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If $m$ is odd, the group cohomology of the dihedral group $D_m$ of order $2m$ is given by $$H^n(D_m;\mathbb{Z}) = \begin{cases} \mathbb{Z} & n = 0 \\ \mathbb{Z}/(2m) & n \equiv 0 \bmod 4, ~ n > 0 \\ \mathbb{Z}/2 & n \equiv 2 \bmod 4 \\ 0 & n \text{ odd} \end{cases}$$ This is a nice application of the Lyndon-Hochschild-Serre spectral sequence. The calculation uses the assumption, that $m$ is odd, in an essential way. If $m$ is even, several complications will arise ... therefore my question is:

Question. What is the group cohomology of $D_m$ for even $m$?

I have found here the group cohomology of $D_4$, which already looks quite "wild" as compared to the odd case.

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  • $\begingroup$ Bonus question: Is there a spectral sequence proof? $\endgroup$ – Martin Brandenburg May 23 '15 at 2:07
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This can be found in the proof of Theorem 5.2 of the following paper:

Handel, David. "On products in the cohomology of the dihedral groups." Tohoku Mathematical Journal, Second Series 45, no. 1 (1993): 13-42.

In particular, for $m$ even and $n>0$, we have: $$ H^n(D_m;\mathbb{Z}) \;=\; \begin{cases} (\mathbb{Z}/2)^{(n-1)/2} & \text{if }n\equiv 1\pmod 4 \\ (\mathbb{Z}/2)^{(n+2)/2} & \text{if }n\equiv 2\pmod 4 \\ (\mathbb{Z}/2)^{(n-1)/2} & \text{if }n\equiv 3\pmod 4 \\ (\mathbb{Z}/m) \oplus (\mathbb{Z}/2)^{n/2} & \text{if }n\equiv 0\pmod 4 \end{cases} $$

The cited paper also gives a presentation for the cohomology ring. It is $$ H^*(D_m;\mathbb{Z}) = \mathbb{Z}[a_2, b_2, c_3,d_4]/I $$ where $I$ is the ideal generated by $2a_2$, $2b_2$, $2c_3$, $md_4$, $(b_2)^2 + a_2b_2 + (m^2/4)d_4$, and $(c_3)^2 + a_2d_4$. (Here the subscripts denote the degree, e.g. $d_4 \in H^4(D_m,\mathbb{Z})$.)

Edit: Incidentally, when $m\geq 3$ is odd Handel computes the cohomology ring as $$ H^*(D_m;\mathbb{Z}) = \mathbb{Z}[a_2,d_4]/I $$ where $I$ is the ideal generated by $2a_2$ and $md_4$.

Edit 2: See this answer for the homology of $D_{2m}$.

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  • $\begingroup$ Thank you for this perfect answer! $\endgroup$ – Martin Brandenburg May 23 '15 at 2:03

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