let G be a finite group and let R be a field, show that RG is not an integral domain I know this is true because the group ring of the Quaternions Q$_8$ with the reals $\mathbb{R}$ contains zero divisors.  Also, in the case of finite cyclic groups this would also be true.  
I know that my general group ring will be the set of sums of the form: 
a$_1$g$_1$ + a$_2$g$_2$ + ... + a$_n$g$_n$, where a$_i$ is a field element and g$_i$ is a group element. 
To show that the ring of these elements is not an ID requires showing that there are zero divisors in this ring.  Here's what I had in mind: 
Since groups are closed under their operations and have inverses, for some composition of non-zero group elements I am forced to get zero.  I know this is probably the direction I should be thinking in, I'm just having a hard time putting it into formal language.  
 A: (Before we begin, let's note that $G$ has to be nontrivial, otherwise the group ring is isomorphic to the field, and is an integral domain.)
One good utility to have when working with group rings is this observation:

If $H$ is a finite subgroup of order $n$ in a group $G$, and we write $s(H)=(\sum_{h\in H} h)$, then $s(H)^2=n\cdot s(H)$ in any group algebra $R[G]$.

Now, the integer $n$ is either a unit of $R$ or a zero divisor depending on its relationship to the characteristic of $R$. This leads to two cases:


*

*If the integer $n$ is a unit in $R$, then you can divide both sides by $n^2$ to get $(\frac{s(H)}{n})^2=\frac{s(H)}{n}$. That is, $\frac{s(H)}{n}$ is an idempotent.

*Otherwise, there is some integer $m$ such that $m\cdot s(H)^2=m\cdot n\cdot s(H)=0$.
You will see the utility of the first bullet once you prove this: an idempotent element of an integral domain must be $0$ or $1$.
The second one I think you will understand how to use right away.
Now, take a nontrivial subgroup $H$ of your $G$ and apply what lies above. Just make sure you convince yourself why the interesting quantities above are not zero or the identity, and you have a complete proof.
Conveniently, this also extends the conclusion slightly to possibly infinite groups which have a nontrivial finite subgroup.
