# mod p (odd) cohomology of dihedral groups

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.

• Is it not sufficient to apply the UCT to $H^\ast(D_{2n},\mathbb{Z})$? The latter groups are known. – 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.

• 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.. – Denis Nardin Feb 4 '19 at 17:27
• 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 $\sigma b = -b$. – Mike Miller Feb 4 '19 at 17:50
• @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. – Denis Nardin Feb 4 '19 at 22:32
• 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? – Mike Miller Feb 4 '19 at 22:48
• @MikeMiller Aaand there was still a residue mistake (I assumed that the Sylow subgroup wouldn't be trivial). Fixed now :) – Denis Nardin Feb 4 '19 at 22:54