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Question: Given a group $G$, let $f(G)$ be the smallest dimension of any of its nontrivial irreducible representations over $\mathbb C$. For a positive integer $n$, let $a_n$ be the largest value of $f(G)$ for any group $G$ of order $\leq n$. How quickly does $f(G)$ grow?

Progress: I can come up with a few groups that have $f(G)$ large:

  1. $G=A_n$ has no nontrivial representations of dimension $<n-1$ for $n\geq 7$. This gives a sequence of groups for which $$f(G)\gtrsim \frac{\log |G|}{\log\log |G|}.$$
  2. It's been proven that $G=PSL_2(p)$ has no nontrivial representation of dimension $<\frac{p-1}{2}$. This gives $$f(G)\gtrsim |G|^{1/3}.$$

In addition, as the largest representation of a group $G$ is bounded by $\sqrt{|G|}$, $f(G)$ is certainly at most $\sqrt{|G|}$. I'm wondering whether it can get close, but can't see how to construct any groups that beat $PSL_2(p)$. There are groups of order $n$, such as $\mathrm{Aff}(\mathbb F_p)$, that have a representation of dimension $O(\sqrt n)$, but they seem to also have smaller representations.

Note: In an earlier iteration of this question, I had $S_n$ instead of $A_n$. My intended meaning of "nontrivial" is "not the trivial representation," so $f(S_n)$ is in fact $1$ due to the presence of the sign representation. A corollary of this is that, if $[G,G]\neq G$, then $f(G)=1$.

If all representations of dimension $1$ were considered trivial, then the example of $\operatorname{Aff}(\mathbb F_p)$ would show that $a_n\sim \sqrt{n}$, as this group is of order $p^2-p$ and has $p-1$ irreducible representations of dimension $1$ and one irreducible representation of dimension $p-1$.

Carl Schildkraut
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    Can you explain what you mean by nontrivial? For example $S_n$ has the sign representation coming from it's abelianization, and that's irreducible of dimension 1... – Steve D Mar 09 '20 at 05:38
  • You can improve the upper bound by using a lower bound on the number of irreps. – Oscar Cunningham Mar 09 '20 at 12:35
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    @SteveD I suppose you can replace $S_n$ with $A_n$ to essentially get rid of this problem, since by Clifford theory the dimension of the irrep of $A_n$ indexed by a partition $\lambda$ is either the same as the corresponding representation of $S_n$ or $1/2$ that big (which occurs if and only if $\lambda=\lambda^t$ is symmetric). – Stephen Mar 09 '20 at 12:44
  • @Stephen: sure, but it's important to get clarification from the OP on this. If "nontrivial" means what we all think, then this question becomes a lot different (since $a_n$ is $1$ an awful lot). – Steve D Mar 09 '20 at 17:47
  • @SteveD Oops, I forgot about the sign representation. I'll clarify in my question. – Carl Schildkraut Mar 09 '20 at 21:49
  • I think it follows from this paper that $a_n$ grows more slowly than $\sqrt n$. https://www.ams.org/journals/proc/2008-136-06/S0002-9939-08-09147-8/ – Jalex Stark Mar 11 '20 at 19:38
  • @JalexStark Yeah, this is quite an interesting paper -- the fact that it deals with the largest representation instead of the smallest one is unfortunately quite bad for the purposes of this problem. This, in particular, doesn't show that $a_n$ is $o(\sqrt n)$, just that the constant cannot be $1$. – Carl Schildkraut Mar 13 '20 at 01:42
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    Ah, darn. Thanks for looking into it! – Jalex Stark Mar 14 '20 at 02:32
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    My remark above shows that $a_n$ is $o(\sqrt n)$. For any $\varepsilon>0$ the number of irreps of a group of order $n$ is $\Omega\left(\frac{\log_2(n)}{\log_2(\log_2(n))^{3+\epsilon}}\right)$. So $a_n=O\left(\sqrt{\frac{n\log_2(\log_2(n))^{3+\epsilon}}{\log_2(n)}}\right) = o(\sqrt n)$. – Oscar Cunningham Mar 18 '20 at 09:58
  • @OscarCunningham your parentheses are very weird. do you mean $(\log_2(\log_2(n)))^{3+\epsilon}$? – mathworker21 Apr 03 '20 at 21:39
  • @mathworker21 Yes. – Oscar Cunningham Apr 03 '20 at 22:07
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    This question was answered over at https://mathoverflow.net/questions/400864/is-operatornamepsl2-q-the-most-quasirandom-group – Sean Eberhard Nov 05 '21 at 12:39

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