1 Representation theory of finite groups 1.2 Formalities 1.4 Schur’s Lemma

1.3 Example: dihedral groups

We list the elements of the dihedral group $D_{n}$ as

\{r^{k},sr^{k}:k=0,\ldots,n-1\}.

We aim to show that Table 1 gives the complete list of representations of $D_{n}$ , for $n$ odd. We leave the case of $n$ even as an exercise (there are two more one-dimensional representations in this case).

Table 1: Representations of

D_{n}

Label	Dimension	$\rho(r)$	$\rho(s)$
$\rho_{k}$ , $1\leq k<n/2$	$2$	$\begin{pmatrix}e^{2\pi ik/n}&0\\ 0&e^{-2\pi ik/n}\end{pmatrix}$	$\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$
$\mathbf{1}$	$1$	$1$	$1$
$\epsilon$	$1$	$1$	$-1$

Proof.

Take $(\rho,V)$ to be an irreducible complex representation of $D_{n}$ . Let $v\in V$ be an eigenvector for $\rho(r)$ with eigenvalue $\lambda$ (which must be an $n$ th root of unity since $\rho(r)^{n}=1$ ). Let $w=\rho(s)v$ . The key calculation is:

	$\displaystyle\rho(r)w$	$\displaystyle=\rho(r)\rho(s)v$
		$\displaystyle=\rho(rs)v$
		$\displaystyle=\rho(sr^{-1})v$
		$\displaystyle=\rho(s)\rho(r)^{-1}v$
		$\displaystyle=\rho(s)(\lambda^{-1}v)$
		$\displaystyle=\lambda^{-1}w.$

We also have $\rho(s)w=\rho(s)^{2}v=v$ and so $\left\langle v,w\right\rangle$ is a subrepresentation of $V$ . As $V$ is irreducible, we see that $V=\left\langle v,w\right\rangle$ .

Case 1

Suppose that $\lambda\neq\lambda^{-1}$ . Then $v$ and $w$ are eigenvectors of $\rho(r)$ with distinct eigenvalues, and so are linearly independent. Thus $\dim V=2$ . In the basis $v, w,$ the representation is

	$\displaystyle\rho(r)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle\lambda&0\\ \hidden@noalign{}\hfil\textstyle 0&\lambda^{-1}\end{pmatrix}$
	$\displaystyle\rho(s)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1\\ \hidden@noalign{}\hfil\textstyle 1&0\end{pmatrix}.$

If $\lambda=e^{2\pi ik/n}$ for $1\leq k<n/2,$ then we get the representations in the first line of the table. Otherwise, $\lambda=e^{-2\pi ik/n}$ for some $1\leq k<n/2$ and we instead take the basis $w, v$ to get $\rho_{k}$ again.

Case 2

Suppose that $\lambda=\lambda^{-1}$ . Then $\lambda=1$ as $n$ is odd. Since

\rho(r)(v+w)=\rho(s)(v+w)=v+w

we see that $v+w$ spans a subrepresentation of $V$ . If $v+w\neq 0,$ then $V=\left\langle v+w\right\rangle$ is the trivial representation. Otherwise, $\rho(s)v=w=-v$ and we get the representation $\epsilon$ .

Strictly speaking, we have only shown that if $V$ is an irreducible representation then it is given by matrices as in the table. However, it is easy to see that the matrices given in each row of the table do in fact define representations of $D_{n}$ : one only has to check that $\rho(r)^{n}=\rho(s)^{2}=1$ and $\rho(r)\rho(s)=\rho(s)\rho(r)^{-1}$ . ∎