# Group cohomology: $H^1$ of $\mathbb Z/4 \mathbb Z$

Consider the group $G = \mathbb{Z}/4\mathbb{Z}$ and its subgroup $H = 2\mathbb{Z}/4\mathbb{Z}$. Consider the obvious injection $\mathbb{Z} \to \mathbb{Z}[G]: 1 \mapsto N_G = \sum_{\sigma \in G} \sigma$. Let $A$ be the cokernel of this map, which has an obvious structure of $G$-module. Now consider the map $A \to A \otimes \mathbb{Z}[H]$ induced by $\mathbb{Z} \to \mathbb{Z}[H]: 1 \mapsto N_H$. This map induces a homomorphism of cohomology groups $H^1(G,A) \to H^1(G,A \otimes \mathbb{Z}[H])$.

What are the kernel and cokernel of this homomorphism?

General results about the cohomology of cyclic groups do not seem to give the result; I tried to make the calculations explicit, but I keep walking in circles. Can anyone help?

Edit: suppose I want to do this for $G = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ and $H$ an index $2$ subgroup instead (everything I wrote makes still sense). Then $G$ is not cyclic; how to tackle the question in this case?

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Do you expect the kernel and cokernel to have any nice description at all? Honestly, what you wrote is not motivational enough for me to sit down and describe your kernel and cokernel explicitely... It looks somewhat random :) – Mariano Suárez-Alvarez Oct 25 '11 at 23:35
I know, but I hope that the cokernel will be zero or at least very "small". – Evariste Oct 25 '11 at 23:38

Using the formula $H^1(G,M)=\{m\in M:N_Gm=0\}/\{gm-m:m\in M\}$ (with $g$ a generator of $G$), you can completely describe that map. One you do this, this has nothing to do with group cohomology...

The cokernel is obviously $$\frac{\{x\in A\otimes\mathbb Z[H]:N_Gx=0\}}{\{gx-x:x\in A\otimes\mathbb Z[H]\}+\{y\otimes N_H:y\in A,\;N_Gy=0\}},$$ and the kernel has a similiar, slightly more awkward, description.

In your situation, this is a concrete abelian group. Can you describe this quotient?

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