Let $G$ be a finite group. Is it possible that there are "many" one-dimensional representations, and "very few" high-dimensional irreducible representations?
Originally, I thought it's impossible for the following reason. If $\chi_1, \chi_2$ are two one-dimensional representations such that $\chi_2 \chi_1^{-1} \not= 1$, and let $\rho$ be a high-dimensional irreducible representation. Then $\chi_1 \rho$ and $\chi_2 \rho$ are two nonequivalent irreps. When I tried to prove this formally, I feel it's not the case.
Also, I find a finite group $G$ which has $2^{k-1}$ one-dimensional irreps, and two $2^{k-1}$-dimensoinal irreps, which is defined as follows. Let $G$ be generated by $g_1, g_2, \ldots, g_{2k-1}, -1$ such that $g_i^2 = -1$ and $g_i g_j = -g_j g_i$.
It seems that I already see the answer. But I hope to see some better explanation about it.
Best Answer
Let $p$ be an odd prime and $G$ the semi-direct product of the additive group $ {\mathbb F}_p= {\mathbb Z}/p{\mathbb Z}$ and the group $({\mathbb Z}/p{\mathbb Z})^*$ of the units in the finite field of $p$ elements, acting by multiplication on ${\mathbb F}_p$. The group $G$ has $p-1$ characters (one dimensional representations) and one irreducible representation of dimension $(p-1)$. Hence in the regular representation of $G$, it occurs with multiplicity $p-1$. Therefore, in the regular rep, these reps amount to $p-1+(p-1)^2=card(G).\quad $ Consequently, there are $p-1$ characters, and only ONE higher dimensional irrep for $G$. The order of $G$ can be arbitrarily large.
The irreducible representation of dimension $p-1$ is of the form $\rho= Ind_H^G(\chi)$ where $H={\mathbb F}_p$ and $\chi$ is a non-trivial character of $H$. By Mackey's theorem (easy to prove in this special case), $\rho$ is irreducible.
Suppose that $\rho$ is an irrep of dimension greater than one (exists since $G$ is not abelian). The restriction to $H$ must then contain a non-trivial character $\chi$ for $H$. Conjugation by elements of ${\mathbb F}_p^*$ then implies that $all$ non-trivial characters of $H$ occur in $\rho$. Hence $\rho$ has dimension at least $p-1$. Since we have already seen that there is no room for anything else, the dimension of $\rho$ must be $p-1$ and $\rho =Ind _H^G (\chi)$.