# If $r = \frac{1}{|G|} \sum_{g \in G} \chi(g^{-1})g$, then $r^2 = 1/(\chi(1)) r$ in the group algebra

Suppose we have a finite group $$G$$ and an irreducible character $$\chi$$ of $$G$$. Now, define in $$\mathbb{C} G$$ (the group algebra / group ring of $$G$$), the element $$r = \frac{1}{|G|} \sum_{g \in G} \chi(g^{-1})g.$$

How do we prove that $$\displaystyle r^2 = \frac{1}{\chi(1)} r$$?

We know that $$\mathbb{C} G = \bigoplus_{i=1}^n \left( \bigoplus_{j=1}^{\dim(U_i)} U_j\right),$$ where $$U_1, U_2, \cdots, U_n$$ are all the non-isomorphic irreducible $$\mathbb{C} G-$$modules ($$G-$$vector spaces). Moreover, we also see that $$r \cdot \mathbb{C} G = \{r \cdot v \ \mid \ v \in \mathbb{C} G\} = \bigoplus_{j=1}^k U,$$ where $$U$$ is the irreducible $$\mathbb{C} G-$$submodules of $$\mathbb{C} G$$ with character $$\chi$$, but I don't see how I can link this to the question.

I think that it is easier to show that $$e=\chi(1)r$$ is such that $$e^2=e$$.