# Formula for idempotents in $\mathbb{C}G$

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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As Georges wrote the other day, you should never write the direct product of rings as $\oplus$: there is $\prod$ for that! – Mariano Suárez-Alvarez Nov 11 '12 at 4:15
But don't $\oplus$ and $\prod$ coincide if there are finitely many rings? Also, who is Georges? – Bey Nov 12 '12 at 2:16

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 $1$».
You're saying $e_i\cdot V_i=V_i$ and $e_i\cdot V_j=0$ for $i\neq j$, yes? Can you elaborate on how this helps give the right hand side? – Bey Nov 12 '12 at 2:22
I did not say it gives the right hand side. It provides a characterization (being the unique central idempotent which acts that way on irreps) for the $1$ in the factor of the Wedderburn decomposition, so it allows you to check that the element in the right hand side is what you want: you need to check that it is a central idempotent acting in the correct way on irreps, and you can do that. – Mariano Suárez-Alvarez Nov 12 '12 at 5:25