I am searching a simple example of a finite group $G$, so that the Jacobson radical $J(FG)$, of group algebra $FG$ is commutative, where $F$ is a finite field. I know example for that if $G$ is any finite abelian group, or dihedral group when characteristic of field is $2$ and $p$-groups, or when group ring is semi simple. Please provide me any other example of such finite group $G$ for which corresponding Jacobson radical is commutative. Thank you.
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$\begingroup$ Does that mean that any two elements in the Jacobson radical commute? If yes, then a $p$-group has this property if and only if it is commutative in characteristic p since the algebra is local. This problaby also holds in general when the algebra is not semisimple. $\endgroup$– MareCommented Jun 6, 2023 at 13:52
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$\begingroup$ @Mare yes same as definition of commutative ring. $\endgroup$– neelkanthCommented Jun 6, 2023 at 14:19
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$\begingroup$ And I am searching except p-groups . $\endgroup$– neelkanthCommented Jun 6, 2023 at 14:20
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$\begingroup$ If algebra is not semi simple then it does not hold in genreal, otherwise all time Jacobson Radical will be commutative. $\endgroup$– neelkanthCommented Jun 6, 2023 at 14:22
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1 Answer
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Here is a family of examples. Let $F$ have characteristic two, and let $G$ be a finite group of order $2n$ with $n$ odd. Then $J(FG)$ squares to zero, and is therefore commutative.
Another (similar) family of examples is as follows. Let $F$ have characteristic $p$, let $H$ be a $p'$-group, and let $t$ be an automorphism of $H$ of order $p$ which fixes only the identity element. Let $G$ be the semidirect product $H \rtimes \langle t\rangle$. Then every non-principal block of $FG$ has defect zero, and the principal block is commutative, so $J(FG)$ is commutative.
answered Jun 10, 2023 at 16:55
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$\begingroup$ More generally, for $J(FG)$ to be commutative, probably all blocks have to have one of three types: (1) commutative, (b) defect zero, or (iii) defect one with $p=2$. If this is true, a classification should be pretty easy. $\endgroup$ Commented Jun 10, 2023 at 17:19
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1$\begingroup$ I actually think this question is slightly more interesting than other MO users have taken it for. $\endgroup$ Commented Jun 11, 2023 at 23:02
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$\begingroup$ yes sir, I was afraid of receiving downvotes. $\endgroup$ Commented Jun 12, 2023 at 0:03
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1$\begingroup$ Well, I thought it was useful and upvoted the question and the answer. $\endgroup$ Commented Jun 12, 2023 at 20:46