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Let $G$ and $H$ be finite simple groups.

I expect that if $G$ and $H$ are not isomorphic, then their cohomology groups with integral coefficients are not all isomorphic, that is, $H^*(G,\mathbb{Z})$ and $H^*(H,\mathbb{Z})$ are not isomorphic graded abelian groups. Is there any proof of this?

Notice, I do not want to compute those cohomology groups (and as far as I know it hasn't been done yet completely), just to show that they are not allisomorphic.

YCor
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Fat ninja
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  • Cohomology with which coefficients? – YCor Apr 19 '19 at 18:44
  • @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. – Fat ninja Apr 19 '19 at 18:47
  • With integer coefficients, I guess the cohomology is identically zero... "over a finite field"... a fixed finite field? you should be more precise. – YCor Apr 19 '19 at 19:05
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    @YCor There is no finite acyclic group, see https://mathoverflow.net/questions/291786/acyclic-finite-groups – user43326 Apr 19 '19 at 19:12
  • @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. – YCor Apr 19 '19 at 19:26
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    @YCor Integer cohomology of finite cyclic groups are not zero – Fat ninja Apr 19 '19 at 19:30
  • 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 $H^2(Z/nZ,Z)\neq 0$. – YCor Apr 19 '19 at 19:33
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    @YCor : more precisely, $H^2(\mathbb{Z/nZ},\mathbb{Z}) = \mathbb{Z/nZ}$ – Maxime Ramzi Apr 19 '19 at 20:20
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    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. – Derek Holt Apr 19 '19 at 21:53
  • 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. – user43326 Apr 20 '19 at 04:22
  • @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 (https://www.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. – Fat ninja Apr 20 '19 at 06:58
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    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. – YCor Apr 20 '19 at 07:11
  • @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. – Fat ninja Apr 20 '19 at 07:23
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    @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. – Derek Holt Apr 20 '19 at 09:50
  • 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. – Ben Wieland Apr 20 '19 at 20:01
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    Given that the integral cohomology groups are difficult to calculate, this is the sort of question that is extremely difficult to counterexample. – IJL Apr 22 '19 at 17:55

1 Answers1

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This is a remark rather than an answer to your question. If you remove the word `simple' it is easy to find such pairs of finite groups. The first examples I learned were (I think) constructed by Atiyah. Each of the two finite groups has a normal subgroup that is cyclic of order three and quotient dihedral of order 8. In each case the dihedral group acts non-trivially on the $C_3$, but in one case the kernel of the action is $C_2\times C_2$ whereas in the other case the kernel of the action is $C_4$. The Lyndon-Hochschild-Serre spectral sequence gives a complete calculation of the integral cohomology ring, and it is just $H^*(C_3)^{C_2}\otimes H^*(D_8)$ in each case. Somehow cohomology cannot see that the centralizers of the $C_3$ subgroups are different (which is how one can see that the groups are not isomorphic).

I computed the integral cohomology rings of a family of $3$-groups of nilpotence class two in my thesis, and I observed that for each $n\geq 5$ there are a pair of groups of order $3^n$ with isomorphic integral cohomology rings. I also gave a more conceptual proof that there are pairs of groups of order $p^n$ for each odd $p$ and $n\geq 5$ that are not isomorphic but have isomorphic integral cohomology groups. My articles are '$3$-groups are not determined by their integral cohomology rings' JPAA Vol 103 (1995) 61-79 and '$p$-groups are not determined by their integral cohomology groups' Bull London Math Soc Vol 27 (1995) 585-589.

None of these arguments are any help for finite simple groups though.

IJL
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