# Group cohomology cannot be defined until we define what topology and action is?

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 $$E$$ ?

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 $$H^1(G, \mathbb{F}_9^{\times,tr})=\operatorname{Hom}(G,\mathbb{F}_9^{\times,tr})\cong\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/8\mathbb{Z})\neq0$$.