# How to construct groups and large dimension representations? How about faithful ones?

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.

• 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. – YCor Feb 12 at 22:00
• @YCor That is correct. I edited it, just meant an average argument – user135743 Feb 12 at 22:41
• Yep, well it's easy only when you take the number of conjugacy classes as granted. – YCor Feb 12 at 22:52
• 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. – Jeremy Rickard Feb 13 at 8:14
• @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 $n = \sqrt{[G:Z(G)]}.$ – Geoff Robinson Feb 13 at 13:35

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 $$\sqrt{m}$$.