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Let $G$ be a finite abelian group and let $\widehat{G}$ be the group of group homomorphisms $\chi:G\to\mathbb{C}^*$ (i.e. the dual group of $G$, or the group of characters on $G$). Then there is an isomorphism $\varphi:G\to \widehat{G}$.

We have the following proposition: Given a fixed $\chi\in\widehat{G}$, we have $$\sum_{g\in G} \chi(g) = \begin{cases}|G| & \chi = 1_{\widehat{G}},\\0 & \text{otherwise.}\end{cases}$$ and given a fixed $g\in G$, we have $$\sum_{\chi\in \widehat{G}} \chi(g) = \begin{cases}|\widehat{G}| & g = 1_G,\\0 & \text{otherwise.}\end{cases}$$


I can prove both statements separately but all texts I have found about it state that due to the fact that $G$ and $\widehat{G}$ are isomorphic, then it is enough to prove just one of them and the other one is a consequence of the first. My question is: how is the isomorphism applied here?

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    $\begingroup$ I think you need to understand the natural isomorphism between $\widehat{\widehat{G}}$ and $G$ to see the equivalence of these two statements. $\endgroup$
    – Derek Holt
    Commented Apr 13, 2023 at 19:25
  • $\begingroup$ @DerekHolt care to elaborate? Isn't that isomorphism implied by that of $G$ and $\widehat{G}$ or is the fact that it's natural (not sure what that means) significant here? $\endgroup$
    – Darth Geek
    Commented Apr 13, 2023 at 19:37
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    $\begingroup$ The isomorphism is defined by $g \leftrightarrow \hat{\hat{g}}$, where $\hat{\hat{g}}(\chi) = \chi(g)$. $\endgroup$
    – Derek Holt
    Commented Apr 13, 2023 at 19:41

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To make your second set of equations look the first set, each $g \in G$ defines a character ${\rm ev}_g \colon \widehat{G} \to \mathbf C^\times$ as "evaluate at $g$": $$ {\rm ev}_g(\chi) = \chi(g) $$ for all $\chi$ in $\widehat{G}$. That each ${\rm ev}_g$ is a group homomorphism is due to the definition of multiplication in $\widehat{G}$ as pointwise products: $$ {\rm ev}_g(\chi\psi) = (\chi\psi)(g) = \chi(g)\psi(g) = {\rm ev}_g(\chi){\rm ev}_g(\psi). $$ Thus ${\rm ev}_g$ is a character on $\widehat{G}$.

If $g = 1$, then ${\rm ev}_1$ is the trivial character on $\widehat{G}$. Using the cyclic decomposition for finite abelian groups (or other methods), one can show for $g \not= 1$ in $G$ that there is $\chi \in \widehat{G}$ such that $\chi(g) \not= 1$. Writing that as ${\rm ev}_g(\chi) \not= 1$, we see that if $g \not= 1$ in $G$, then ${\rm ev}_g$ is a nontrivial character on $\widehat{G}$.

For each $g \in G$, let's evaluate $$ \sum_{\chi \in \widehat{G}} \chi(g). $$ If $g = 1$ then each term in the sum is $1$, so the sum is $|\widehat{G}|$. We want to show if $g \not= 1$ that the sum is $0$. Rewrite the above sum as $$ \sum_{\chi \in \widehat{G}} {\rm ev}_g(\chi), $$ which is the sum of a character ${\rm ev}_g$ over the elements of a finite abelian group $\widehat{G}$. If $g \not= 1$ then ${\rm ev}_g$ is a nontrivial character on $\widehat{G}$, so the above sum is $0$ by the general result that the sum of a nontrivial character over the elements of a finite abelian group is $0$. That is, in the equation $$ \sum_{x \in A} \chi(x) = 0 \ {\rm if } \ \chi \not= {\mathbf 1}_A, $$ let $A = \widehat{G}$ and let the character on $A$ be ${\rm ev}_g$ for $g \in G$ with $g \not= 1$. In this way, your second character sum formula can be regarded as a special case of your first character sum formula by using $\widehat{G}$ in place of $G$.

Here I am not explicitly relying on an isomorphism between $G$ and $\widehat{G}$, but there is a natural isomorphism $G \to \widehat{\widehat{G}}$ given by the mapping $g \mapsto {\rm ev}_g$, and this isomorphism is related to the argument above.

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