Let $G$ be a finite group acting transitively on a finite set $\Omega$. Let $K$ be field such that its characteristic divides $|\Omega|$.
Is it true that $K\Omega$ is not semisimple?
I think this is true. I've been trying to show that $ KG x = K x $ where $$x = \sum_{\omega \in \Omega} b_\omega $$ has no complement, but I've not had much success.
I want to show that IF it has a complement then the complement must be the submodule $$ W = \{ \ \sum_{\omega} \lambda_\omega b_\omega : \sum \lambda_\omega = 0 \ \}$$ I can complete the proof from this i think.
Note that $W = \ker \psi $ where $\psi: b_\omega \mapsto x $.
Yes, you're right:
The $KG$-linear maps $K\Omega\to Kx$ are precisely the scalar multiples of the map given by $b_\omega\mapsto x$ for all $\omega\in\Omega$: You can either check this directly by noting that the image of any $b_\omega$ determines the whole map by $G$-linearity, or alternatively firstly write $\Omega=G/H$ for a subgroup $H\subseteq G$, with $G$ acting by left multiplication, secondly note that $K\Omega\cong K(G/H)=\text{Ind}_H^G K$ is induced from the trivial $H$-module, and thirdly and finally apply the adjunction property to deduce $\text{Hom}_{KG}(K\Omega,Kx)\cong\text{Hom}_{KH}(K,Kx)\cong K$.
Once you know that the $G$-linear maps $K\Omega\to Kx$ have the above form, you conclude that the compositions $Kx\hookrightarrow K\Omega\to Kx$ are the multiples of the composition $x\mapsto\sum\limits_{\omega\in\Omega} b_\omega\mapsto |\Omega| x$, which vanishes if $|\Omega|=0$ in $K$. Hence, $Kx\hookrightarrow K\Omega$ does not split, and therefore $K\Omega$ is not semisimple as a $KG$-module.