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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.)

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1 Answers1

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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.

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rschwieb
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  • please help me https://math.stackexchange.com/questions/2995452/description-of-the-group-1jfg-where-jfg-is-jacobson-radical-of-the-gro – neelkanth Nov 12 '18 at 15:48
  • Yes, this "conjecture" is true, see Passman's "Infinite group rings", Dekker, New York, 1971, Theorem 16.6. A lot more (but not everything!) is known, e.g. if $G$ is a Frobenius group complemented by the $P$-Sylow [so $P$ and its conjugates are disjoint except at $1$]. – user46358 Aug 12 '19 at 13:42