Past exam of the groups and representations course of the University of Oxford / 2018 / 1 / c / Solution

Put $\alpha =e^{2\pi i/3}$, so ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$. Since the group is Abelian, every conjugacy class is a singleton. The character table isFor the character inner product, complex conjugation givesFor $j=k$, all three summands are one, giving unit norm. For $k-j\equiv 1(\mathrm{mod}3)$ the sum is $1+\alpha +{\alpha }^2=0$; for $k-j\equiv 2(\mathrm{mod}3)$ it is $1+{\alpha }^2+\alpha =0$. This checks every distinct pair, proving