# Exponent of $1+J(GF),$ $J$ stands for Jacobson radical.

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.

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