Consider the group algebra $F[G]$, where $F$ is a finite field of characteristic $2$ and $G=\operatorname{SL}(2,3)$ i.e. group of $2\times2$- matrices over the integers modulo $3$.

After some calculation I proved that $J(FG)^8=0$, and that $J(FG)$ is nonabelian and hence $(1+J(FG))^8=1$, where $G^n=\{g^n\mid g\in G\}$. My target is to tell the exact exponent of the group $1+J(FG).$ Now it is given that the index of nilpotency of $J(FG)$ is strictly greater than $4$ and strictly less than $8$.

I am thinking like this :

Because $1+J(FG)$ is nonabelian, its exponent can't be $2.$ Now can I say that if exponent of $1+J(FG)$ is $4$ i.e. $(1+J(FG))^4=1$, then $J(FG)^4=0$ by using the chararacteristic of field ? i.e. is index of nilpotency of $J(FG)$ is $\leq 4$, which is not possible as index of Nilpotency is given strictly bigger than $4$. So exponent of $1+J(FG)$ is $8$. Thanks.

  • $\begingroup$ What is $n$ in your expression for $(1+J(FG))^8$? $\endgroup$ – Servaes Jan 18 at 18:30
  • $\begingroup$ 8.............. $\endgroup$ – neelkanth Jan 18 at 18:40
  • $\begingroup$ @Servaes I told the meaning of $G^n$ .... $\endgroup$ – neelkanth Jan 18 at 18:42
  • $\begingroup$ I see, I misunderstood the expression as being one big expression. Allow me to make a small edit to clarify. $\endgroup$ – Servaes Jan 18 at 18:43
  • $\begingroup$ Yes ...please improve the question for convenience... thanks $\endgroup$ – neelkanth Jan 18 at 18:45

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