It is well-known that the group algebra $F[G]$ is a direct sum of irreducible $G$-modules. The proof in my text book is as follows
Write $F[G] = \oplus_{i=1}^n V_i$, where $V_i$ is a set of irreducible submodules of $F[G]$ (Maschke).
Let $M$ be an irreducible G-module, then there holds as vector spaces
$$ M \cong \text{Hom}_{F[G]}(F[G],M) = \text{Hom}_{F[G]}(\oplus_{i=1}^n V_i,M) \cong \oplus_{i=1}^n \text{Hom}_{F[G]}(V_i,M) $$
Beceause $M \neq \{ 0 \}$ there exists an $i$ such that $\text{Hom}_{F[G]}(V_i,M) \neq \{0\}$. It follows by Schur's lemma that $V_i$ is isomorphic with $M$.
The proof is verry elegant, but my attemt to prove this is to construct a submodule $V$ of $F[G]$ which is isomorphic to $M$. Let $\{w_1,\ldots, w_n \}$ be a base of $M$. I want to define $V$ to be the span of linearly independant elements of $F[G]$.
We write those elements as
$$\sum_{j=1}^{|G|} a_{ij} \: g_j,$$
where $G= \{ g_1,\ldots, g_{|G|} \}$.
We define the isomorphism $\phi: M \rightarrow V$ as the natural linear map such that
$$\phi(w_i) = \sum_{j=1}^{|G|} a_{ij} \: g_j.$$
Our goal is to find the $a_{ij}$ such that $\phi$ is indeed an isomorphism.
We have to show that $\phi(g_k \cdot w_i) = g_k \phi(w_i)$ for each $k: 1, \ldots , |g|$.
Let $\lambda^k_{ij} \in F$ such that $$g_k \cdot w_i = \sum_{l=1}^n \lambda^k_{il} w_l. $$
Then $\phi(g_k \cdot w_i) = g_k \phi(w_i)$ is equivalent to
$$ \phi \left( \sum_{l=1}^n \lambda^k_{il} w_l \right) = \sum_{j=1}^{|G|} a_{ij} \: g_kg_j $$
$$ \Longleftrightarrow$$
$$ \sum_{l=1}^n \lambda^k_{il} \left( \sum_{j=1}^{|G|} a_{lj} \: g_j \right) = \sum_{j=1}^{|G|} a_{ij} \: g_kg_j. $$
Define the matrices $\Lambda^k = [\lambda^k_{ij}], A = [a_{ij}]$, then this can be written as
$$ \Lambda^k A
\begin{bmatrix}
g_{1} \\
g_{2} \\
\vdots \\
g_{|G|}
\end{bmatrix}
= A \begin{align}
\begin{bmatrix}
g_k g_{1} \\
g_k g_{2} \\
\vdots \\
g_k g_{|G|}
\end{bmatrix}
\end{align}. $$
The action of $g_k$ on $G$ can also be given by a matrix $L^k$, such that
$$ \Lambda^k A = A L^k. $$
How can we prove there exists such an $A$? In fact there exists $\dim M$ such $A's$, because this is the amount of times $M$ appears in $F[G]$.