Step to prove that if |G| is infinite or |G| is divisible by char(k) then k[G] is not semisimple.

Question

I know that there are plenty of resources proving that if |G| is infinite or |G| is divisible by char(k) then k[G] is not semisimple, but I was instructed to try it in a particular way.

I am down to the point where I need to show that the short exact sequence $0\rightarrow \text{ker}(\epsilon)\rightarrow k[G]\rightarrow k\rightarrow 0$ does not split where $\epsilon$ is the augmentation homomorphism $\epsilon:k[G]\rightarrow k$ defined by $$\sum_{g\in G, \text{ almost all } \lambda_g=0} \lambda_g g\mapsto \sum_{g\in G}\lambda_g.$$

I understand the other ways of going about this proof but I find that I am pretty weak with working with short exact sequences, so I want to continue trying this method. A hefty hint would be greatly appreciated.

Answer 1

If the short exact sequence splits, then there's a $k[G]$-module homomorphism $\psi:k\to k[G]$ such that $\epsilon\circ\psi$ is the identity on $k$. Let $\alpha=\psi(1)\in k[G]$. Then $\epsilon(\alpha)=1$ so $\alpha\ne0$. Also as $\psi$ is a $k[G]$ module homomorphism, $\alpha g=\alpha$ for all $g\in G$, so that if $\alpha=\sum_{g\in G}a_g g$, then all the $a_g$ are equal. That's impossible for $G$ infinite.

Suppose then that $G$ is finite. Then $\alpha=a_1\sum_{g\in G}g$ and $1=\epsilon(\alpha)=a_1|G|$. If $k$ has characteristic $p$ and $p\mid |G|$, again this is impossible.