Cohomology of groups.

Question

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, $d:G \rightarrow M$: $d(g)=gm-m$).

$H^1(G,M):= H^1(Hom^_G(F,M))$. 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

score 3 · Accepted Answer · answered Sep 19 '13 at 11:37

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.

Cohomology of groups.

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