The group ring $R[G]$ is a unital $R$-algebra with basis $G$, and each $g\in G$ is a unit of $R[G]$

For each fixed $g\in G$, the set map $u_g:G\to R[G]$, $u_g(h):=[gh]$, extends uniquely by [L2] to an $R$-linear map ${\lambda }_g:R[G]\to R[G]$ with ${\lambda }_g([h])=[gh]$ for every $h\in G$.

For each fixed $y\in R[G]$, the set map $v_y:G\to R[G]$, $v_y(g):={\lambda }_g(y)$, extends uniquely by [L2] to an $R$-linear map $m_y:R[G]\to R[G]$. Define the product by $xy:=m_y(x)$.

By construction, $[g][h]=m_{[h]}([g])={\lambda }_g([h])=[gh]$. The map $x\mapsto xy$ is $R$-linear because $m_y$ is, and if $x={\sum }_{g\in F}{r}_g[g]$ then $xy={\sum }_{g\in F}{r}_g{\lambda }_g(y)$, so the $R$-linearity of each ${\lambda }_g$ makes $y\mapsto xy$ linear as well. Thus the product is $R$-bilinear.

For fixed $y,z\in R[G]$, the maps $x\mapsto (xy)z$ and $x\mapsto x(yz)$ are $R$-linear by step 3.1, so it is enough to compare them on basis elements $[g]$. For fixed $g,z$ the maps $y\mapsto ([g]y)z$ and $y\mapsto [g](yz)$ are likewise linear, so it is enough to compare them on basis elements $[h]$. Repeating once more in the variable $z$ reduces associativity to basis triples, where $([g][h])[k]=[(gh)k]=[g(hk)]=[g]([h][k])$ by associativity in $G$. Hence the product on $R[G]$ is associative.

The same bilinear reduction shows that $[e]$ is a two-sided identity, because $[e][g]=[eg]=[g]$ and $[g][e]=[ge]=[g]$ for every basis element. Likewise $[g][g^{-1}]=[e]=[g^{-1}][g]$, so each $[g]$ is a unit with inverse $[g^{-1}]$.

The map $\eta :R\to R[G]$ defined by $\eta (r)=r[e]$ is additive and satisfies $\eta (rs)=(r[e])(s[e])=rs[e]$ and $\eta (1_R)=1_R[e]=[e]$ by steps 3.1 and 4.2. For every basis element $[g]$, one has $\eta (r)[g]=r[g]=[g]\eta (r)$; bilinearity extends this equality to every element of $R[G]$.

If $\ast$ is any other $R$-bilinear product with $[g]\ast [h]=[gh]$, then for $x={\sum }_{g\in F}{r}_g[g]$ and $y={\sum }_{h\in E}{s}_h[h]$ bilinearity forces $x\ast y={\sum }_{g\in F}{\sum }_{h\in E}{r}_g{s}_h[gh]$, which is exactly the product already constructed in steps 1.1-3.1. Therefore the multiplication is unique, and with steps 4.1-5.1 it makes $R[G]$ a unital $R$-algebra as in [L3].