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[edited in response to some corrections by Geoff Robinson and F. Ladisch]

Throughout, all my groups are finite, and all my representations are over the complex numbers.

If $G$ is a group and $\chi$ is an irrreducible character on it, of degree $d$, say that $\chi$ has quite large degree if $d\geq (4/5)|G|^{1/2}$.

This condition has arisen in some work I am doing concerning (Banach algebras built on) restricted direct products of finite groups, and while our main concern was to find some examples with this property, I am curious to know if anything more can be said.

(I've had a quick look at the papers of Snyder (Proc AMS, 2008) and Durfee & Jensen (J. Alg, 2011), but these don't seem to quite give what I'm after. However, I may well have missed something in their remarks.)

Examples. The affine group over a finite field with $q$ elements has order $q(q-1)$ and a character of degree $q-1$, which has "quite large degree" for $q\geq 3$. The affine group of the ring ${\mathbb Z}/p^n{\mathbb Z}$ has order $p^{2n-1}(p-1)$ and a character of degree $p^{n-1}(p-1)$, which has quite large degree for $p\geq 3$. Being fairly inexperienced in the world of finite groups, I can't think of other examples (except by taking finite products of some of these examples).

Some simple-minded observations from a bear of little brain. A group can have at most one character of quite large degree (this is immediate from $|G|=\sum_\pi d_\pi^2$). If such a character $\chi$ exists, it is real (hence rational) valued (edit: as pointed out by Geoff Robinson in comments, $\chi$ must be equal to its Galois conjugates, and hence rational valued), and its centre is trivial (consider its $\ell^2$-norm). In particular $G$ can't be nilpotent. Moreover, since $\chi\phi=\chi$ for any linear character $\phi$, a bit of thought shows that $\chi$ vanishes outside the derived subgroup of $G$.

Now, let ${\mathcal C}$ be the class of all groups $G$ that possess a character of quite large degree clearly this is closed under taking finite products. Let ${\mathcal S}$ denote the class of all solvable groups.

Question 1. Does there exist a finite collection ${\mathcal F}$ of groups such that every group in ${\mathcal C}$ is the product of groups in ${\mathcal F}\cup{\mathcal S}$?

Question 2. Can we bound the derived length of groups in ${\mathcal C}\cap{\mathcal S}$?

Question 3. If the answers to Q1 and Q2 are negative, or beyond current technology, would anything improve if we replaced $4/5$ with $1-\epsilon$ for one's favourite small $\epsilon$?

Question 4. It may well be the case that hoping for a description of the class ${\mathcal C}$ in terms of familiar kinds of group is far too naive and optimistic. If so, could someone please give me some indications as to why? Perhaps it reduces to a set of known open problems?

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Have you seen dpmms.cam.ac.uk/~wtg10/quasirandomgroups.pdf ? –  Qiaochu Yuan Jun 23 '11 at 0:12
    
Qiaochu: I take it you mean the example of PSL(2,q)? Well, that has order $O(q^3)$ and the character table tells me that its largest character degree is $q+1$, so I don't think it's going to help here. I should also note that WTG seems to be considering groups where all nonlinear irreps have relatively large degree, while in the example of the affine group over $Z/p^nZ$, there are always irreps of degree $p-1$, yet there is a character of quite large degree. –  Yemon Choi Jun 23 '11 at 1:24
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This is not closed under taking finite products: for example, $S_3$ has a character of quite large degree ($2\geq (4/5)\sqrt{6}$), but $S_3 \times S_3$ has not (since $4 < (4/5)\sqrt{36}$). –  Frieder Ladisch Jun 23 '11 at 18:27
    
Thank you Geoff and F. -- some of the errors you point out were due to my haste in transcribing and "translating" from my notes, but some were just due to plain ignorance/muddle-headedness on my part. I will edit the question to take note of your corrections –  Yemon Choi Jun 23 '11 at 22:57
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2 Answers

A good potential supply of this sort of character is provided by Frobenius groups. These are finite groups $G$ of the form $G = KH$, where $K \cap H = 1$ and $K \lhd G$, and furthermore $C_{G}(x) \leq K$ for all non-identity elements $x \in K$. By a Theorem of Thompson the group $K$ is necessarily nilpotent.

One Frobenius group not mentioned in your examples is given by $H \cong {\rm SL}(2,5)$ and $K$ elementary Abelian of order $121$, which admits a regular action (on non-identity elements) by $H$. The semidirect product gives an example of a finite group $G$ which is not solvable, but has an irreducible character of quite large degree. Frobenius complements which are not solvable are very rare though.

In general, the irreducible characters $\chi$ of a (general) Frobenius group $G$ which do not contain $K$ in their kernels have degrees of the form $|H|\mu(1)$, where $\mu$ is an irreducible character of $K$. For such a $\chi$ to have quite large degree, we clearly need $|H|\mu(1)^2 \geq 0.64 |K|$. Notice also that $|K| \equiv 1$ (mod $|H|$). If $|H| < |K|-1$, then we obtain the contradiction $\mu(1)^2 > 1.28 |K|$. Thus the only Frobenius groups with irreducible characters of quite large degree are those in which the complement $H$ acts transitively on non-identity elements of the kernel $K$. This forces the kernel $K$ to be elementary Abelian. Anyway, it looks as though you won't get many new examples from Frobenius groups (later edit-as is made explicit by Noah Snyder's comment below. Previous version did not say what I meant in any case).

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The classification of doubly transitive Frobenius groups follows from Zassenhaus's 1935 classification of "near fields" as a doubly transitive Frobenius group is always the semidirect product $k \rtimes k^*$ for $k$ a near field. –  Noah Snyder Jun 23 '11 at 18:54
    
OK, thanks. I suppose I have seen that at some time. –  Geoff Robinson Jun 23 '11 at 19:00
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There's an example on page 8 of my paper which can give you a group of size $q^3(q-1)$ and an irrep of dimension $q(q-1)$ for any prime power $q$. This gives a factor of $\sqrt{\frac{q}{q-1}}$.

In my paper and Durfee-Jensen (there's also a preprint of Isaacs and a preprint of Larsen-Malle-Tiep which were written between those two papers), we're looking at representations whose dimensions are much much closer to $\sqrt{|G|}$. So those results can't be directly applied to your more general question. Nonetheless, the "moral" result of those papers is that "having a large character should mean you're in the case studied by Gagola and Kuisch-van der Waall." One could certainly hope that this result is true for a weaker definition of "large character" than we used. However, you should expect that proving this for a weaker definition of "large character" is going to get harder.

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Isaacs's preprint points out a nice description of the example I mentioned here. It's just 3x3 upper triangular matrices over F_q whose diagonal entries are all 1 except for the lower right corner which can be any nonzero scalar. –  Noah Snyder Jun 23 '11 at 19:01
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