Silly question about group rings Let $R$ be a ring and $G$ be a finite group, and $RG$ be the group ring. 
What does it mean to say that $|G|$ is invertible in $R$? Since $|G| \in \mathbb{N}$, $|G|$ is not an element of $R$, does it mean that $|G|$ is invertible is same as there exists an element $x \in R$ s.t. $x|G| = 1$? Does it mean that $|G|$ is invertible in $R$ if there exist an element of $R$ which if added $|G|$ times gives $1$?
But if it means this, what does $\frac{1}{|G|} \sum_{\alpha \in RG} \alpha$ mean? How could $|G|$ divide an element of a ring?
 A: As $R$ is a ring, it contains a multiplicative identity $1_R$. As such, there is a unique ring homomorphism $\phi : \mathbb{Z} \to R$ given by $\varphi(n) = n\cdot 1_R$; note, $n\cdot 1_R$ does not denote the multiplication of two elements of $R$, rather it is shorthand for the sum of $n$ copies of $1_R$ if $n$ is positive, or $|n|$ copies of $-1_R$ (the additive inverse of $1_R$) if $n$ is negative. 
The element $n\cdot 1_R$ is sometimes denoted by $n$, but to distinguish it from the element of $\mathbb{Z}$, the notation $\overline{n}$ is also used; the source you are using follows the former convention. Therefore, the statement "$|G|$ is invertible in $R$" really means $|G|\cdot 1_R$ is invertible in $R$. That is, $|G|\cdot 1_R$ has a multiplicative inverse, which is an element $x \in R$ such that $x(|G|\cdot 1_R) = (|G|\cdot 1_R)x = 1_R$. Note that
$$1_R = (|G|\cdot 1_R)x = (\underbrace{1_R + 1_R + \dots + 1_R}_{|G|\ \text{summands}})x = \underbrace{1_Rx + 1_Rx + \dots + 1_Rx}_{|G|\ \text{summands}} = |G|\cdot x$$
so as you correctly said in your post, the invertibility of $|G|$ implies that there is an element (namely $x$) which, when added to itself $|G|$ times, becomes $1_R$.
In the formula you wrote, $\frac{1}{|G|}$ is used to denote the multiplicative inverse of $|G|\cdot 1_R$ (namely $x$); the motivation for this notation is the identity $|G|\cdot x = 1_R$. So the formula could be rewritten as $x\sum_{\alpha\in RG}\alpha$ which is well-defined if $RG$ is finite.
A: Yes. ${}{}{}{}{}{}{}{}{}{}{}{}{}$
A: For example, if $G=S_3$ and $R=\mathbb{Z}_{19}$ then $|G|=6$ is invertible in $R$ (in fact, $\bar 6$ , the image of $6$ under the canonical morphism $\mathbb{Z}\to R$, is ), since $6 \cdot 16 = 96 = 1 (\!\!\!\mod 19)$ that is, $\bar 6 \cdot \overline{ 16} = \bar 1 $ in $
\mathbb{Z}/19$.
