Orthogonality of characters for powers of a character

Solution 1:

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.