a representation which is not completely reducible

If $G$ is a finite group, and $k$ is a field whose characteristic does divide the order of the group, then Maschke's theorem fails. For example let $V$ be the regular representation of $G$ , which can be thought of as functions from $G$ to $k$ , with the $G$ action $g\cdot\vp(g')=\vp(g^{-1}g')$ . Then this representation is not completely reducible.

There is an obvious trivial subrepresentation $W$ of $V$ , consisting of the constant functions. I claim that there is no complementary invariant subspace to this one. If $W'$ is such a subspace, then there is a homomorphism $\vp:V\to V/W'\cong k$ . Now consider the characteristic function of the identity $e\in G$

and $\ell=\vp(\delta_e)$ in $V/W'$ . This is not zero since $\delta$ generates the representation $V$ . By $G$ -equivarience, $\vp(\delta_g)=\ell$ for all $g\in G$ . Since $$\eta=\sum_{g\in G}\eta(g)\delta_g$$ for all $\eta\in V$ , $$W'=\vp(\eta)=\ell\left(\sum_{g\in G}\eta(g)\right).$$ Thus, $$\ker\vp=\{\eta\in V|\sum_{\in G}\eta(g)=0\}.$$ But since the characteristic of the field $k$ divides the order of $G$ , $W\leq W'$ , and thus could not possibly be complementary to it.

For example, if $G=C_2=\{e,f\}$ then the invariant subspace of $V$ is spanned by $e+f$ . For characteristics other than $2$ , $e-f$ spans a complementary subspace, but over characteristic 2, these elements are the same.