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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.

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Perhaps it is relevant that $1/K_G$ is the probability two randomly selected elements in $G$ commute. –  Steve D Apr 12 '10 at 5:41
    
The dependency you mentioned is (well, one possibility): If d = d(G) then G has an abelian subgroup of index at most (d!)^2 (This is from Isaacs book, so newer and better results might be known) –  Tobias Kildetoft Apr 12 '10 at 9:52
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4 Answers 4

up vote 9 down vote accepted

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 $d(G)=2^n$.

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Thanks! that's a very concrete family of examples. I have a follow-up question (which I'll probably post separately) where unfortunately the extraspecial groups won't give me a counterexample. But even in that case, this is an extremely helpful example for background context. –  Yemon Choi Apr 12 '10 at 19:02
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Let b(G) be the maximum degree of an irreducible character of the finite group G.

This answer is mostly to address the solvable, but not nilpotent case. I already up-voted the other two excellent answers: Jim Humphreys mentioned how the groups of Lie type behave (showing that KG≈q2N can be enormous compared to b(G)≈qN), and Bugs Bunny gave the very clear example of the family of extraspecial p-groups showing the opposite extreme (with KG = p(2n+1)/(p(2n)+2) < p, and b(G) = pn).


One idea is to bound b(G) using important characteristic subgroups:

For any finite group G, b(G) ≤ √[G:Z(G)] and b(G) divides [G:Z(G)].

If G is solvable, then Gluck's conjecture is that √[G:Fit(G)] ≤ b(G), and this has been verified for solvable G such that G/Φ(G) has an Abelian Sylow 2-subgroup or G such that G″ = 1. (If G is non-abelian simple, then Fit(G)=1, and so the bound cannot hold).


Another idea is to bound b(G) using abelian subgroups:

If A is an abelian subgroup of G, then b(G) ≤ [G:A]. If A is abelian and subnormal, then in fact b(G) divides [G:A].

If G is solvable, then min(√lg([G:A])) ≤ lg(b(G)). If G is solvable of odd order, then min([G:A]1/6) ≤ b(G). If γ(G) is abelian, then min([G:A]1/4) ≤ b(G).


Versions of most of these results are in Isaacs's book (page 28, 30, 84, 190, 212, 216), but I tried to include the 21st century versions where I found them.

I would also caution against thinking "having an abelian subgroup of small index" as "close to abelian". Perhaps there is some truth to it, but the p-groups that are furthest from being abelian (maximal class, that is, having the largest nilpotency class possible amongst groups of the same order) also tend to have the largest abelian subgroups, often index p. Poor extra-special groups are nilpotency class 2 and nearly abelian (they are understood as symplectic vector spaces), yet their abelian subgroups are tiny, |A| ≤ p√[G:Z]. In other words extremely not-abelian groups may have large abelian subgroups, and extremely abelian groups may not.

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+1. These are really useful thoughts & pieces of information. If I could accept this answer as well, I would; in some sense it is a more complete answer to some of the vaguer questions which motivated my original post. (However, Bugs Bunny's answer is technically the one which answers the original question.) Thanks again for this! –  Yemon Choi Apr 12 '10 at 19:05
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As a matter of interest, Bob Guralnick and I proved in "On the commuting probability in finite groups", Journal of Algebra 300 (2006) 509–528, that what you call $K_{G} \to \infty$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup of $G$ (this particular fact does not require the classification of finite simple groups, though several results from that paper do). (We were actually working with the reciprocal of your $K_{G}$). Hence, in insisting that $K_{G_n}$ is bounded, you are only allowing finitely many possibilities for the isomorphism type of $G_n/F(G_n)$. Now by a Theorem of Ito, the largest degree of an irreducible character of $G$ divides $[G:A]$ whenever $A$ is an Abelian normal subgroup of $G$. These facts together seem to indicate that the ``meat" of your question is contained in the difference between $F(G)$ and the maximal order Abelian normal subgroups of $G$. Of course, while there is a unique maximal nilpotent normal subgroup of any finite group, there is not usually a unique maximal Abelian normal subgroup of a finite group. In the above result that Guralnick and I proved, it is not possible to replace $[G:F(G)]$ by the index of an Abelian normal subgroup. Also, note that the largest degree of an irreducible character of $G$ is at most $[G:F(G)]$ times the largest degree of an irreducible character of $F(G)$ (and is at least as great as the largest degree of an irreducible character of $F(G)$). Since your question bounds $[G:F(G)]$, it can be made more precise that your question is really about nilpotent groups.

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The question is intriguing but possibly open-ended. It's a good idea to look at natural infinite families of finite groups to see what is possible: for example, the symmetric groups $S_n$ of order $n!$ where the number of classes or characters is given by the partition function; or the closely related alternating groups. For finite simple groups of Lie type or close relatives over an arbitrary finite field of order $q$ (split or quasisplit), the asymptotics are well understood by work of Chevalley, Steinberg, Deligne-Lusztig: If $N$ is the number of positive roots and $r$ the rank of the ambient algebraic group, the largest character degree is on the order of $q^N$ while the order of the group is roughly $q^{r+2N}$ and the number of classes is roughly $q^r$. Such a family of groups for a fixed prime but increasing prime power $q$ is pretty far from the asymptotic behavior asked for. At the same time the groups are far from being abelian. What about $p$-groups, which are abundant and can vary from being abelian to being highly nonabelian while always being solvable?

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