec 15 '14 at 17:07
• @GeoffRobinson, this is interesting. Indeed, given that this phenomenon is so general, the rate of convergence becomes interesting when considering the "non-abelian-ness" mentioned in the original question. The Liebeck-Pyber paper is very strong in this regard, but of course such a notion makes no sense in connection with the sporadic groups. – Nick Gill Dec 15 '14 at 17:12