Let $G:=\{g_1:=1_G,g_2,\dots,g_n\}$ be a finite multiplicative group of order $n\in \Bbb Z^+$ and $R$ be a ring with $1_R$. We will write $RG$ for the group ring of $G$ over $R$.
I am studying some theory for group rings from the book "An Introduction to Group Rings", by Milies and Sehgal and I stuck on the following point.
It says that the map $i:R\longrightarrow RG,\ g\longmapsto i(g):=1_Rg $ is an embedding. Thus, $G \lesssim RG$.
And I tried to prove that. Without any difficulty, $i$ is a group homomorphism, since $$\forall g,h\in G:\ i(gh)=1_R(gh)=(1_Rg)(1_Rh)=i(g)i(h).$$ Now let's compute the kernel of $i$. It is, $$\ker i=\{g\in G:\ i(g)=0_{RG}\}=\{g\in G:\ 1_Rg=0_R1_G\}.$$ Is $0_{RG}=0_R1_G$ and how can we prove that $\ker i =\{0_G\}$ (and thus $i$ is injective)?
Moreover we can consider the map $\nu:R\longrightarrow RG,\ r\longmapsto \nu (r):=r 1_G$ and similarly $$\ker \nu = \{ r\in R:\ \nu(r)=0_{RG} \}= \{ r\in R:\ r1_G=0_R1_G \}.$$ But here, $r1_G=0_R1_G \iff r=0_R$ and we are done. Right?
Thank you.