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 >.<.
 A: First of all, before we compute $\sigma^2$, where
$\sigma = \displaystyle \sum_{i = 1}^n g_i, \tag 1$
we observe that $\sigma$ is in the center of $G$; that is,
$\sigma \in Z(\Bbb Q G); \tag 2$
this follows from the group ring identity
$g_j \left ( \displaystyle \sum_{i = 1}^n g_i \right ) = \left ( \displaystyle \sum_{i = 1}^n g_i \right ) g_j, \forall g_j \in G, \tag 3$
which follows from
$g_j \left ( \displaystyle \sum_{i = 1}^n g_i \right ) = \displaystyle \sum_{i = 1}^n g_j g_i \tag 4$
and
$\left ( \displaystyle \sum_{i = 1}^n g_i \right ) g_j = \displaystyle \sum_{i = 1}^n g_i g_j, \tag 5$
and we have
$\displaystyle \sum_{i = 1}^n g_j g_i = \sum_{i = 1}^n g_i g_j, \tag 6$
which binds since each side is simply a sum of all $g \in G$, being as left or right multiplication by an element simply 
permutes the members of $G$ (which we take as well-known); (3) then follows from (4), (5), and (6) acting together in collusion.  And from (3) it follows by linearity that  
$g \left ( \displaystyle \sum_{i = 1}^n g_i \right ) = \left ( \displaystyle \sum_{i = 1}^n g_i \right ) g, \forall g \in G, \tag 7$
that is, (2).
We proceed to calculate $\sigma^2$:
$\sigma^2 = \left ( \displaystyle \sum_{i = 1}^n g_i \right ) \left ( \displaystyle \sum_{j = 1}^n g_j \right ) = \left ( \displaystyle \sum_{i = 1}^n g_i  \left ( \displaystyle \sum_{j = 1}^n g_j \right ) \right ) = \displaystyle  \sum_{i = 1}^n \left ( \displaystyle \sum_{j = 1}^n g_i g_j \right ); \tag 8$
again, since $g_i$ merely permutes the $g_j$, we see that
$\displaystyle \sum_{j = 1}^n g_i g_j = \sum_1^n g_j = \sigma,\tag 9$
whence
$\sigma^2 = \displaystyle \sum_1^n \sigma = n \sigma. \tag{10}$
Now let
$e = \dfrac{\sigma}{n}; \tag{11}$
then
$e^2 = \dfrac{\sigma^2}{n^2} = 
\dfrac{n \sigma}{n^2} = \dfrac{\sigma}{n} = e, \tag{12}$
so $e$ is idempotent; furthermore, in the light of (2),
$e \in Z(\Bbb Q G) \tag{13}$
as well.
Now consider the two-sided principal ideals
$(\Bbb Q G) e, (\Bbb Q G)(1 - e) \subset \Bbb Q G; \tag{14}$
each is a sub-ring with unit of $\Bbb Q G$, the unit of $\Bbb Q G$ is easily seen to be $e$, since
$(re)e = re^2 = re, \; \forall re \in (\Bbb Q G)e; \tag{15}$
also, since
$(1 - e)^2 = 1 - 2e + e^2 = 1 - 2e + e = 1 - e, \tag{16}$
that is, $1 - e$ is itself idempotent, it is the unit of $(\Bbb Q G)(1 - e)$:
$(r(1 - e)(1 - e) = r(1 - e)^2e^2 = r(1 - e), \; \forall r(1 -e) \in (\Bbb Q G)(1 - e); \tag{17}$
and if $t \in \Bbb Q G$,
$t = t1 = t(e + (1 - e)) = te + t(1 - e) \in (\Bbb Q G)e + (\Bbb Q G)(1 - e), \tag {18}$
i.e.,
$\Bbb Q G = (\Bbb Q G)e + (\Bbb Q G)(1 - e); \tag{19}$
finally, 
$(\Bbb Q G)e \cap (\Bbb Q G)(1 - e) = \{0\}, \tag{20}$
since if
$re = s(1 - e), \tag{21}$
then
$s(1 - e) = re = re^2 = ree = s(1 - e)e = s(e - e^2) = 0; \tag{22}$
thus at last we may write
$\Bbb Q G = (\Bbb Q G)e \oplus (\Bbb Q G)(1 - e), \tag{23}$
the direct sum of $(\Bbb Q G)$ and $(\Bbb Q G)(1 - e)$.
Since 
$a \in (\Bbb Q G)e, b \in (\Bbb Q G)(1 - e) \Longrightarrow ab = 0, \tag{24}$
we see that there are a panoply of zero divisors in $\Bbb Q G$.
A: 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 
$$(\frac{1}{n} \sigma)^2=??$$
