TITLE: Jacobson radical of a group algebra
QUESTION [7 upvotes]: I have not too much experience of playing with rings. So the question could be elementary but for me it is not till now.
Let $K$ be a field of characteristic $p$ and $G$ a finite group of order divisible by $p$. Can we determine the Jacobson radical of the group algebra $K[G]$? 
If this is difficult for arbitray finite group (with $p||G|$) then taking simplest example - $G=\langle x|x^p=1\rangle$, can we determine $J(K[G])$?
(The thing I know is that if characteristic of a field does not divides $|G|$ or if it is zero, then the group algebra is semi-simple so I can ensure that Jacobson radical of the group algebra is zero. I am considering complementary side of this fact.)

REPLY [5 votes]: It's well known that if you have a field $F$ of characteristic $p>0$ and a finite group $G$, then $F[G]$ is local (the maximum ideal is the augmentation ideal, and is also the Jacobson radical) iff $G$ is a $p$-group. This covers your case of the cyclic $p$-group.
If $G$ has a normal $p$-Sylow subgroup, I'd conjecture that he Jacobson radical is the kernel of the projection onto the groupring of he group mod the p-subgroup. It is certainly is contained in there, but I'm not totally sure it is equal.