# Orthogonality of characters for powers of a character

charactersgroup-theorynumber theoryorthogonality

Suppose $$G$$ is a finite cyclic group, and let $$\chi: G \to \mathbb{C}^*$$ be a character of order $$n$$. That is, $$\chi^n$$ is the identity homomorphism. I came across the following relation in a paper but I cant quite prove it for myself:

$$\frac{1}{n}\sum_{i=1}^n\chi^i(a)\overline{\chi}^i(b) = \begin{cases} 1, &\chi(a) = \chi(b)\\ 0, &\chi(a) \neq \chi(b) \end{cases}$$
The first case is easy to see – if $$\chi(a) = \chi(b)$$ then every term in the sum is $$1$$. But I cant quite see the case for $$\chi(a) \neq \chi(b)$$. I assume this is very very closely related to the typical orthogonality relations for characters. Any thoughts?

Edit: I thought of a solution different to the posted answer so I thought I would also share.

If $$\chi(a) \neq \chi(b)$$ then the sum is
$$\frac{1}{n}\sum_{i=1}^n z^i$$
where $$z$$ satisfies $$z^d = 1$$ but $$z \neq 1$$. Thus, $$z$$ is a primitive $$k$$-th root of unity for some $$k|d$$. So we can rewrite the sum as
$$\frac{1}{n}\frac{d}{k}\sum_{i=1}^k z^i.$$
This type of sum over a $$k$$-th root of unity is $$0$$.

#### Best Answer

Since $$\chi$$ has order $$n$$, it is trivial on the subgroup $$G^n$$ of $$n$$th powers in $$G$$. For the purposes of this problem we can replace $$G$$ by $$G/G^n$$, on which $$\chi$$ is still a character. Therefore we can assume all elements of $$G$$ have trivial $$n$$th power and thus their orders divide $$n$$ (in terms of the original group $$G$$, these orders are really in the quotient group $$G/G^n$$).

Ignoring the $$1/n$$ outside the sum, we are looking at a sum of a character $$\chi$$ over the subgroup of $$G$$ (really, subgroup of $$G/G^n$$) that is generated by $$ab^{-1}$$ repeated $$m$$ times, where $$m$$ is $$n/{\rm order \, of \,} ab^{-1}$$. That order divides $$n$$ because of the reduction step we made. If $$\chi(a)$$ and $$\chi(b)$$ are not equal then $$\chi(ab^{-1}) \not= 1$$, so we are summing a nontrivial character $$\chi$$ over a finite abelian group $$m$$ times. Thus the sum is $$0$$. It is not necessary to assume $$G$$ is cyclic.