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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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    I know, but I hope that the cokernel will be zero or at least very "small".2011-10-25

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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?