I am searching a simple example of a finite group $G$, so that the Jacobson radical $J(FG)$, of group algebra $FG$ is commutative, where $F$ is a finite field. I know example for that if $G$ is any finite abelian group, or dihedral group when characteristic of field is $2$ and $p$-groups, or when group ring is semi simple. Please provide me any other example of such finite group $G$ for which corresponding Jacobson radical is commutative. Thank you.



Example of a group algebra with commutative Jacobson radical
Here is a family of examples. Let $F$ have characteristic two, and let $G$ be a finite group of order $2n$ with $n$ odd. Then $J(FG)$ squares to zero, and is therefore commutative.


Another (similar) family of examples is as follows. Let $F$ have characteristic $p$, let $H$ be a $p'$-group, and let $t$ be an automorphism of $H$ of order $p$ which fixes only the identity element. Let $G$ be the semidirect product $H \rtimes \langle t\rangle$. Then every non-principal block of $FG$ has defect zero, and the principal block is commutative, so $J(FG)$ is commutative.


