I've been trying to find the cohomology for the trivial module for $\operatorname{PSL}_2(r^n)$ over $\mathbb{F}_p$ for $2 \neq p \neq r$ and have managed to reduce this to the cohomology of a maximal torus $D_{r \pm 1}$ (where $|D_{2n}| = 2n$), dependent on which is divisible by $p$, but am struggling to find a reference for $\operatorname{H}^i(D_{2n}, \mathbb{F}_p)$ for $i > 2$ though this must almost certainly be known.

I can see by computations in magma that the answer should be $0$ for $i \equiv 1, \, 2 \mod 4$ and $\mathbb{F}_p$ otherwise.

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    $\begingroup$ Is it not sufficient to apply the UCT to $H^\ast(D_{2n},\mathbb{Z})$? The latter groups are known. $\endgroup$ – Chris Gerig Feb 4 '19 at 20:17

EDIT: Thanks a lot to Mike Miller for pointing out in the comments significant simplifications to the proof I wrote

First suppose that $p$ does not divide $n$. Then we have $$H^*(D_{2n};\mathbb{F}_p)=\begin{cases}\mathbb{F}_p & \textrm{ if }n=0\\ 0 &\textrm{ otherwise}\end{cases}\,.$$ Otherwise, let us suppose that $p$ divides $n$.

Let $C_n$ be the subgroup of $D_{2n}$ consisting of rotations. This is a normal cyclic subgroup of index 2. Hence, it has cohomology of the form $$H^*(C_n;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/C_n;\,H^*(C_n;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/C_n$ has order prime to $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(C_n;\mathbb{F}_p)^{D_{2n}/C_n}\,.$$ Note that $D_{2n}/C_n\cong \mathbb{Z}/2$ acts on $C_n$ by sending $z$ to $z^{-1}$. To conclude then it's enough to determine the action of $\mathbb{Z}/2$ on $a$ and $b$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_n→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $C_n$ in $S^1$, and the action of $\mathbb{Z}/2$ extends to $S^1$. By the Hurewicz theorem we have an isomorphism $$\mathrm{Hom}(\pi_2BS^1;\mathbb{F}_p)\cong H^2(BS^1;\mathbb{F}_p)\cong H^2(BC_n;\mathbb{F}_p)$$ compatible with the action of $\mathbb{Z}/2$. In particular the action is nontrivial (since the action on $\pi_2BS^1\cong \pi_1S^1\cong\mathbb{Z}$ sends $1$ to $-1$), so $\sigma b= - b$.

Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

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    $\begingroup$ I am oscillating between being convinced that this answer is correct and being convinced that it is complete bunk. Right now I think it's correct and I've got to leave now, so I'll offer it to you clemence.. $\endgroup$ – Denis Nardin Feb 4 '19 at 17:27
  • $\begingroup$ Everything is correct. Your argument can be streamlined by working with the LHSS for $\Bbb Z/n \to D_{2n} \to \Bbb Z/2$ directly instead of passing to a $p$-Sylow, then reducing to the case of $S^1$, which has an automorphism extending negation on $\Bbb Z/n$. The induced map of this automorphism (complex conjugation) on $S^1$ is complex conjugation on $\Bbb{CP}^\infty$, so $\sigma b = -b$. $\endgroup$ – Mike Miller Feb 4 '19 at 17:50
  • $\begingroup$ @MikeMiller Thanks! I was trying to avoid using the fiber sequence $C_p\to S^1\to S^1$ to give a more "algebraic" proof, but you're right, the argument flows much better if we embrace it. $\endgroup$ – Denis Nardin Feb 4 '19 at 22:32
  • $\begingroup$ I thought your answer was fine as it was, I just wanted to add something for anybody passing by :) Some small points: there is a surviving $C_p$ somewhere. Because $S^1$ is a non-discrete space as well as a group, can I suggest writing $H^2(BS^1; \Bbb F_p)$ on the second-to-last displayed line (and maybe similarly for $BC_n$) to avoid confusion? $\endgroup$ – Mike Miller Feb 4 '19 at 22:48
  • $\begingroup$ @MikeMiller Aaand there was still a residue mistake (I assumed that the Sylow subgroup wouldn't be trivial). Fixed now :) $\endgroup$ – Denis Nardin Feb 4 '19 at 22:54

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