$G$ is finite group, prove that $\frac{1}{|G|}\sum_{g\in G} U_g$ is idempotent in the group-ring $R[G]$

Question

$G$ is finite group, prove that $\frac{1}{|G|}\sum_{g\in G} U_g$ is idempotent in the group-ring $R[G]$

I was trying to use induction on |G|, and show that each $U_g$ appears $n$ times in $(\sum_{g\in G} U_g)^2$. A second approach was to distribute $G$ to $G \setminus \langle g\rangle$ and then work with $R[G\setminus\langle g\rangle]$. But I couldn't find a solution.

Answer 1

Let $s=\sum_{g} U_g$. Then $$s^2= (\sum_{g} U_g)(\sum_{h} U_h) = \sum_{g} \sum_{h} U_gU_h = \sum_{g} \sum_{h} U_{gh} = \sum_{g} \sum_{k} U_{k} = \sum_{g} s = |G|s $$

The main point is that, for fixed $g \in G$, we have that $gh$ runs through all elements of $G$ as $h$ runs through the elements of $G$.

Whether you can then divide by $|G|^2$ to get what you want depends on the gcd of $|G|$ and the characteristic of $R$.

Answer 2

You can even say more: if $H$ is any finite subgroup of $G$ such that $|H|$ is a unit in $R$, then $e=\frac{1}{|H|}\sum_{h\in H}$ is an idempotent. (Of course, yours is a special case when $G$ is finite.)

After observing that $h'(\sum_{h\in H}h)=\sum_{h\in H}h$ for every $h'\in H$, it's clear that $(\sum_{h\in H}h)(\sum_{h\in H}h)=|H|\sum_{h\in H}h$, and dividing both sides by $|H|^2$ you have the proposed idempotent.

Also of note is the fact that this idempotent generates the left annihilator of the right augmentation ideal for $H$ in $R[G]$.