Let $E$ be a $G$-module. $1$st Group cohomology $H^1(G,E)$ is defined as a set of representative of continuous map $G \to E$ which satisfies cocycle relations modulo boundary relation.

This definition seems to depend on how $G$ acts on $E$.

For example, $F=Gal( \overline{ \Bbb{Q}}/ \Bbb{Q})$, $E= \overline{ \Bbb{Q}}$. Natural action is usual way galois group acts on $E$ But we can also consider another action, for example, trivial one.

To say minor detail. Is group cohomology $H^1(G,E)$ concept determined after we fix the action $G$ to $E$, and topology of $G$ and $E$ ?


1 Answer 1


You are correct, the cohomology groups are very much dependent on the action of $G$ on $E$.

For an example, let $G=\operatorname{Gal}(\mathbb{F}_9/\mathbb{F}_3)\cong \mathbb{Z}/2\mathbb{Z}$. Denote by $\mathbb{F}_9^{\times,tr}$ and $\mathbb{F}_9^{\times,nat}$ the abelian group $\mathbb{F}_9^{\times}$ with trivial or natural $G$-action respectively. Then $H^1(G,\mathbb{F}_9^{\times,nat})=0$ by Hilbert's Satz 90. But we also have $H^1(G, \mathbb{F}_9^{\times,tr})=\operatorname{Hom}(G,\mathbb{F}_9^{\times,tr})\cong\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/8\mathbb{Z})\neq0$.


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