A small doubt in group rings.

Question

Let RG be a group ring then if $r \in R$ and $g \in G $ then why $rg=gr$ in RG?

What does the author means here. Why does these embeddings implies $rg=gr$

Answer 1

I guess the author used the standard defnition $$\left(\sum_{g\in G}a_gg\right)\left(\sum_{g\in G}b_gg\right):=\sum_{g\in G}\left[\left(\sum_{h\in G}a_hb_{h^{-1}g}\right)g\right]$$ From which it follows that $R=R\cdot e$ commutes with $G=1\cdot G$ through the identification he stated.

Actually, one may say that the multiplication above is the one and only making those identification commute between each other.