Show that $G$ is a basis of group ring $RG$ over $R$. Show that $G$ is a basis of group ring $RG$ over $R$.
Comments: That $G$ is through direct generator, if $\alpha \in RG$ then $\alpha = \sum_{g \in G} a_gg$. 
I am not able to show that $RG$ is a linearly independent set.
The definition for group ring is: the set $RG$, whose elements are $\alpha = \sum_{g \in G} a_gg$ and this set is a ring defining 
$$\sum_{g \in G}a_gg + \sum_{g \in G}b_gg = \sum_{g \in G}(a_g+b_g)g$$
$$(\sum_{g \in G}a_gg)(\sum_{h \in G}b_hh) = \sum_{g,h \in G}(a_gb_h)(gh)$$
The set is a $R-$ module defining 
$$\lambda (\sum_{g \in G}a_gg)= \sum_{g \in G}(\lambda a_g)g$$ for $\lambda \in R.$

By definition, $RG$ is the $R$-module freely generated by the set $G$, which means that by construction $G$ is a basis of $RG$.
