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.
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 $\ker(\phi)\subseteq J(F[G])$.