# Groups with no small nontrivial representation

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 $$ for $$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 $$p-1$$.

• 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 9 at 5:38
• You can improve the upper bound by using a lower bound on the number of irreps. – Oscar Cunningham Mar 9 at 12:35
• @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 9 at 12:44
• Ah, darn. Thanks for looking into it! – Jalex Stark Mar 14 at 2:32
• 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 at 9:58