# Formula of a the group of character of a group

If I have a finite abelian group $$G$$ and I consider the group of the characters $$G^*$$, i.e the group of morphisms $$\phi: G\to \mathbb{C}^*$$ , then is it true that

$$\sum_{\chi \in G^*}\chi(h)=0$$

for each fixed $$h\in G- \{1\}$$ ?

Is it true that

$$\sum_{\chi \in G^*}\chi(h)^s=0$$

for each fixed $$s and for each fixed $$h\in G-\{1\}$$?

The first formula is trivially true when $$G$$ is cyclic because for each fixed $$h\in G$$ we have that $$\chi(h)$$ is a $$o(h)-$$ root of the unity so

$$\{\chi(h)\in \mathbb{C}^* : \chi\in G^*\}$$

is the set of $$o(g)-$$ root of the unity and we know that the sum of all $$n-$$root of unity is always zero.

• What is the problem? – Federico Fallucca Jun 17 at 18:39

The first equation isn't true if $$h$$ is the identity of $$G$$.
On the other hand, if $$h$$ is not the identity then there is some character $$\chi_0\in G^*$$ such that $$\chi_0(h)\neq 1$$ (see, e.g., Theorem 3.3 of these notes). Now set $$S=\sum_{\chi\in G^*}\chi(h).$$ Then, since $$G^*$$ is a group, we have $$\chi_0(h)S=S$$. So $$S=0$$ since $$\chi_0(h)\neq 1$$.
A similar adjustment will provide conditions for when the second identity holds (in particular, if $$h^s\neq 1$$ then there is some $$\chi_0\in G^*$$ such that $$\chi_0(h^s)\neq 1$$).