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Let $G$ be a finite group with $|G|=n$. Label the irreps $V_1,\ldots , V_t$ of $G$ over $\mathbb{C}$; let $d_i$ denote the degree of $V_i$. By Maschke's theorem we have $\mathbb{C}G\cong \bigoplus_{i=1}^tM_{d_i}(\mathbb{C})$. Let $e_i$ denote the element of $\mathbb{C}G$ whose image in $\bigoplus_{i=1}^tM_{d_i}(\mathbb{C})$ is the identity in the $i$ component, zeroes elsewhere. Each $e_i$ is idempotent and $e_i=\frac{\chi_i(1)}{n}\sum_{g\in G}\chi_i(g^{-1})\,g\,.$

Certainly $e_i=\sum_{g\in G}z_g\,g$ for suitable $z_g\in \mathbb{C}$, but how do you show that $z_g=\chi_i(1)\,\chi_i(g^{-1})/n$ ?

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    But don't $\oplus$ and $\prod$ coincide if there are finitely many rings? Also, who is Georges?2012-11-12

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There is exactly one central idempotent which acts by the identity on the $i$-th irrep and by zero on all the others: the $1$ of the factor in Wedderburn's decomposition of $\mathbb CG$ corresponding to the $i$-th irrep.

(Prove this and) use this to show that the right hand side of your equation is «that $1$».

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    I've got it now; thank you!2012-11-12