If $k$ is algebraically closed and $\mathrm{char}k\nmid \mathrm{\mid }G\mathrm{\mid }$, then $k[G]\cong {\prod }_{i=1}^r{M}_{n_i}(k)$

By [L1], the ring $k[G]$ is semisimple. It is also nonzero because the basis element $[e]$ is its identity, so [L2] gives $k[G]\cong {\prod }_{i=1}^r{M}_{n_i}(D_i)$ for positive integers $r,n_i$ and division rings $D_i$.

For each factor, let $S_i$ be the simple module supplied by [L3]; under the ring isomorphism of step 1.1 it becomes a simple left $k[G]$-module, and then [L4] turns it into an irreducible representation of $G$ over $k$. By [L5] and [L6], every $k[G]$-module endomorphism of $S_i$ is scalar multiplication by an element of $k$.

Identify $S_i$ with the column module $D_i^{n_i}$. If $f:S_i\to S_i$ is $M_{n_i}(D_i)$-linear, write $e_1,\dots ,e_{n_i}$ for the standard column basis and let $d\in D_i$ be the first entry of $f(e_1)$. Since the matrix units $E_{aa}$ kill every coordinate except the $a$-th and $E_{a1}{e}_1=e_a$, linearity forces $f(e_a)=e_{a}d$ for every $a$. Hence $f(x)=xd$ for every column $x$, and composition corresponds to multiplication in the opposite order. So ${\mathrm{End}}_{k[G]}(S_i)\cong D_i^{\mathrm{o}\mathrm{p}}$. Step 2.1 therefore gives $D_i^{\mathrm{o}\mathrm{p}}\cong k$, hence $D_i\cong k$ because a field is canonically isomorphic to its opposite ring. Replacing each $D_i$ in step 1.1 by $k$ yields the claimed decomposition.