The Maschke's theorem say that\
Let $G$ be a finite group and $F$ a field whose characteristic does not divide $\mid G \mid$. Then every $FG$-module is completely reducible (I'm using the notation of Isaac's in the book Character Theory of finite groups).
The converse of this theorem is true, but I can not prove it.\
I would like your help, if possible.
Thank you.
Below I outline the idea of proof.
Best Answer
If $kG$ were completely decomposable, then there would exist an ideal $I$ for which $kG=kv\oplus I$. I claim this would imply $$I=\left\{\sum_{g\in G} c_gg:\sum c_g=0\right\}.$$ First, we show that everything in $I$ is of this form. Suppose $x=\sum_{g\in G} c_gg\in I$. This implies that $\sum_{h\in G} h\cdot x\in I$ as well, so that (using standard sum rearrangment and reindexing tricks) $$ \sum_{h\in G} h\cdot\sum_{g\in G} c_gg=\sum_{g\in G} c_g\sum_{h\in G}hg=\sum_{g\in G} c_g\sum_{h\in G} h=\left(\sum_{g\in G} c_g\right)v\in I $$ But $\left(\sum_{g\in G} c_g\right)v$ is in $kv$ as well. Since $kv\cap I=\{0\}$, this means that $\sum c_g=0$.
This proves $I\subseteq \left\{\sum c_gg:\sum c_g=0\right\}.$ The reverse inclusion must hold as well, since the both of these are vector subspaces with a dimension of $|G|-1$.
However, it cannot be the case that $kG=kv\oplus I$, since $kv\subset I$! Specifically, the sum of the coefficients of $v=\sum g_i$ is $\sum_{i=1}^{|G|}1=|G|=0$, since $p$ divides $|G|$. Thus, $kG$ cannot be completely decomposable.