How would you work out the orders of the irreducible representations of the dihedral group \(D_n\)?
Why is the sum of squares of dimensions equal to the order of the group?
Solution
Write \[ D_n=\langle r,s\mid r^n=s^2=1,\ srs=r^{-1}\rangle, \qquad |D_n|=2n. \] Its abelianization has order \(2\) when \(n\) is odd and \(4\) when \(n\) is even. Hence there are respectively \(2\) or \(4\) one-dimensional irreducible complex representations.
The remaining irreducibles are two-dimensional. For \[ \chi_k(r)=e^{2\pi i k/n} \] on \(\langle r\rangle\cong C_n\), the induced representation \[ \operatorname{Ind}_{\langle r\rangle}^{D_n}\chi_k \] is irreducible unless \(k=0\) or, when \(n\) is even, \(k=n/2\). Moreover \(k\) and \(n-k\) give isomorphic induced representations. Thus the number of two-dimensional irreducibles is \[ \frac{n-1}{2}\quad(n\text{ odd}), \qquad \frac{n-2}{2}\quad(n\text{ even}). \] Hence \[ 2\cdot1^2+\frac{n-1}{2}\cdot2^2=2n \] when \(n\) is odd, while \[ 4\cdot1^2+\frac{n-2}{2}\cdot2^2=2n \] when \(n\) is even.
The general identity \[ \sum_i(\dim V_i)^2=|G| \] comes from the regular representation: \[ \mathbb C[G]\cong\bigoplus_i V_i^{\oplus\dim V_i}. \] Taking dimensions gives the formula.