Character Sums and Characteristic Functions Let $\mathbb{F}_q$ be the finite field with $q$ elements with $q$ odd. Consider the subgroup $H$ of $(\mathbb{F}_q)^{\times}$ consisting of squares: $H=\{x^2\,:\,x\in(\mathbb{F}_q)^{\times}\}$. We can express $$1+\chi_2(t)=\begin{cases} 2 & t\in H\\ 0 &t\notin H\end{cases},$$ where $\chi_2$ is the unique multiplicative character of order 2. We've managed to write a function vanishing outside $H$. I am trying to generalize this result to all subgroups $H$ of $(\mathbb{F}_q)^{\times}$. For instance, suppose that $H=\{x^k:x\in (\mathbb{F}_q)^{\times}\}$ where $k\mid q-1$. Then $$\sum_{\mbox{ord} (\chi)\mid k }\chi(t)=\phi(k)$$ provided $t\in H$. Here, the sum is taken over all multiplicative characters of $(\mathbb{F}_q)^{\times}$ whose orders divide $k$. Unfortunately,  I can't imagine why this function would vanish for $t\notin H$. Any suggestion on how to generalize the $k=2$ case would be greatly appreciated.
 A: Ok, nevermind. What you do is form the sum $1+\chi(t)+\cdots +\chi^{d-1}(t),$ where $\chi$ is any multiplicative character of order $d$ with $d\mid q-1$. This sum takes the value $d$ on the subgroup $H$ and vanishes outside of $H$.
A: The given answer (+1) is adequate for this purpose. You can also think of it as follows.


*

*Assume that $G$ is a finite abelian group, and that $\hat G$ is its group of characters. It is well known and easy to prove that for any $x\in G$
$$
\sum_{\chi\in\hat G}\chi(x)=\begin{cases}|G|,&\text{if $x=1_G$}\\ 0,& \ \text{otherwise.}\end{cases}
$$

*If, additionally, $H\le G$ is a subgroup, then recall that any character of $G/H$ can also be viewed as a character of $G$ be precomposing the character with the natural projection $G\to G/H$. Furthermore, the set (it's actually a group) of characters of $G$ we get this way consists precisely of those characters of $G$ that become trivial when restricted to $H$. Therefore for any $x\in G$
$$
\sum_{\chi\in\widehat{G/H}}\chi(x)=\begin{cases}|G|/|H|,&\text{if $x\in H$}\\ 0,& \ \text{otherwise.}\end{cases}
$$

