# $\frac{\chi(e)}{|G|}\sum_{h\in G}\chi(hg)\overline{\chi(h)}=\chi(g)$

I want to prove that $\frac{\chi(e)}{|G|}\sum_{h\in G}\chi(hg)\overline{\chi(h)}=\chi(g)$. This is needed to prove that $\frac{\chi(e)}{|G|}\sum_{h\in G}\chi(h^{-1})h\in K[G]$ is an idempotent. I don't know which is easier to prove.

Sorry, forgot to mention $\chi$ is an irreducible character of $G$.

I think it is easier to prove your second sum is an idempotent. By the orthogonality property of characters, it acts as the identity on an irreducible module with character $\chi$, and as zero on any irreducible module with any other character. Applying it twice to any representation is the same as applying it once. Considering the regular representation shows it's an idempotent.