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

Question

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.

Answer 1

HINT

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

See for example in Lam's book "A First Course in Noncommutative Rings" pg 135.