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Let $p$ be a prime number and H be a finite group with $|H|=p-1$ and consider $\varphi: H \times Z_{p^k} \rightarrow Aut(Z_{p^k})$ as a non-trivial action of $H \times Z_{p^k}$ on $Z_{p^k}$ such that $Z_{p^k} \subset Ker(\varphi).$

How can I compute the first cohomology group $H^1(H \times Z_{p^k}, Z_{p^k})$ for the above action? I want to know this group is trivial or not?

Update: Is there any relation between $H^1(H \times Z_{p^k}, Z_{p^k})$ and product of $H^1(H, Z_{p^k})$ and $H^1(Z_{p^k}, Z_{p^k})$ in general?

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Are you familiar with the inflation restriction exact sequence?

If $N \unlhd G$ and $G$ acts on module $A$, then

$$0 → H^1(G/N, A^N) → H^1(G, A) → H^1(N, A)^{G/N} → H^2(G/N, A^N) →H^2(G, A)$$

is exact. Taking $N=H$ in your example, we have $A^N=0$, and $|N|$ and $|A|$ are coprime (I am assuming that $Z_{p^k}$ denotes a cyclic group of order $p^k$), so $H^1(N,A)=0$, and hence $H^1(G,A)=0$.

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  • $\begingroup$ thank you. I didn't know about this exact sequence. $\endgroup$
    – user97635
    Aug 4, 2014 at 11:36
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    $\begingroup$ The Lyndon-Hochschild-Serre spectral sequence can be used to conclude in this case that $H^n(G,A)=0$ for all $n \ge 1$, but that is more advanced and technical. $\endgroup$
    – Derek Holt
    Aug 4, 2014 at 12:21

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