While looking for the group cohomology of dihedral groups, I came across this post, which gives the cohomology groups $H^{k}(D_{2n},\mathbb{Z})$. We can see that \begin{align} H^{k}(D_{2n},\mathbb{Z}) &= \mathbb{Z_2}^{k/2} \times \mathbb{Z}_{n}, & k \text{ is a multiple of 4} \\ H^{k}(D_{2},\mathbb{Z}) &= \mathbb{Z_2}^{k/2} \times \mathbb{Z}_{2}, & k \text{ is a multiple of 4} \\ H^{k}(D_{2n},\mathbb{Z}) &= H^{k}(D_{2},\mathbb{Z}), & \text{ otherwise}. \end{align}
Suppose $D_{2n}$ is generated by the reflection $s$, with $s^2 = 1$, and the rotation $r$, with $r^{2n} = 1$, such that $s r s = r^{-1}$. Then it has a subgroup $D_2 = \{1,s,r^n,sr^n\}$. My question is whether the group cohomology of $D_{2n}$ can indeed be understood as 'mostly' coming from that of its $D_2$ subgroup. To be precise,
Can $D_{2n}$ be understood as a group extension involving $D_2$ and $\mathbb{Z}_n$? I have not been able to find such a representation.
Is the $H^k(D_2,\mathbb{Z})$ subgroup of $H^k(D_{2n},\mathbb{Z})$ basically induced by the $D_2$ subgroup of $D_{2n}$?
Can the extra $\mathbb{Z}_n$ factor in the cohomology of $D_{2n}$ be related to the quotient $\mathbb{Z}_n = D_{2n}/D_2$?
I have some knowledge of the Lyndon-Hochschild-Serre spectral sequence, but I don't know enough about group cohomology to answer these questions rigorously.
