Understanding Maschke's Theorem https://en.wikipedia.org/wiki/Maschke%27s_theorem#Proof
I'm trying to understand the need for the condition 'K's characteristic does not divide the order of G' in the statement of the theorem. Where does this condition play a role in the proof?
 A: Suppose that $|G|=0$ in $k$ but $kG$ is semisimple. Then the augmentation map $$\epsilon:kG\to k:g\mapsto 1$$ splits because it is surjective. Let $s:k\to kG$ be a splitting and let $s(1)=\sum_{g\in G}\lambda_g g$. It is a $kG$-module map, thus $$\sum_{g\in G}\lambda_g g=s(1)=s(h1)=hs(1)=\sum_{g\in G}\lambda_g hg$$ for every $h\in G$. But that implies that $\lambda_g=\lambda_h$ for every $g,h\in G$ (because $G$ is a basis of $kG$). Let's call this common scalar $\lambda$. Then $s(1)=\lambda \sum_{g\in G} g$ and $$1=\epsilon s(1)=\epsilon( \lambda \sum_{g\in G} g)=\lambda |G|=0$$ which is, of course, a contradiction.
A: If  $char(F)$ divides $|G|$, then you can show that $\sum_{g\in G}g$ is a nonzero central nilpotent element, so it is in the Jacobson radical.  The Jacobson radical of a semisimple ring is zero of course.
A: If $p = \operatorname{char} k > 0$  divides $\# G$, then we can't construct the averaging projection $\frac{1}{\#G} \sum_g$. For a concrete example of what can go wrong, consider the representation of $G = \mathbb{Z}_p$ on $V = \mathbb{F}_p^2$ that sends a generator $g$ to
\begin{align*}
g \to \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}.
\end{align*}
Its only eigenvalue is $1$, but the action of $G$ on $V$ is nontrivial. The problem is that the representation $V'  = \mathbb{F}_p \oplus 0 \subset V$ doesn't split, as it would if we could construct a projection as in the case where $p$ does not divide $\#G$.
