It is actually a follow-up question of a mathoverflow question. I don't quite understand the answer there.
I tried to compute the group cohomology of $H^n(\mathbb{Z}_4,\mathbb{Z})$ via the Lyndon-Hochschild-Serre spectral sequence as a sanity check. Since we have the following short exact sequence $1\rightarrow\mathbb{Z}_2\rightarrow\mathbb{Z}_4\rightarrow\mathbb{Z}_2\rightarrow1$, I got the second page $E^2_{p,q} = H^p(\mathbb{Z}_2,H^p(\mathbb{Z}_2,\mathbb{Z})$ by knowing that:
(1) The group cohomology $H^n(\mathbb{Z}_2,\mathbb{Z})=H^n(\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$ when $n$ is a positive even integer.
(2) The first $\mathbb{Z}_2$ in the short exact sequence has no action on the second $\mathbb{Z}_2$ since it is a central extension.
\begin{array}{|c c c c c c} 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}_2& 0 &\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}_2& 0 &\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}_2& 0 &\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}& 0 &\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 \\ \hline \end{array}
It is obvious that the spectral sequence collapses on the second page. Even though we have an extension problem here, we still know from the spectral sequence the sizes of the group cohomology groups. For example, $|H^4(\mathbb{Z}_4,\mathbb{Z})|=8$ and $|H^6(\mathbb{Z}_4,\mathbb{Z})|=16$. However, this result contradicts with the fact that $H^n(\mathbb{Z}_4,\mathbb{Z})=\mathbb{Z}_4$ when $n$ is a positive even integer, say $n=4, 6$.
It seems that something went wrong to my understanding of the spectral sequence! Thanks!
I made a mistake here. Please take a look at the comment below.