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Fall 2016 - Problem 4

representation theory

Let $D$ be a dihedral group of order $2p$ with normal cyclic subgroup $C$ of order $p$ for an odd prime $p$ . Find the number of $n$ -dimensional irreducible representations of $D$ (up to isomorphisms) over $\C$ for each $n$ , and justify your answer.

Solution.

Write $D = \gen{r, f \st f^2 = r^p = frfr = 1}$ . Suppose $\func{\rho}{D}{V}$ is an irreducible representation for $D$ , and let $v \in V$ be an eigenvector for $\rho\p{r}$ with eigenvalue $\lambda \in \C$ . Then

v = \rho\p{r}^p v = \lambda^p v,

so $\lambda$ is a $p$ -th root of unity. Moreover,

\rho\p{fr^a} v = \lambda^a \rho\p{f}v,

so the dimension is determined by $\rho\p{f}$ . Indeed, there are two cases: if $\rho\p{f}v \in \span{v}$ , then $\span{v}$ is the entire representation since $V$ was irreducible, and hence $V$ is one-dimensional. On the other hand, if $\rho\p{f}v \notin \span\p{v}$ , then $V \simeq \span\set{v, \rho\p{f}v}$ is two-dimensional.

There are $k = \frac{p + 1}{3}$ representations with dimensions $n_1, \ldots, n_k$ , one for each conjugacy class of $D$ , and these must satisfy $n_1^2 + \cdots + n_k^2 = 2p$ . We showed above that $n_i \leq 2$ for each $i$ , so let $N_1, N_2$ be the number of one- and two-dimensional irreducible representations, respectively. Then

\begin{cases} N_1 \cdot 1^2 + N_2 \cdot 2^2 = 2p \\ N_1 + N_2 = \frac{p + 3}{2}. \end{cases}

This yields $N_1 = 2$ and $N_2 = \frac{p - 1}{2}$ .