# Maschke's theorem and possible generalization?

One way to state maschke's theorem is that for a finite group $$G$$ and a field $$k$$ the following holds. If we see $$k[G]$$ as a left-module over itsself then $$k[G]$$ is semi-simple if and only if $$\text{char}(k) \nmid |G|$$. Now we want to generalize it as follows. Let $$G$$ act transitively on a fnite set $$X$$ and let $$F(X)$$ be the free vector space of $$X$$ over $$k$$. Then we can see $$F(X)$$ as a left-module over $$k[G]$$ via the group action. It is easy to show that if $$F(X)$$ is semi-simple then $$\text{char}(k)\nmid |X|$$. However i am not sure wether the converse holds or not. Does someone know a counterexample or a proof?

Let $$G=A_5$$ acting on transitively on $$15$$ points, Then the permutation module over $${\mathbb F}_4$$ (a splitting field in characteristic $$2$$) has two indecomposable components of dimension $$5$$ with composition factors of dimensions $$2$$, $$1$$, and $$2$$.
• Thanks for your answer. In general if $F(X)$ is not semi-simple it follows by maschke's theorem that $\text{char}(k) \mid |G|$. Now the interesting case is when exactly do we have $\text{char}(k)\nmid |X|$ aswell? (So when does it not follow trivialy from $F(X)$ being semi-simple?) – kevkev1695 Nov 29 '19 at 15:28
• @kevkev1695,note that here $\operatorname{char}(k)=2$ but $|X|=15$. – AnalysisStudent0414 Nov 29 '19 at 15:34
• Indeed. The question becomes whenever this exactly happens. If $\text{char}(k)\nmid |X|$ and $\text{char}(k) \mid |G|$ then clearly $\text{char}(k) \mid |G|/|X|$ which is the Order of a stabilizer $G_x$ of the group action. In particular it follows that $k[G_x]$ is not semi-simple over itsself. Can we work with that? – kevkev1695 Nov 29 '19 at 15:42