So I'm working on Abstract Algrebra (Dummit & Foote).
Let $R$ be a commutative ring with identity $1$ and let $G=\{g_1, ..., g_n\}$ be a finite group. Prove that $$I=\{\sum_{i=1}^n ag_i | a\in R\},$$ i.e. the $R$-multiples of the element $g_1+...+g_n$, is an ideal in the group ring $RG$.
On the one hand,
$(1).$ $I$ is non-empty since $$g_1+...+g_n=\sum_{i=1}^n 1g_i\in I.$$
$(2).$ $I$ is closed under subtraction since $$\sum_{i=1}^n ag_i-\sum_{i=1}^n bg_i=\sum_{i=1}^n (a-b)g_i.$$
$(3).$ $I$ is closed under multiplication by all the elements in $RG$, so we need to prove that $$(\sum_{i=1}^n b_ig_i)\cdot (\sum_{i=1}^n ag_i)=\sum_{k=1}^n(\sum_{g_ig_j=g_k} ab_j)g_k\in RG.$$ How can one show that $$\sum_{g_ig_j=g_k} ab_j=\sum_{g_ig_j=g_s} ab_j$$ for any $1\leq k,s\leq n$?
Alternatively, $I$ must be the kernel of some ring homomorphism. It seems to be trivial but I just cannot find it.
Any help is appreciated!