I'm trying to understand how to construct the Lyndon-Hochschild-Serre spectral sequence for the cohomology (with integer coefficients) of the central extension $G$ of a group $Q$ by a group $N$, given a representative cocycle of $H^2(Q,N)$ corresponding to such an extension. I will use the example of $\mathbb Z_4$, which is a nontrivial central extension of $\mathbb Z_2$ by $\mathbb Z_2$. I have tried to explain my reasoning below.

So I start with the exact sequence

$0 \rightarrow \mathbb Z_2 \overset{i}{\rightarrow} \mathbb Z_4 \overset{p}{\rightarrow} \mathbb Z_2 \rightarrow 0$

and compute the $E_2$-page using $E_2^{p,q} = H^p(\mathbb Z_2,H^q(\mathbb Z_2,\mathbb Z))$. In the case of the trivial central extension, where the action of every group on its respective coefficient module is trivial, we get a page which vanishes for $q$ odd and is just $\mathbb Z_2$, except for $E_2^{0,0} = \mathbb Z$, when $q$ is even. This sequence stabilizes on the $E_2$-page giving the results expected by using, say, the Kunneth formula.

However, this cannot be the right $E_2$-page for $\mathbb Z_4$, because the higher cohomology groups would then be too large. Therefore we must have a nontrivial action of $\mathbb Z_2$ on the coefficient modules $H^q(\mathbb Z_2,\mathbb Z)$. This is where I am stuck. I have two questions:

1) What is the above action, and is there a systematic way to see it for large $q$?

2)Is there a systematic way to compute differentials in the $E_2$-page given the maps $i,p$ and a representative cocycle for this extension?

  • $\begingroup$ In this specific case, notice that $H^q(\mathbb{Z}_2,\mathbb{Z})$ is either 0 or $\mathbb{Z}_2$. Both groups have trivial automorphism groups, so the action must be trivial. In this specific case, you know the answer because you can calculate the cohomology of $\mathbb{Z}_4$ directly. The Kuenneth formula is good when you have a direct product of groups which is not the case here. In fact, you will have many nonzero differentials at the $E_2$-page. In this specific case you can understand what should the image and kernel of the differentials be, just by knowing the cohomology of $\mathbb{Z}_4$ $\endgroup$
    – Ehud Meir
    Aug 14, 2018 at 13:09

1 Answer 1


The action of the quotient on the cohomology groups of the normal subgroup is the trivial action, because the normal subgroup is central. (Think of group cohomology as a functor of the group: the conjugation action of $Q$ on $N$ induces the action of $Q$ on $H^*(N;\mathbb{Z})$.)

What happens for the spectral sequence that you are thinking of is that there are non-trivial differentials. In this particular example, all the non-trivial differentials are $d_3$. In the $E_2$ page you have that $E_2^{i,j}$ is either zero or order two (except for $E_2^{0,0}$ which is infinite cyclic), and the pairs $(i,j)$ for which the group has order two are those for which $j=0$ and $i>0$ is even, or $j>0$ is even and $i$ is arbitrary.

The differential $d_3$ cancels lots of these order two groups in pairs, leaving groups of order two in $E_4^{i,j}$ only in the cases when either $j=0$ and $i>0$ is even or when $j=2$ and $i$ is even. There are lots of tricks for computing differentials. In this case the easiest thing it probably to use the ring structure on each $E_i^{*,*}$, together with the known values of $H^k(\mathbb{Z}_4;\mathbb{Z})$ for $k=0,1,2$.

  • $\begingroup$ Thanks! I'm not very familiar with understanding the ring structure of cohomology- but this is what I needed. I'll go back and work on it. $\endgroup$ Aug 24, 2018 at 15:37

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