First homology group of an abelian group

Question

Let $G$ be an abelian group and $R$ a commutative ring , with trivial G-action on $R$, I am trying to see that $H_1(G,R) \cong H_1(G)\otimes R \cong G\otimes R$, so by definition if we wake a projective resolution $F$ of $\mathbb{Z}$ over $\mathbb{Z}[G]$, we have that $H_1(G,R)=H_1(F\otimes_G R)$, and now i have tried using something like the universal coefficient theorem but got nowhere. Thanks in advance.

Answer 1

The point of the Universal Coefficient Theorem for Group homology is that it is just the UCT for modules over $\mathbf{Z}$ adapted to the case of $\mathbf{Z}[G].$

To make this work, note that you can start with a projective $\mathbf{Z}[G]$ resolution of $\mathbf{Z}$. So you have $$P_\bullet\to \mathbf{Z}$$ as $\mathbf{Z}[G]$ modules. Then you can tensor this with $-\otimes_{\mathbf{Z}[G]}R$ and note that you get a new complex $ P_\bullet\otimes_{\mathbf{Z}[G]}R$. The cohomology is defined as $$H_{*}(G,R)=H_{*}(P_\bullet\otimes_{\mathbf{Z}[G]}R) $$

However note that in case $R$ is a trivial $G$-module, we have $$P_\bullet\otimes_{\mathbf{Z}[G]}R\cong (\mathbf{Z}\otimes_{\mathbf{Z}[G]} P_\bullet) \otimes_{\mathbf{Z}}R.$$

So you can apply the UCT for chain complexes of $\mathbf{Z}$ modules here which gives $$H_{i}(G,R)=H_{i}(G,\mathbf{Z})\oplus \textrm{Tor}_1^{\mathbf{Z}}(H_{i-1}(G,\mathbf{Z}),R).$$

The general result now is that $H_1(G,R)=G/[G,G]\otimes R$ whenever $R$ is a trivial $G$ module, since the Tor term vanishes as $H_0(G,\mathbf{Z})=\mathbf{Z}$ is flat over itself. The point is that the UCT for Group Homology only works in the case $R$ is a trivial $\mathbf{Z}[G]$ module, for then the coinvariant contribution of the $G$ action only comes from the projective complex.

Edit. Note that group homology is defined as the left derived functors $L_i(\mathbf{Z}\otimes_{\mathbf{Z}[G]}(-))R.$ This is the same as taking the left derived functors $L_i((-)\otimes_{\mathbf{Z}[G]}R)\mathbf{Z}.$