## Irreducibility of representations of degree two

Posted: February 8, 2011 in Representations of Finite Groups
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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. $\Box$