Semisimplicity of group rings with finite groups Let $K$ be a field and $G$ a finite group such that $\text{ch}(K)\nmid |G|$. We have defined the group ring $$K[G]:=\left\{\sum_{g\in G}k_g\chi_g:k_g\in K,k_g\neq0\text{ for only finitely many }g\in G\right\}$$ via the addition of functions and the convolution product, i.e. the $K$-bilinear extension of $\chi_g\ast\chi_h=\chi_{gh}$. Here $\chi_g$ means the characteristic function for $g\in G$.
With the assumption $\text{ch}(K)\nmid |G|$ I want to prove the

Claim. $K[G]$ is semi-simple (as a ring), i.e. the left $K[G]$-module $_{K[G]}K[G]$ is a direct sum of simple submodules.

Remark. The conditions 1) $_{K[G]}K[G]$ is semi-simple, 2) $_{K[G]}K[G]$ is a sum of simple submodules and 3) for every submodule $N\subset _{K[G]}K[G]$ there is a submodule $L$ such that $_{K[G]}K[G]=N\oplus L$ are all equivalent.
We have to work with the
Hint. Let $V$ be a $K[G]$-module and $U$ a submodule of $V$. Since $V$ is a $K$-vectorspace, we can find a subvectorspace $W$ of $V$ such that $V=U\oplus W.$ Let $P$ be the projection onto $U$ and $$\widetilde{P}:=\frac{1}{|G|}\sum_{g\in G}\rho(g)P\rho(g^{-1}),$$ where $\rho(g)$ is the action of $g$ on $V$. Now show that $\widetilde{P}$ is a $K[G]$-module homomorphism and $\widetilde{P}^2=\widetilde{P}$.
Now assume that I can show the two points in the hint. How does the claim follow?
 A: If you let $U'=\ker\widetilde{P}$, those two points will show that $\widetilde{P}$ is a projection along $U'$, so that $V=U\oplus U'$ as vector spaces. One checks again that $U'$ is stable under the $G$-action, hence is actually a $K[G]$-submodule, and thus $V=U\oplus U'$ is actually a decomposition into $K[G]$-submodules. 
This shows any finite dimensional representation of $K[G]$ is completely reducible, which is one of the definitions of $K[G]$ being semisimple.
Perhaps you are aware, but this is Maschke's Theorem, details of the above can be found in Etingof's Represenation Theory, particularly Proposition 3.5.8 and Theorem 4.1.1.
A: If $M$ is an $R$-module and $p \colon M \to M$ is an idempotent endomorphism (i.e. $p^2 = p$) then $M = \operatorname{im} p \oplus \ker p$ (this is a straightforward calculation).
(The mapping $p \mapsto (\operatorname{im} p, \ker p)$ does in fact give a bijection between the idempotent endomorphisms of $M$ and the pairs $(L,N)$ of submodules $L, N \subseteq M$ with $M = L \oplus N$.)
Given any submodule $U \subseteq V$ the hint shows how to construct an idempotent endomorphism $\tilde{P} \colon V \to V$ with $\operatorname{im} \tilde{P} = U$.
Then $V = \ker \tilde{P} \oplus \operatorname{im} \tilde{P} = \ker \tilde{P} \oplus U$.
Hence every submodule of $V$ has a direct complement.
If $V$ is finite-dimensional then by iterating this process we can decompose $V$ into a direct sum of simple modules.
This does in particular apply to ${}_{K[G]} K[G]$ itself, since $G$ is finite.
