Conceptual doubt in a theorem in Group rings

Question

Let R be a ring and G be a group and $H \unlhd G$ , where RG is the group ring. Define $\widehat{H}= \sum_{h \in H } h$. And $e_H$ defined below is central idempotent. Now I was understanding the following proof..

Now I have some problem in understanding the group homomorphism $\phi : G \to Ge_H$ defined $g \to ge_H$ has $ker(\phi)= H$.

Conceptual doubts-

1) First I guess I am not clear on how $Ge_H$ is defined as a group. According to me $Ge_H$ is a a subset of Group ring $RG$ containing element of type $\frac{1} {|H|}(1_R.gh_1+1_Rg.h_2+ \ldots +1_Rg.h_k+\ldots)\ \forall\ g \in G$. So if it is a group it should be group under + in $RG$ so let $ x=\frac{1} {|H|}(1_R.gh_1+1_Rg.h_2+ \ldots +1_Rg.h_k+\ldots)\ \text{and }\ y= \frac{1} {|H|}(1_R.g'h_1+1_Rg'.h_2+ \ldots +1_Rg'.h_k+\ldots)\$$ be two different elements of $Ge_H$. Then $x+y= \frac {1} {|H|}(1_R.gh_1+1_Rg.h_2+ \ldots +1_Rg.h_k+\ldots)+(1_R.g'h_1+1_Rg'.h_2+ \ldots +1_Rg'.h_k+\ldots)) $, so why is closure satisfied? What is the correct operation on $Ge_H$ ?
(Clearly I am doing some mistake here, I guess I am not using correct operation, but this the operation in RG is considered a group, so what is I am getting wrong here? Please help me understand it.)

2) If $h \in H$ then $\phi(h)=he_H=e_H$. So $e_H$ should be the identity of $Ge_H$ (If I can see it as a group under some appropriate operation). How?

3) Also what does $ {|H|} $ is invertible in a ring means. Am I right that it means if there is some element $a$ in R s.t. when $a$ is added |H| times it gives 1, because |H| is a integer and Ring can have elements as matrices so obviously, |H| does not have to be in R , and this must be the meaning of being invertible in R.

Please clear these basic doubts of mine. I will be highly thankful. I am studying on my own and these things are not cleared in the book. Untill these are clear, I am stuck.

Answer 1

$Ge_H$ is a group under multiplication! However, it usually won't be a subgroup of the group of units $R[G]^\times$.

First of all, it is multiplicatively stable: if $x,x'\in Ge_H$, then there are $g,g'\in G$ with $x=ge_H$ and $x'=g'e_H$, thus $x\times x'=ge_H\times g' e_H=(gg')e_H\in Ge_H$ since $e_H$ is central idempotent.

Associativity is clear, since multiplication is already associative on $R[G]$.

It (clearly) has a unit $e_H$, and inverses: if $x\in Ge_H$, then there exists $g\in G$ such that $x=ge_H$, and $x'=g^{-1}e_H$ satisfies $xx'=x'x=e_H$.

Therefore $(Ge_H,\times)$ satisfies all the axioms of a group, and the map $G\to Ge_H$ is clearly a morphism of groups. You can easily see that its kernel is $H$ using the fact that $e_g,g\in G$ is an $R$-basis of $R[G]$, and that for $g\notin H$, $gH\neq H$ (actually $gH\cap H=\emptyset$).

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Now for what it means that "$|H|$ (a positive integer) is invertible in $R$", first remember the following: for any unital ring $R$, there is a unique morphism of unital rings $u:\mathbb Z\to R$. Here is why:

1. For any group $(G,\cdot)$ and any element $g\in G$, there exists a unique morphism of groups $\phi=\phi_g:(\Bbb Z,+)\to (G,\cdot)$ such that $\phi(1)=g$.
2. Apply this to $(G,\cdot)=(R,+)$ and $g=1_R$, and you get that there is a unique morphism of groups $u:\Bbb Z\to R$ suc that $u(1)=1_R$.
3. Verify by hand that $u$ is actually a morphism of unital rings (it's simple enough).
4. If $v:\Bbb Z\to R$ is any other morphism of unital rings, then by definition, it must map the multilicative unit of $\Bbb Z$ to that of $R$, i.e. $v(1)=1_R$. But $v$ is also a morphims of groups $(\Bbb Z,+)\to(R,+)$, and thus by the first point, $v=u$.

Thus, any integer $n\in\Bbb Z$ defines a canonical element of $R$, namely $u(n)$. If $n$ is positive, then this element is $\underbrace{1_R+\cdots+1_R}_{n\text{ times}}$. This element of $R$ may or may not be invertible in $R$. If it is, we say that $n$ is invertible in $R$, even though we actually mean that $u(n)$ is invertible in $R$.