Timeline for Cohomology of simple finite groups remembers the group?
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2019 at 18:12 | answer | added | IJL | timeline score: 8 | |
| Apr 22, 2019 at 17:55 | comment | added | IJL | Given that the integral cohomology groups are difficult to calculate, this is the sort of question that is extremely difficult to counterexample. | |
| Apr 20, 2019 at 20:01 | comment | added | Ben Wieland | You're not going to prove a theorem about simple groups without the classification... You can strengthen groups to rings to DGA. Slightly stronger is asking that the group rings be isomorphic. There is a famous example of isomorphic group rings which is not simple, so I doubt a simple example exists (of this stronger case)... I think that you can recover the size of a group from its cohomology (as DGA). So what simple groups have the same size? There is an infinite family of pairs and a sporadic pair. I wouldn't be surprised if one of them has homology iso as groups. | |
| Apr 20, 2019 at 9:50 | comment | added | Derek Holt | @user43326 It seems to me that you only find the meaning clear because you have rejected a number of possible interpretations that lead to trivial solutions. The meaning of a well formulated question should be clear without having to do that. | |
| Apr 20, 2019 at 7:29 | history | edited | YCor | CC BY-SA 4.0 |
clarified the question
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| Apr 20, 2019 at 7:23 | comment | added | Fat ninja | @YCor let's juct stick with the edit made by user43326, that $H^n(G,\mathbb{Z})\cong H^n(H,\mathbb{Z})$ $\forall n\in \mathbb{Z} _{\geq 0}$ as abelian groups. | |
| Apr 20, 2019 at 7:11 | comment | added | YCor | But you'll never reach a precise formulation if you systematically refuse to use quantifiers and refuse to say isomorphism as what. Writing $H^*(G,\mathbf{Z})\cong H^*(G,\mathbf{Z})$ has at least 4 possible meanings since it can mean isomorphic as abelian group, graded abelian group, algebra, graded algebra. Given the previous formulation however, I guess that you mean isomorphism as graded group. | |
| Apr 20, 2019 at 6:58 | comment | added | Fat ninja | @DerekHolt Thanks for the interest. Of course the isomorphism couldn't be induced by a homomorphism $f:G\to H$ because otherwise it would imply $f$ is an isomorphism (jstor.org/stable/2042568 - it's about homology groups, actually, but I think the similar argument can be applied). So I just want $H^*(G,\mathbb{Z})\cong H^*(H,\mathbb{Z})$ as user43326 wrote. | |
| S Apr 20, 2019 at 6:52 | history | suggested | user43326 | CC BY-SA 4.0 |
Removed ambiguity.
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| Apr 20, 2019 at 4:22 | comment | added | user43326 | Well, the meaning seems clear to me (OP can't certainly be talking about the cohomology group in one particular degree, and wouldn't be implying that the isomorphism should be induced by a map). In any case I edited the post so that it would be clearer. | |
| Apr 20, 2019 at 4:18 | review | Suggested edits | |||
| S Apr 20, 2019 at 6:52 | |||||
| Apr 19, 2019 at 22:10 | review | Close votes | |||
| Apr 20, 2019 at 15:49 | |||||
| Apr 19, 2019 at 21:53 | comment | added | Derek Holt | I agree with YCor that you need tro make this question more precise. What exactly does "The cohomology groups with integral coefficients are not isomorphic" mean? for example. It is not reasonable to expect readers to have to spend time figuring out exactly what you are asking. | |
| Apr 19, 2019 at 20:20 | comment | added | Maxime Ramzi | @YCor : more precisely, $H^2(\mathbb{Z/nZ},\mathbb{Z}) = \mathbb{Z/nZ}$ | |
| Apr 19, 2019 at 19:33 | comment | added | YCor | Ah I see... for nontrivial finite cyclic groups $Z/nZ$ we have a central extension $1\to Z\to Z\to Z/nZ\to 1$, so $H^2(Z/nZ,Z)\neq 0$. | |
| Apr 19, 2019 at 19:30 | comment | added | Fat ninja | @YCor Integer cohomology of finite cyclic groups are not zero | |
| Apr 19, 2019 at 19:26 | comment | added | YCor | @user43326 I explicitly meant the cohomology with coefficients in $\mathbf{Z}$ vanishes, not the homology. Just think that for $G$ simple abelian, $H^1(G,\mathbf{Z})=0$ while $H_1(G,\mathbf{Z})\neq 0$. I know that nontrivial finite groups are not acyclic. | |
| Apr 19, 2019 at 19:12 | comment | added | user43326 | @YCor There is no finite acyclic group, see mathoverflow.net/questions/291786/acyclic-finite-groups | |
| Apr 19, 2019 at 19:05 | comment | added | YCor | With integer coefficients, I guess the cohomology is identically zero... "over a finite field"... a fixed finite field? you should be more precise. | |
| Apr 19, 2019 at 18:47 | comment | added | Fat ninja | @YCor thanks, I edited. Of course it's important to consider cohomology with coefficients in a finite field to approach the original problem, I guess. | |
| Apr 19, 2019 at 18:45 | history | edited | Fat ninja | CC BY-SA 4.0 |
added 26 characters in body
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| Apr 19, 2019 at 18:44 | comment | added | YCor | Cohomology with which coefficients? | |
| Apr 19, 2019 at 18:34 | history | asked | Fat ninja | CC BY-SA 4.0 |