The group algebra is not semisimple if characteristic divides group order.

Question

I'm studying a proof that if a prime $p$ has $p\mid |G|$ and $k$ is a field of characteristic $p$, then the group algebra $kG$ is not semisimple.

My issue is that there is an assertion in the first line that I don't understand - given this assertion, I am fine with the rest of the proof.

The proof goes like this:

1. If $kG$ were semisimple, the trivial module $k$ would appear exactly once as a summand in a decomposition of $kG$ into simple $kG$ modules. This must be something special about the group algebra, because e.g. $k\oplus k$ doesn't satisfy this and is semisimple, but I don't know where this comes from.

2. By the Artin-Wedderburn theorem, any composition series of $kG$ has exactly one factor isomorphic to $k$.

3. The augmentation ideal $\Sigma$ (the kernel of the augmentation map sending each group element to 1) has $kG/\Sigma\cong k$.

4. $\sigma=\sum_{g\in G}g$ lies in the augmentation ideal since $k$ is of characteristic $p$. So, $k\sigma$ is a submodule of $kG$ contained in the augmentation ideal.

5. Refining $kG\supset \Sigma \supset k\sigma\supset 0$ to a composition series will give at least two factors isomorphic to the trivial module, a contradiction.

I understand how, assuming (1), the proof works, and I'm sure I'm missing something blatantly obvious. Why is (1) true?

Answer 1

As for the question at hand, I can't think of anything better than what Matthew Towers has already suggested in the comment. What I suggest below is probably closely related to that observation.

Although certain steps in that proof intrigue me, and are probably worthwhile learning, it seems like there are far simpler ways to prove it. If I were proving this I would just note that:

$x=\sum_{g\in G}g$ is

1. nonzero
2. central
3. nilpotent (it squares to zero.)
4. and therefore $x$ is in the Jacobson radical of the ring.

Since semisimple rings have trivial Jacobson radical, the ring is not semisimple.