# How to show that this representation of $D_n$ is irreducible?

Write $D_n = \langle r, s \mid s^2 = r^n = 1, srs= r^{-1} \rangle$ with $n \geq 3$. Define a representation $\rho$ on the generators as $\rho(r) = \begin{pmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{pmatrix}$ and $\rho(s) = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$. Note that $\omega$ is an $n$th root of unity. How do you show that this extends to an irreducible representation of $D_n$?

I know that I need to show that there is no non-zero subspace of $\mathbf{C}^2$ which gets mapped to itself, but I'm having trouble making progress with this definition.

• The nonzero subspace would be a one-dimensional space, namely the span of a vector. What directions does $\rho(s)$ fix? Does $\rho(r)$ fix either of those? – Bob Jones Feb 28 '18 at 18:52
• @BobJones is that equivalent to finding an eigenvector of $\rho(s)$? – Drew Brady Feb 28 '18 at 18:54
• An eigenvector gives you a one-dimensional space fixed by $\rho(s)$, yes? It should also be clear that the only one-dimensional subspaces fixed by $\rho(s)$ are eigenvectors. – Bob Jones Feb 28 '18 at 18:56
• The invariant subspaces of $\rho(r)$ are its eigenspaces. These are spanned by $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$, respectively. So, the only hope for reducibility is if one of these spaces is $\rho(s)$-invariant. That doesn't happen, so you are done. – David Hill Feb 28 '18 at 18:57

Call your basis vectors (of $\mathbb{C}^2$) $\bf{x}$ and $\bf{y}$, so e.g. $\rho(r)({\bf x}) = \omega \bf{x}$ and $\rho(r)({\bf y}) = \omega^{-1}\bf{y}$. Now, a non-zero proper subspace of $\mathbb{C}^2$ must be 1-dimensional, so it must be spanned (as a vector space) by some single vector, say ${\bf v} = \lambda {\bf x} + \mu {\bf y}$.
For the subrepresentation generated by ${\bf v}$ to be 1-dimensional, $\bf v$ would need to be an eigenvector of every matrix in $\rho(D_n)$. Show that it can't be.
• Or, if you prefer thinking in terms of matrix notation, you can keep doing that: take ${\bf v} = \begin{pmatrix} \lambda\\ \mu\end{pmatrix}$. – Billy Feb 28 '18 at 18:57