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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.

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$\chi_1 \rho$ and $\chi_2 \rho$ may be equivalent. It may be the case that whenever $\chi_1(g) \neq \chi_2(g)$ we have $\rho(g) = 0$ (where I am referring to characters here). – Qiaochu Yuan Mar 2 '13 at 5:04
If "very few" is allowed to be zero, then of course there are the abelian groups. – Gerry Myerson Mar 2 '13 at 5:11
Although it is not directly connected to your question, you might like to see the answers to a question I asked some time ago:… – Yemon Choi Mar 2 '13 at 5:24
Let me mention that the answers below generalize quite a bit, to the case of Frobenius groups $K\rtimes H$. The irreducible reps of these groups come in two flavors: reps of $H$, and reps induced up from $K$. So if you choose $H$ to be abelian, you get reps of dimension 1 in the first case, and then every other rep has dimension $\ge |K|$. – Steve D Mar 2 '13 at 15:40
For the specific purposes of this question, you would want $H$ to be big and abelian, so you want a $K$ with large automorphism group. It is then natural to first consider the case where $K$ is elementary abelian. For example, the Frobenius group $(C_2\times C_2)\rtimes C_3$ - which is just the alternating group $A_4$ - has three linear characters, and one of degree 3. – Steve D Mar 2 '13 at 15:45
up vote 22 down vote accepted

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 thunits 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 $\rho =Ind _H^G (\chi)$.

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There is no reason to restrict your field to have $p$ elements. This works equally well in any finite field, of order $p^n$. – Steve D Mar 2 '13 at 16:01

For a more elementary approach, avoiding induction and Mackey theory, you might try a concrete construction. Realize Aakumadula's group $G$ as a $2 \times 2$ matrix group over $\mathbb{F}_p$ (say for $p$ an odd prime) consisting of all $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$. Here $a, b$ run over respectively the multiplicative group and the additive group of the field. This realizes a semidirect product $A \ltimes B$ having normal Sylow $p$-subgroup $B$ consisting of matrices with $a=1$, acted on by its (cyclic) automorphism group $A$ of order $p-1$ (the diagonal group acting by conjugation).

Check first that the commutator subgroup is just $B$, so its index $p-1$ in $G$ counts the number of linear characters (those complex irreducible characters of degree 1). Since the sum of squares of degrees adds up to the group order $p(p-1)$, the problem is to see that there is only one more irreducible character (of degree necessarily $p-1$). As Aakumadula suggests, you might construct this directly by induction from a nontrivial linear character of $B$. (But then you'd have to check irreducibility.)

On the other hand, another very classical fact is that the number of (distinct) irreducible characters equals the number of conjugacy classes in $G$. By linear algebra, conjugates must have the same eigenvalues. Sylow theory shows that elements of order $p$ (with both eigenvalues 1) are all in the normal subgroup $B$, and by conjugation with $A$ these $p-1$ elements are all conjugate. Along with the trivial class you have so far just 2 classes. But then it's easy to check that each of the $p-2$ elements with fixed $a \neq 1$ is conjugate in $G$ to precisely the $p$ elements sharing the same eigenvalues $a, 1$. Now you have $p$ classes, with no more possible eigenvalues (or group elements) to consider.

P.S. As the answers (and comment about arbitrary finite fields) indicate, the question can be approached narrowly or more broadly. What works best depends heavily on what one already knows. The approach I've sketched is deliberately elementary, restricted to the most basic knowledge of character theory, linear algebra, groups and rings. Even here there are lots of shortcuts and variants. But what is the motivation other than curiosity?

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Construction of the different representations, including the nonlinear irreducible representation without using induction, is carried out on the first $2+\varepsilon$ pages at – KConrad Mar 4 '13 at 1:30

Slightly in the other direction, if $p$ is an odd prime and $P$ is a finite $p$-group which has a non-linear irreducible complex character $\chi,$ then $P$ has at least $p-1$ irreducible characters of $\chi(1),$ namely the algebraic conjugates of $\chi.$ I won't spell out the details, but consider an element outside the kernel of $\chi$ which is represented as a central element of order $p$ in the representation. To complete that picture, for each positive integer $n$ and each prime $p$ (including $p=2$), there do exist $p$-groups of order $p^{2n+1}$ which have $p-1$ irreducible characters of degree $p^{n}$ and $p^{2n}$ linear characters ( these are the extra-special groups).

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There is another way to produce such examples. Take $V$ to be an $n$ dimensional ${\mathbb F}_2$ vector space, let $V^*$ be its dual and consider the Heisenberg group $H$ of $(n+1)\times (n+1)$ matrices with ${\mathbb F}_2$ entries, which consist of unipotent upper triangular matrices, all of whose upper off diagonal entries are zero except those in the first row or the $n+1)$-th column. Thus we have an exact sequence $$1 \rightarrow {\mathbb F}_2 \rightarrow H \rightarrow V\oplus V^* \rightarrow 1. $$ It is easily seen that there is one irreducible representation $\rho$ of the Heisenberg group $H$, of dimension $2^n$ where the centre ${\mathbb F}_2$ acts non-trivially. All the other irreducible representations are one dimensional (and are characters of the quotient group $V\oplus V^*$.

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Am I right in thinking that these are a concrete realization of $2$-extraspecial groups (as mentioned in Geoff Robinson's answer)? – Yemon Choi Mar 3 '13 at 8:50
I do not know. But, this fact about Heisenberg group is part of a more general theory, and hence I thought I should record it. In some sense, this makes it "expected", since it is known (in many ways) that the Heisenberg group has only one higher dimensional representation. – Venkataramana Mar 3 '13 at 8:56
@Yemon Choi, I think you are right; the Heisenberg group is the same as the ones mentioned by Geoff Robinson, it seems. At least the number of irrep.s add up correctly! – Venkataramana Mar 3 '13 at 10:58
I also think the OP's example is the same as this one (except that the two irreps that the OP has claimed maybe equivalent). – Venkataramana Mar 3 '13 at 14:40
A small correction: the Heisenberg group has only one higher-dimensional representation for every nontrivial central character. This explains why this construction requires the base field to have 2 elements. – Victor Protsak Jul 31 '15 at 5:42

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