# Description of the group $1+J(FG),$ where $J(FG)$ is jacobson radical of the group ring $GF.$

My group is $$G=(\mathbb{Z}_3\times\mathbb{Z}_3)\rtimes\mathbb{Z}_3$$ 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 $$(1+J(FG))^{3^3}=1,$$ 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 $$a,b\in G$$. In particular, if you select $$a,b$$ such that $$ab\neq ba$$ this is true.
Then $$a=a-1+1$$ and $$b=b-1+1$$ are both in $$1+J(FG)$$ and they don't commute.
• my last question is about exponent of group $1+J(FG)$...thanks ... – neelkanth Nov 12 '18 at 16:36
• Can i say its exponent must be $27$ as $3$ and $9$ order groups are always abelian.? – neelkanth Nov 12 '18 at 16:39
• So $1+J(FG)$ is a non abelian group of exponent $27?$ – neelkanth Nov 12 '18 at 16:39
• @neelkanth You asked if it was abelian or not, and I'm saying it doesn't appear to be. You do see that the map $x\to 1+x$ is one to one, right? And it is into hard to count the elements of the augmentation ideal. In this case it's a subspace of codimension $1$. Does that answer your question? – rschwieb Nov 12 '18 at 17:08