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Let $G$ be a finite group and $F$ be a field of prime characteristic $p$. Suppose further that $G$ has a normal $p$-Sylow subgroup $N$.

Question: Is it true that the Jacobson radical of the group algebra $J(F[G])$ is the kernel of the homomorphism $\phi:F[G]\to F[G/N]$ induced by the projection $G\to G/N$?

It would be true if the following conjecture is true:

Conjecture: $\ker(\phi)$ is a nilpotent ideal.

By Maschke's theorem $F[G/N]$ is semisimple, so $\ker(\phi)\supseteq J(F[G])$ in any case. And the converse would be true if the kernel were a nilpotent ideal. I think the conjecture is probably obvious if $N$ is contained in the center of $G$ because then $(x-1)R$ is nilpotent for each $x\in N$, and I think elements of the form $(1-x)$, $x\in N$ probably generate the kernel.

I arrived at this question while working on another question about the Jacobson radical of $F_3[S_3]$, but there it was easier to conclude that the kernel of this homomorphism is a nilpotent ideal due to some special properties of a generator of the kernel. I'm not sure how one would make progress in the general case. Maybe the things that happen for $F_3[S_3]$ also happen for $N$ and $G$ as described, and it is just not as obvious.

This is all consistent in the special case where $G$ is a finite $p$-group, where it is well-known that $F[G]$ is local, the radical being the augmentation ideal.

Community
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rschwieb
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1 Answers1

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Yes, it's true.

You already know that $J(F[G])\subseteq\ker(\phi)$.

Let $S$ be a simple $F[G]$-module. By Clifford's theorem, the restriction of $S$ to $F[N]$ is semisimple. Since $N$ is a $p$-group, the only simple $F[N]$-module is the trivial module. So $N$ acts trivially on $S$. Therefore the action of $G$ on $S$ factors through $G/N$, and so $\ker(\phi)$ annihilates $S$. Since $J(F[G])$ is the intersection of the annihilators of all simple $F[G]$-modules, this proves that $\ker(\phi)\subseteq J(F[G])$.

Jeremy Rickard
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  • Nice! This is part of group representations that I never reached. If we have a non-normal $p$-subgroup, these ideas would apply to the group ring for the normalizer of the $p$-subgroup: do you think we would still be able to conclude that the elements of the kernel there are contained in $J(F[G])$ (although perhaps the two sets become unequal in this situation)? – rschwieb Dec 08 '16 at 18:07
  • @rschwieb No, if by "the kernel there" you mean the kernel of $F[N_G(P)]\to F[N_G(P)/P]$. That would imply that if $P$ is a Sylow $p$-subgroup of $N_G(P)$ then $J(F[N_G(P)])\subseteq J(F[G])$, so that every simple $F[G]$-module restricts to a semisimple $F[N_G(P)]$-module. But, for example, you could take $p=2$ and $G=S_3$, so $P=N_G(P)=C_2$, where there is a $2$-dimensional simple $F[G]$-module that restricts to the regular module for $F[P]$. – Jeremy Rickard Dec 08 '16 at 19:37
  • OK, great. Thanks for the simple example (and one very close to the one I was already thinking about, too.) I am not at all familiar with these patterns of reasoning and it is very instructive. – rschwieb Dec 08 '16 at 19:51
  • @JeremyRickard 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 15:47