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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.

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    $\begingroup$ @YCor There is no finite acyclic group, see mathoverflow.net/questions/291786/acyclic-finite-groups $\endgroup$
    – user43326
    Commented Apr 19, 2019 at 19:12
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    $\begingroup$ @YCor Integer cohomology of finite cyclic groups are not zero $\endgroup$
    – Fat ninja
    Commented Apr 19, 2019 at 19:30
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    $\begingroup$ @YCor : more precisely, $H^2(\mathbb{Z/nZ},\mathbb{Z}) = \mathbb{Z/nZ}$ $\endgroup$
    – Maxime Ramzi
    Commented Apr 19, 2019 at 20:20
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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Apr 19, 2019 at 21:53
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    $\begingroup$ @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. $\endgroup$
    – Derek Holt
    Commented Apr 20, 2019 at 9:50

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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.

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