# $H^1$ of $\Bbb Z$ as a trivial $G$-module is the abelianization of $G$ [duplicate]

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!

## marked as duplicate by user26857, Brandon Carter, Lord_Farin, Micah, rschwiebMay 8 '13 at 16:25

• Here is a tiny remark. I think the nice way to think about this result is that $H^\ast(G,\mathbb Z)$ computes the integral cohomology of $BG$, where $BG$ is the unique space up to homotopy with $\pi_1(BG) = G$, $\pi_k(BG) =0$ for $k > 1$. Then the assertion follows by Hurewicz theorem. – Dan Petersen Mar 21 '12 at 12:32
• It seems that nobody has pointed out that it is $H_1(G,\mathbb{Z})$ which is isomorphic to the abelianization of $G$, not $H^1$. If $G$ is any finite abelian group, then $H^1(G,\mathbb{Z})=\mathrm{Hom}(G,\mathbb{Z})=0$ since $\mathbb{Z}$ is torsion-free. – Keenan Kidwell Apr 12 '12 at 19:42
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.
• Thanks! This is easier than what I thought. It should have been $-(h-1)$ though. – user27126 Mar 21 '12 at 8:29