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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 $s<ord(h)$ 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.

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  • $\begingroup$ What is the problem? $\endgroup$ – Federico Fallucca Jun 17 at 18:39
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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 $\chi_0(h^s)\neq 1$).

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  • $\begingroup$ Thank you very much $\endgroup$ – Federico Fallucca Jun 17 at 18:50

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