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If $k$ is a field of characteristic $p>0$ and $G$ is a finite group of order divisible by $p$ then $k[G]$ is not a semi-simple ring.

My failed attempt: Since $G$ is finite and $p$ divisdes $|G|$ then $G$ has an element of order $p$. I wanted to use the cyclic subgroup generated by this element to create an ideal of $k[G]$ and show that it could not be produced by an idempotent element but I got stuck.

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Hint:

Square $\sum_{g\in G}g$, and observe that it is central.

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    Is $(\sum_{g \in G} g)^2 = |G| \sum_{g \in G} g$? (Because the sum runs over all elements in $G$ and if I multiply the sum by a single element in $G$ from left, the sum remains unchanged). And is central the same as being in the center of $k[G]$? – 2017-01-12
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    Oh! And since $\mathrm{char}(k) \mid |G|$ then $(\sum_{g \in G} g)^2 = 0$. – 2017-01-12
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    @user375190 yes, you understood my meaning of central correctly. I also agree with your computation of the square. And you see why this all means the ring is not semisimple? – 2017-01-13
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    I haven't written down what I'm going to say on a paper yet, but I guess the fact that it's central makes $R\alpha$ an ideal of $R=k[G]$ where $\alpha=\sum_{g\in G} g$. If it's indeed an ideal, then if $R=k[G]$ were semi-simple, $R\alpha$ should be generated by some idempotent element $e$, i.e. $R\alpha = Re$. But squaring them both gives us $Re=0$ but $R\alpha$ is not $0$ by definition of $k[G]$. Is that it? – 2017-01-13
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    @user375190 What you did is a solid approach. Actually what I had in mind was that this creates a nonzero nilpotent ideal (something which semisimple rings do not have.) – 2017-01-13
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    @user375190 nilpotent ideals lie in the Jacobson radical, and the Jacobson radical of a semisimple ring is zero. Obviously you can't have nonzero idempotents in a nilpotent ideal. – 2017-01-13
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    @user375190 Nilpotent elements are **not** necessarily in the radical of a ring (but the central nilpotent elements are.) Nilpotent ideals being contained in the Jacobson radical is an elementary fact: if you view $J(R)$ as the intersection of primitive ideals, they are necessarily prime ideals, and so $T^k=0\subseteq P\implies T\subseteq P$ etc – 2017-01-13