# A question related to group cohomology and spectral sequences

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.

• I made a mistake here. Actually, $H^n(\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$ also for odd $n$. Therefore, the spectral sequence does not stabilize on this page and cannot be that simple. Apparently, my counting of the size of the group cohomology was wrong. – Herman Chu Aug 20 '18 at 18:06
• Please don't put "question" into the title. Imagine how the main page would look if everyone did this. This is a question & answer site; all posts are (or should be) questions. – joriki Aug 20 '18 at 18:11
• Why is it obvious the spectral sequence collapses at the second page? – Pedro Tamaroff Aug 24 '18 at 10:55
• I made a mistake when I ask the question. The ground cohomology of $H^n(\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$ for all $n$. Therefore the spectral sequence that I drew in the question is not quite right and the spectral sequence does not collapse on the second page. Sorry for confusion! – Herman Chu Aug 24 '18 at 18:27