Show that $ker(\phi)=H$ Ler $R$ be a ring and $G$ be a group, $RG$ be the group ring and $H$ be a normal subgroup of $G$. So if |$H$| is invertible in $R$, then setting $e_H=\frac {1} {|H|} \widehat {H}$ where $\widehat{H}=\sum_{h \in H} h$, I have to show that $\phi : G \to Ge_H$ defined by $g \to ge_H$ is a group epimorphism and $ker(\phi)=H$.
Now my problem is in showing $ker(\phi)=H$. As if I let $h \in H$ then $h$ goes to $he_H=\frac {h} {|H|} \widehat {H}$ but $h. \widehat {H}= \widehat{H}$ so $he_H=e_H$. Why is it identity?

If $g\notin H$ then $H\cap gH=\varnothing$. You can check that the coefficient of elements of $H$ in $ge_H$ is zero, but the coefficient of elements of $H$ in $e_H$ is not zero, so they are not equal.
More precisely, remember the formal definition of group ring: it is the set of all functions $f:G\to R$ with finite support. $ge_H$ sends the elements of $H$ to zero, but $e_H$ sends the elements of $H$ to nonzero element of $R$. 
