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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 >.<.

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    $\begingroup$ Note that for every $g\in G$, $g\sigma=\sigma$ because $x\mapsto gx$ is a bijection $G\to G$. $\endgroup$ – SMM Sep 12 '18 at 21:25
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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 $\Bbb Q G$.

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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 $$(\frac{1}{n} \sigma)^2=??$$

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  • $\begingroup$ Wow what a great comment! But how do we know that $((1/n) \sigma) \neq e$? And also how does this lead us to understand that $\mathbb{Q}G$ can be written as a nontrivial direct sum and hence has "many" zero divisors? $\endgroup$ – Math is hard Sep 14 '18 at 20:47
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    $\begingroup$ @MichaelVaughan $\frac{1}{n} \sigma=e$ contradicts the linear independence of $G$ over $\mathbb Q$. Also, if $f$ is any idempotent in any ring $R$ then it is well known that $$R= Rf \oplus R(1-f)$$ Indeed $x=xf+x(1-f)$ and if $x \in Rf \cap R(1f)$ then $$x \in Re \Rightarrow xf=x \\ x \in R(1-f) \Rightarrow x(1-f)=x \\ x=x(1-f)-xf=x-x=0$$ $\endgroup$ – N. S. Sep 15 '18 at 0:23
  • $\begingroup$ @MichaelVaughan If $R= T \oplus S$ as rings , then $$(x,0) \cdot (0,y)=(0,0)$$ so there are "many" zero divisors. $\endgroup$ – N. S. Sep 15 '18 at 0:24

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