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Exercise 2 §III.1 'Cohomology of Groups' K.S. Brown:

Let $G$ be a group and $M$ a $G$-module. Show that $H^1(G,M)$ is isomorph to the group of derivation from $G$ to $M$ modulo the subgroup of principal derivations (principal derivation means: m \in M be fix, $d:G \rightarrow M$: $d(g)=gm-m$).

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How has $H^1(G,M)$ been defined there? –  Tobias Kildetoft Sep 19 '13 at 11:03
    
$H^1(G,M):= H^1(Hom^*_G(F,M))$. I $F$ being a free complex and the $n$-th term of the complex $Hom^*(F,M)$ is defined by: $Hom^*_G(F,M)^n=Hom_G(F_n,M)$. –  R2D2 Sep 19 '13 at 11:07

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You have the bar resolution $B_\bullet$ of $\mathbb Z$ as a $G$-module, which is a $G$-projective resolution. Use it to compute $H^1(G,M)$: apply the functor $\hom_G(\mathord-,M)$ to $B_\bullet$ and write down exactly what it means for a $1$-cochain in that complex to be a $1$-cocycle, and what are the $1$-coboundaries.

If you know what the bar resolution is, this is just a matter of chasing down definitions.

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