Question on Group ring $\mathbb{Q}G$ where $G$ is a finite group Let $G=\{e=g_1,g_2,.....g_n\}$ be a finite group of order n and let $\mathbb{Q}G$ be the group ring. Let $\sigma=\sum_1^ng_i$. Prove that $\sigma^2=n\sigma$ and deduce that $\mathbb{Q}G$ has a nontrivial idempotent $e$. Deduce that $\mathbb{Q}G$ can be written as nonrivial direct sum and hence has "many" zero divisors.
Okay, so this one is looking pretty tricky. As far as showing $\sigma^2=n\sigma$, I guess I should square that sum and see what happens. I'm a little nervous about distribution two $n$ component polynomials, so maybe there is a way I can think about it involving permutations or something? And relaly I just need help on this, i've thought about it and went over my notes and am not seeming to be getting anywhere >.<.

Hint For each $g \in G$ the function $f: G \to G, f(x)=gx$ is a bijection. Use this to deduce that 
$$g\sigma = \sigma$$
From here you get 
$$\sigma^2= \sum_{j=1}^n g_j \sigma=  \sum_{j=1}^n  \sigma$$
which gives the first claim.
Finally, what is 
$$(\frac{1}{n} \sigma)^2=??$$
