The problem is inspired by eigenvalue bounds of random Cayley graphs on $SL_2(q)$.

Definition. An infinite series of finite groups $S$ is α-rich if the dimension of the smallest nontrivial representation of $G$ on $\mathbb{C}$ is $\Omega(|G|^\alpha)$ for every $G\in S$.

For example, the series of groups $SL_2(q)$ is $\frac{1}{3}$-rich.

Question. Are there any series of $α$-rich groups with $\alpha>\frac13$?

Known. One only needs to consider simple groups, and by CFSG, it's just going through Lie groups.

Along the lines of David E Speyer, a permutation representation on $O(|G|^\frac13)$ points rules out the possibility for the group to beat $SL_2(q)$.

This holds for all classic groups of type $A_n (n\geq2)$, $B_n(n\geq3)$, $C_n(n\geq2)$, $D_n(n\geq4)$, $^2D_n(n\geq3)$, $^2A_n(n\geq4)$: just consider the action of these groups on the 1-dimensional subspaces of their defining vector space. $^2A_2$ is not in the list, but $^2A_2(q)$ has a representation on $q^2-q+1$ dimensions, hence ruled out.

The same argument rules out $F_4$, $E_n$ and $^2E_6$: All of them are covered in a paper given by Derek Holt. None of them beats $SL_2(q)$.

$G_2$ and $^2G_2$ do not beat $SL_2(q)$, by Derek Holt's answer.

For $^2B_2(q)$, see Wikipedia: it almost beats $SL_2$, but it has two characters of dimension $O(q^{3/2})$.

For $^3D_4(q)$, see Deriziotis, D. I., and G. O. Michler. "Character table and blocks of finite simple triality groups $^3D_4(q)$." Transactions of the American Mathematical Society 303.1 (1987): 39-70.: there's a character of dimension $q(q^4-q^2+1)$.

The last case $^2F_4$ is found in Die unipotenten Charaktere fur die GAP-Charaktertafeln der endlichen Gruppen vom Lie-Typ. M. Claßen-Houben; Diplomarbeit, RWTH Aachen; 2005. According to the conventions of the paper, the group has size $q^{52}$, and there's a character of dimension $q^2Φ_{12}Φ_{24}$, which is $O(q^{14})$.

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    $\begingroup$ If $H$ is a quotient of $G$, then any nontrival representation $V$ of $H$ is also a representation of $G$, and $\tfrac{\log \dim V}{\log |G|} \leq \tfrac{\log \dim V}{\log |H|}$. So the maximal $\alpha$ will occur for simple groups. $A_{n-1}$ has an $n$-dimensional representation, so that is terrible. It sounds like the question is basically to go through the groups of Lie type and see whether any of them beat $1/3$? $\endgroup$ Sep 11, 2019 at 15:04
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    $\begingroup$ Suggestion: Consider the adjoint form of $G_2$ over $\mathbb{Z}$ and look at $G_2(\mathbb{F}_q)$, which has size $\approx q^{14}$. The natural analogue of the $(q+1)$-dimensional permutation representation of $PSL_2(q)$ is the permutation action on $(G/P)(\mathbb{F}_q)$ for either of the two maximal parabolic $P$, which has size $\approx q^5$. We have $\tfrac{5}{14} > \tfrac{1}{3}$. I don't know if $G_2(\mathbb{F}_q)$ has any smaller representations, but probably an expert will show up soon to tell us. $\endgroup$ Sep 11, 2019 at 15:17
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    $\begingroup$ I am pessimistic about this approach now. I messed around with Hecke operators, and computed that the permutation representation on $(G/P)(\mathbb{F}_q)$ has dimension $q^5+q^4+q^2+q^3+q^2+q+1$ and breaks up into three nontrivial summands plus one trivial. I don't know how to compute the dimensional of these nontrivial summands, but it seems likely to me that some of them grow at rate $q^4$ or slower, and $\tfrac{4}{14} < \tfrac{1}{3}$. $\endgroup$ Sep 13, 2019 at 0:56
  • $\begingroup$ Possibly relevant: sciencedirect.com/science/article/pii/0021869374901501 $\endgroup$ Sep 14, 2019 at 17:49
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    $\begingroup$ I have now found a reference for the minimal degrees for all exceptional groups of Lie Type - see my edited answer. In particular, for ${}^2F_4(q^2)$ we get degree $O(q^{11})$, which is less that $O(q^{14})$. $\endgroup$
    – Derek Holt
    Oct 7, 2019 at 14:37

1 Answer 1


This answer is still not quite complete, but I hope to finish it soon!

The smallest degrees of the faithful permutation representations of the finite simple groups have all been known for a while now. The most convenient reference is probably Table 4 of this paper, although none of the results are original to that paper. The results for the exceptional groups of Lie type were proved in a series of papers of A. V. Vasil'ev.

Anyway, this approach works for $E_7(q)$ (order about $q^{133}$, minimal permutation degree about $q^{27}$), $E_8(q)$ (order about $q^{248}$, minimal permutation degree about $q^{57}$, $^{2}E_6(q)$ (order about $q^{78}$, minimal permutation degree about $q^{21}$), and ${}^3D_4(q)$ (order about $q^{28}$, minimal permutation degree about $q^9$).

However, the approach fails for $G_2(q)$, which has order about $q^{14}$ and minimal permutation degree $(q^6-1)/(q-1)$, $^{2}G_2(q)$ (with $q$ an odd power of $3$), with order about $q^7$ and minimal permutation degree $q^3+1$, and $^{2}F_4(q)$ (with $q$ an odd power of $2$), order about $q^{26}$, minimal permutation degree about $q^{10}$.

So I have been hunting around for results about minimal degrees of representations of $G_2(q)$,$^{2}G_2(q)$, and $^{2}F_4(q)$. For $G_2(q)$ I found them on Page 126 of G. Hiss, Zerlegungszahlen endlicher Gruppen vom Lie-Typ in nicht-definierender Charakteristik, Habilitationsschrift, Aachen 1990. For sufficiently large $q$, these are $q^4+q^2+1$, $q^3+1$, and $q^3-1$ when $q \equiv 0,1,-1 \bmod 3$, respectively. So this is less that $q^{1/3}$, and we are OK.

For $^{2}G_2(q)$, the character tables are computed in this paper. The table is towards the end of the paper, and the smallest character degree is $q^2-q+1$, so again we are OK!

I still need to check $^{2}F_4(q)$.

Added later: i have now located a better reference for minimal degrees of characters of exceptional groups of Lie type, namely Lübeck, Frank, Smallest degrees of representations of exceptional groups of Lie type. Comm. Algebra 29 (2001), no. 5, 2147–2169, available here.

In particular, for sufficiently large $q$, the smallest character degree of ${}^2F_4(q^2)$ has degree $11$ in $q$, which is easily sufficient to prove the required result.

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    $\begingroup$ The case of $^2F_4$ is solved(I will write it up soon). Thanks for your answer! $\endgroup$ Oct 4, 2019 at 12:48

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