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.

  • $\begingroup$ @rschwieb Please help me to solve it .... $\endgroup$ – neelkanth Nov 12 '18 at 15:46

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.

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

  • $\begingroup$ Thanks sir ....actually as you discussed in math.stackexchange.com/questions/2049645/… according to it J(FG ) is same as augmentation ideal as unique sylow subgroup is group itself so right side group being trivial and hence kernel is augmentation ideal. $\endgroup$ – neelkanth Nov 12 '18 at 16:32
  • $\begingroup$ my last question is about exponent of group $1+J(FG)$...thanks ... $\endgroup$ – neelkanth Nov 12 '18 at 16:36
  • $\begingroup$ Can i say its exponent must be $27$ as $3$ and $9$ order groups are always abelian.? $\endgroup$ – neelkanth Nov 12 '18 at 16:39
  • $\begingroup$ So $1+J(FG)$ is a non abelian group of exponent $27?$ $\endgroup$ – neelkanth Nov 12 '18 at 16:39
  • $\begingroup$ @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? $\endgroup$ – rschwieb Nov 12 '18 at 17:08

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