Let $G$ be a group and $\mathbb{Z}$ regarded as a trivial $G$-module. As title, I'm trying to prove that $H^1(G,\mathbb{Z})$ is isomorphic to $G/[G,G]$.
It is easy to see that $H^1$ is isomorphic to $I_G/I_G^2$ where $I_G$ is the kernel of the summing-coefficient map $\mathbb{Z}[G] \to \mathbb{Z}$. It is also clear that there is a map from $G/[G,G] \to I_G/I_G^2$ defined by $s \to s - 1$. I fail to see why this is injective. (I tried to play in the group ring, but couldn't get anywhere). This is being claimed in Cassels-Frohlich directly. Can anyone give me a hand for this?
Thanks!
As an abelian group, $I_G$ is free on $\{(g-1): g \in G\}$, so we may define a homomorphism $I_G \to G/G'$ by $g-1 \mapsto gG'$. The kernel contains $I_G^2$ because of the identity $(g-1)(h-1) = (gh-1)-(g-1)-(h-1)$, so this descends to a map $I_G/I_G^2\to G/G'$ which is inverse to your map $G/G' \to I_G/I_G^2$. Both maps are therefore isomorphisms. See for example Gruenberg's Cohomological Topics in Group Theory, Springer LNM #143.