Cfrgtkky

My group is $$G=(mathbb{Z}_3timesmathbb{Z}_3)rtimesmathbb{Z}_3= {a,b,c:a^3=b^3=c^3=e, ab=ba, ac=ca, cb=abc}.$$ which is non abelian group of order $27.$

Now my problem is whether the group $1+J(FG)$ is abelian or non-abelian and what is its exponent? Here $F$ is any finite field of characteristic $3.$ I only know that $(1+J(FG))^{27}=1,$ by using below proposition given in the book "The Jacobson radical of group algebras" by G.Karpilovsky.

$textbf{Proposition}$. Let $N$ be a normal subgroup of $G$ such that $G/N$ is $p$-solvable. If $|G/N|=np^a$ where $(p,n)=1$ then $$J(FG)^{p^a}subseteq FG.J(FN)subseteq J(FG)$$ In particular, if $G$ is $p$-solvable of order $np^a$ where $(p,n)=1,$ then $$J(FG)^{p^a}=0.$$

Please anyone try to help me . I will be very thankful. Thanks.

It seems to me that since $J(FG)$ is maximal and contains the augmentation ideal, it contains $a-1, b-1$ for any elements $a,bin G$. In particular, if you select $a,b$ such that $abneq ba$ this is true.

Then $a=a-1+1$ and $b=b-1+1$ are both in $1+J(FG)$ and they don't commute.

I'm not sure why anything harder is necessary...