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Below I am referring to complex representations.

We know that if $G$ is a finite group with $m=(G:Z(G))$, then every irreducible representation has size at most $\sqrt{m}$. One cannot hope for this to be exact up to $O(1)$ as $|G|\to \inf$ since there are finitely many groups with a given number of conjugacy classes. I am interested in interested in constructions/proof of existence of $G$, where this bound is as close as possible to being tight.

An easy nonsatisfying lower bound is by finding $G$ with a small number of conjugacy classes $C$, since then by averaging there must be a representation with dimension at least $\sqrt{|G|}/C$ (although it's not obvious how to build these?)

I am also very interested in the above when you can make the representation faithful - I thought these are interesting in particular since $G$ that has a large faithful irreducible representation has a small outer automorphism group.

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    $\begingroup$ Just to be sure, you want estimates on $c(G)=$ sup of dimension of irreducibles of $G$, right? I don't really understand about all this stuff about small number of conjugacy classes: if it's an "easy" lower bound you should be able to give it explicitly. $\endgroup$
    – YCor
    Feb 12, 2019 at 22:00
  • $\begingroup$ @YCor That is correct. I edited it, just meant an average argument $\endgroup$
    – user135743
    Feb 12, 2019 at 22:41
  • $\begingroup$ Yep, well it's easy only when you take the number of conjugacy classes as granted. $\endgroup$
    – YCor
    Feb 12, 2019 at 22:52
  • $\begingroup$ I don't understand the sentence starting "One cannot hope ...". It's not clear to me exactly what you want to bound in terms of what. Obviously, for every $n>0$ there is a group $G$ with $|G|=n$ and an irreducible representation of degree exactly $\sqrt{|G:Z(G)|}$ ... and often there's even a nonabelian one. Such groups are called "of central type", by the way. $\endgroup$
    – Jeremy Rickard
    Feb 13, 2019 at 8:14
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    $\begingroup$ @JeremyRickard : I don't think you said quite what you meant. I think it is true that for every integer $n$ there is finite group $G$ with a faithful complex irreducible character of degree $n$ and with $n = \sqrt{[G:Z(G)]}.$ $\endgroup$
    – Geoff Robinson
    Feb 13, 2019 at 13:35

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In the infinite family $G_p=(\mathbb{Z}/p\mathbb{Z})\rtimes (\mathbb{Z}/p\mathbb{Z})^{\times}$, as $p$ runs over prime numbers, the bound is tight up to $O(1)$, and the representation in question is faithful. Indeed, the centre is trivial, and $\#G_p=p^2-p$, so $p-1\leq \sqrt{m}\leq p$. By Clifford theory, the induction of any faithful character from $\mathbb{Z}/p\mathbb{Z}$ to $G_p$ is a faithful irreducible representation of $G_p$ of dimension $p-1$, which is less than $1$ away from $\sqrt{m}$.

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