This question was homework once upon a time. I have long since handed it in.
"Let $D_{n}$ be the dihedral group with $2n$ elements. Show that every irreducible representation of $D_{n}$ must have dimension less than or equal to two"
The way I did this was to split this into the cases $n$ is even, and $n$ is odd, and find every single irreducible representation and show that each irreducible representation has dimension less than 2. Whilst it was doable, it was quite the marathon.
My question: Was there an easier, less painful way to do this? I.e, without explicitly finding all the irreducible representations?
My attempt for an easier method is as follows:
For the sake of calculation, let us assume that $n$ is odd. Then there are $\frac{n-1}{2}+1+1=m$ conjugacy classes. Then we know that $\sum_{i=1}^{m}dim(V_{i})^2=2n$. For small values of $n$, i.e $n=1,3,5,7$ and $9$ once can easily show that all dimensions must be less than 2. But for $n=11$ there are 7 conjugacy classes and $4^2+1^2+1^2+1^2+1^2+1^2+1^2=22$. So clearly this formula alone is not enough to show that all irreducible representations must be of dimension less than 2. I can't see how one would show this without the explicit representations.
Hint:
$D_n = \langle r, s \mid r^n=s^2=srsr=1\rangle$. Let $V$ be an irreducible representation for $D_n$, and let $v \in V$ be an eigenvector for $r$. Consider the subrepresentation generated by $v$.