Can a finite group have 2D and 3D faithful irreducible representations?

Question

I am looking for finite groups that have a 2D (complex matrix) faithful irreducible representation and a 3D faithful irreducible representation. Up to order 1023 GAP found none. Other combinations of dimensions seem to occur. Are groups with 2D and 3D faithful irreps non-existing?

Answer 1

Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${\rm G}(2,{\mathbb C})$, but I am more familiar with the subgroups of ${\rm GL}(2,q)$ for finite $q$, so I will use that approach.

Let $G$ be a finite irreducible subgroup of ${\rm GL}(2,{\mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.

The irreducible subgroups of ${\rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${\rm SL}(2,3)$ or ${\rm SL}(2,5)$.

The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.

On the other hand, ${\rm SL}(2,3)$ or ${\rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.

Answer 2

Proposition 1: Suppose that $ G $ has a faithful degree $ n $ (continuous) complex irrep whose image contains $ \zeta_n I $, where $ I $ is the identity matrix and $ \zeta_n $ is a primitive $ n $th root of unity. Then every faithful finite dimensional (continuous) complex irrep of $ G $ must have degree divisible by $ n $.

proof: A central element of order $ n $ and determinant 1 must map in every faithful irrep to an order $ n $ scalar matrices with determinant 1. Such a matrix must have the form $ \zeta_n I $ and so will only have determinant $ 1 $ if the degree of the irrep is divisible by $ n $.

Corollary 2: Suppose that $ G $ has a faithful 2D (continuous) complex irrep then every faithful finite dimensional (continuous) complex irrep of $ G $ must have even degree.

proof: Use fact 0 and proposition 1

Corollary 3: No group has both a faithful 2D and 3D (continuous) complex irrep

Fact 0: Every irreducible (closed) subgroup of $ GL(2,\mathbb{C}) $ contains $ \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} $.

pf: This result can be obtained from the classification theorem.