Orthogonality of characters for powers of a character 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 $0$.

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.
