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Apr 16, 2023 at 13:58 vote accept user44312
Apr 10, 2023 at 13:28 vote accept user44312
Apr 10, 2023 at 13:28
Apr 10, 2023 at 12:48 history edited LSpice CC BY-SA 4.0
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Apr 9, 2023 at 4:58 answer added user44312 timeline score: 1
Apr 8, 2023 at 3:56 comment added user44312 Thanks for your idea. @RichardLyons
Apr 8, 2023 at 3:55 comment added user44312 Thanks for your idea. @DaveBenson
Apr 7, 2023 at 15:42 comment added Richard Lyons Perhaps Harris's proof can be modified to prove something like the following. Let $A$ be a finite abelian $p$-group on which a group $X$ acts. Suppose that that for every $X$-invariant subquotient $B/C$ of $A$ (i.e., both $B$ and $C$ are $X$-invariant) such that $B/C$ is elementary abelian, $X$ acts completely reducibly on $B/C$. Then $A$ is the direct sum of $X$-invariant homocyclic subgroups.
Apr 6, 2023 at 22:03 comment added Dave Benson Yes, that's right.
Apr 6, 2023 at 3:49 comment added user44312 My understanding of "FH-modules in the principal block have fixed points" is: Brauer characters in principal block has the trivial Brauer character as its irreducible constituent. So, $|IBr(B_0)|=1$. Am I right? @DaveBenson
Apr 6, 2023 at 2:36 comment added user44312 Thanks for your answer. Sorry, I don't understand your point. Especially, the sentences "$FH$-modules in the principal block have fixed points" and "so things tend to decompose." Could you please explain more? @DaveBenson
Apr 5, 2023 at 11:18 comment added Dave Benson The problem with trying to construct an example is that if $F$ has order coprime to $p$ and $H$ is a $p$-group then $FH$-modules in the principal block have fixed points, and all non-principal blocks have defect zero, so things tend to decompose. Maybe you can turn this into a proof of something.
Apr 5, 2023 at 1:37 history edited user44312 CC BY-SA 4.0
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Apr 4, 2023 at 23:56 history edited user44312 CC BY-SA 4.0
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Apr 4, 2023 at 16:24 history edited Derek Holt CC BY-SA 4.0
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Apr 4, 2023 at 14:10 history asked user44312 CC BY-SA 4.0