Timeline for Uniqueness of the fusion ring for simple finite group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2016 at 20:32 | comment | added | Geoff Robinson | @QiaochuYuan : OK, no problem. | |
| Jan 29, 2016 at 20:01 | comment | added | Qiaochu Yuan | @Geoff: my apologies for the confusion; my comment was directed at David's answer, not at your comments. | |
| Jan 29, 2016 at 19:58 | comment | added | Geoff Robinson | @Qiaochu Yuan: My last comment was to the second comment above, which seems quite clear to me: he seems to be asking whether $A_{c}(G)/{\rm Inn}(G)$ is non-trivial, where $A_{c}(G)$ is the subgroup of automorphisms of $G$ which fix all irreducible characters (at least that is what the question in comments say). The answer to that question is "no", in fact that quotient group need not be Abelian. In a similar vein, to clarify : Huppert's conjecture is different from asking whether the character table determines a simple group (which as you correctly point out in your answer, it does). | |
| Jan 29, 2016 at 19:06 | comment | added | Qiaochu Yuan | No, this is not what the OP is asking. In the comments the OP clarifies that he thinks of fusion rings as rings with basis. | |
| Jan 29, 2016 at 14:21 | comment | added | Geoff Robinson | That is definitely not true for general finite groups unfortunately. I think Burnside already knew this. I think C.H. Sah constructed examples where the quotient of that subgroup of Aut(G) by Inn(G) is not even Abelian. | |
| Jan 29, 2016 at 13:37 | comment | added | David Handelman | We could perhaps do a bit better if the following were true: if an automorphism of a finite group fixes all the characters (that is, induces the identity on the representations), then it must be inner. Is it? Then Out(G) would embed in the automorphism group of $R(G)$ (as a partially ordered ring), etc. | |
| Jan 29, 2016 at 2:36 | history | edited | David Handelman | CC BY-SA 3.0 |
added 3 characters in body
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| Jan 29, 2016 at 0:04 | comment | added | Geoff Robinson | It is a (consequence of a) conjecture of B. Huppert that two non-Abelian finite simple groups which have the same set of complex irreducible character degrees (even ignoring multiplicities) should be isomorphic. I am not sure how widely this has been checked to date though some cases are known. | |
| Jan 28, 2016 at 23:57 | history | answered | David Handelman | CC BY-SA 3.0 |