When is the $k/n$ representation of $D_n$ irreducible, and why?

The $k/n$ representation of the Dihedral group of order $2n$ in $GL(2,\mathbb{C})$ is induced by mapping the rotation element of $D_n$ to the Rotation Matrix $R(\frac{2\pi k}{n})$, and the reflection element of $D_n$ to the matrix $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$. As mentioned in the first hyperlink, this representation is irreducible unless $n$ is even and $k=n/2$. The article also gives a brief justification for this, however I don't immediately understand why this is the case. Thanks in advance, and apologies for my inexperience formatting these posts!

• What part of the argument are you more specifically struggling with? – Ben Sheller Nov 2 '15 at 22:50
• The article says that the fact that this representation is irreducible unless $n=2k$, and that this "can be seen by computing its character norm"...this quoted phrase is what I don't understand. Thanks for the edits, now I see it's actually quite simple to correctly format these posts. – L. T. P. L. Nov 2 '15 at 23:00
• It's more or less clear to me that in the case $n=2k$, this representation is not irreducible. However, I do not understand why: if we take $0<k<n$ with $n$ not equal to $2k$, this representation should be irreducible. – L. T. P. L. Nov 3 '15 at 3:31

Let's suppose we are OK with the case when $n=2k$ (in this case the rotation matrix simplifies significantly and it becomes easy to find an invariant subspace).
Call the representation $\rho$ and let $\chi_\rho$ denote the character of this representation.
In the case $n=2k$ the representation gives rise to an invariant subspace - the rotation matrix will simplify to a diagonal matrix, and then any product of these two diagonal matrices will also be diagonal. This allows us to easily see that the subspace of $V=\mathbb{C}^2$ given by $V'=\{(z,0)\in\mathbb{C}^2\}$ is invariant.
Now let $n\neq2k$. To show irreducibility we wish to prove that the character norm $<\chi_\rho,\chi_\rho>$ is equal to 1.<\chi_\rho,\chi_\rho>=\frac{1}{2n}\sum_{g\in D_n}\chi_\rho(g)\chi_\rho(g^{-1})\implies<\chi_\rho,\chi_\rho>=\frac{1}{2n}[(2cos(\frac{2\pi k}{n}))^2+(2cos(2\frac{2\pi k}{n}))^2\cdots+2cos(n\frac{2\pi k}{n})+(0+0)+(cos\frac{2\pi k}{n}-cos\frac{2\pi k}{n})+...+(cos(n\frac{2\pi k}{n})-cos(n\frac{2\pi k}{n}))]$We now apply the identity$2cos^2\theta=1+cos(2\theta)$(we skip a few substeps) to obtain that$\frac{1}{n}[(1+cos(2\frac{2\pi k}{n}))+(1+cos(2\frac{4\pi k}{n}))+\cdots+(1+cos(2\frac{2n\pi k}{n}))]=\frac{1}{n}[n+cos(\frac{4\pi k}{n})+cos(2\frac{4\pi k}{n})+\cdots+cos(n\frac{4\pi k}{n})]$. Now we use another trigonometric result, namely:$cos(\alpha)+cos(2\alpha)+\cdots+cos(n\alpha)=\frac{cos(\alpha+\frac{n-1}{2}\alpha)sin(\frac{n\alpha}{2})}{sin(\frac{\alpha}{2})}$with$\alpha=\frac{4\pi k}{n}$. [note: this is the moment where we lose the case where$n=2k$, since it would give a zero in this denominator.] It is easy to see now that in our case this sum is zero. Since$sin(\frac{n\alpha}{2})=sin(\frac{4\pi k}{n}\frac{n}{2})=sin(2\pi k)=0$for all$k$. Finally,$<\chi_\rho,\chi_\rho>=\frac{1}{n}n=1\$ and the representation is irreducible.