Irreducibility of representations of degree two

Posted: February 8, 2011 in Representations of Finite Groups
Tags: , ,

Throughout G is a finite group.

Theorem. Let \rho : G \longrightarrow \text{GL}(V) be a degree two representation of G. Then \rho is irreducible if and only if the linear transformations \rho(g), \ g \in G, have no common eigenvector.

Proof. Suppose first that the linear transformations \rho(g), \ g \in G, have a common eigenvector, say v \in V. Then v \neq 0 and thus W=\mathbb{C}v is a one dimensional \mathbb{C}-subspace of V. Also, W \neq V becasue \dim_{\mathbb{C}}V=2. Now let g \in G. Then \rho(g)(v)=\lambda v, for some \lambda \in \mathbb{C}, because we assumed that v is an eigenvector of all \rho(g), \ g \in G. Thus \rho(g)(W)=\rho(g)(\mathbb{C}v)=\mathbb{C} \rho(g)(v)=\lambda \mathbb{C}v \subseteq W. So W is a \mathbb{C}[G]-submodule of V and hence V is not simple, i.e. \rho is not irreducible. Conversely, suppose that \rho is reducible, i.e. V is not simple. Let (0) \neq W \neq V be a \mathbb{C}[G]-submodule of V. Then \dim_{\mathbb{C}}W=1 because \dim_{\mathbb{C}}V=2. So there exists 0 \neq v \in V such that W=\mathbb{C}v. Now, let g \in G. Then \rho(g)(v) \in W because v \in W and W is a \mathbb{C}[G]-module. Therefore \rho(g)(v)=\lambda v, for some \lambda \in \mathbb{C}. So we have proved that v is an eigenvector of all \rho(g), \ g \in G. \ \Box

Example. In Example 1 in this post I gave m degree two representations of D_{2m}, the dihedral group of order 2m. In the remark, following Example 1 in that post, I showed that the 2m matrices of the representation corresponding to an m-th root of unity \zeta are

\begin{pmatrix} \zeta^{k} & 0 \\ 0 & \zeta^{-k} \end{pmatrix} and \begin{pmatrix} 0 & \zeta^{-k} \\ \zeta^{k} & 0 \end{pmatrix},

where 0 \leq k \leq m-1. We now have the following claim:

Claim . If m \geq 3 and \zeta \neq \pm 1, then \rho is irreducible.

Proof. The eigenvectors of \alpha = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} are \begin{pmatrix} \pm 1 \\ 1 \end{pmatrix} and the eigenvectors of \beta = \begin{pmatrix} 0 & \zeta^{-1} \\ \zeta & 0 \end{pmatrix} are \begin{pmatrix} \pm \zeta^{-1} \\ 1 \end{pmatrix}. So \alpha and \beta have no common eigenvector and thus, by the above theorem, \rho is irreducible. \Box

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