Exercise 6.1 in Serre's Representations of Finite Groups I am trying to show that if $p$ divides the order of $G$ then the group algebra $K[G]$ for $K$ a field of characteristic $p$ is not semisimple. Now Serre suggests us to consider the ideal 
$$U = \left\{ \sum_{s \in G} a_s e_s \hspace{1mm} \Bigg| \hspace{1mm} \sum_{s\in G} a_s = 0\right\}$$
of $K[G]$ and show that there is no submodule $K[G]$ - submodule $V$ such that $U \oplus V = K[G]$. Now I sort of have a proof in the case that $G = S_3$ and $K = \Bbb{F}_2$ but can't seem to generalise. 
Suppose there were such a $V$ in this case. Then $V$ is one - dimensional spanned by some vector $v \in K[G]$ such that the sum of the coefficients of $u$ is not zero. If that $v$ is say $e_{(1)}$, then by multiplying with all other other basis elements of $\Bbb{F}_2[S_3]$ and taking their sum we will get something in $U$, contradiction. 
We now see that the general plan is this: If $v \notin U$ is the sum of a certain number of basis elements not in $U$, we can somehow multiply $v$ by elements in $K[G]$ then take the sum of all these to get something in $U$, contradiction.
How can this be generalised to an aribtrary field of characteristic $p$ and finite group $G$?
Thanks.

Here's yet another solution to your problem. Consider the element
$$n=\sum_{g\in G} g$$
We notice that $n^2=\vert G\vert n=0$ since the caracteristic of $K$ divides the order of $G$. Furthermore, $n$ commutes with all elements in $G$, and defines a nilpotent $G$-equivariant endomorphism
$$l_n:K[G]\to K[G],x\mapsto nx$$
Our last observation is that $\mathrm{Ker}(l_n)= U$. Indeed, you can convince yourself of the following formula :
$$\forall x\in K[G],~nx=(\sum_{g\in G} x_g)n$$
where $x=\sum_{g\in G}x_g g$. We can now conclude that $U$ is not complemented as a $K[G]$-submodule, because a $G$-invariant complement would provide a $l_n$ stable complement $V$ such that $$K[G]=\mathrm{Ker}(l_n^2)=\mathrm{Ker}(l_n)\oplus V$$ which is impossible.
