# Group homology in degrees zero and one using the standard resolution

I'm trying to show using the standard resolution that for any group $$G$$, $$H_0(G) = \mathbb{Z} \text{ and } \quad H_1(G) = \frac{G}{[G,G]}.$$

(I already know this holds, as by definition it is $$H_\bullet(G) = H_{\bullet}(X)$$ with $$X$$ a $$K(G,1)$$ space: these are always path connected, and by Hurewicz's theorem we get $$H_1(X) = \pi_1(X)^{ab} = G^{ab}$$. However, this exercise explicitly asks for the computations to be done with the standard resolution and so I want to get to understand it.)

So, we start with the chain complex

$$\cdots \xrightarrow{d_n} C_n \to C_{n-1} \xrightarrow{d_{n-1}} \cdots \to C_1 \xrightarrow{d_1} C_0 \xrightarrow{\varepsilon} \mathbb{Z} \to 0$$

where $$C_n$$ is the free $$\mathbb{Z}$$-module generated by $$G^{n+1}$$ with its $$\mathbb{Z}[G]$$-module structure given by the action $$g(g_0,\dots,g_n) := (gg_0,\dots,gg_n)$$, and the differentials are defined by

$$d_n : (g_0, \dots,g_n) \in C_n \mapsto \sum_{i=0}^n(-1)^i(g_0,\dots,\widehat{g_i},\dots,g_n) \in C_{n-1}.$$

Hence I want to compute the homology in degrees zero and one of

$$\cdots \to \mathbb{Z} \otimes_{\mathbb{Z}[G]} C_2 \xrightarrow{1 \otimes d_2} \mathbb{Z} \otimes_{\mathbb{Z}[G]} C_1 \xrightarrow{1 \otimes d_1}\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_0 \to 0.$$

So far, I have proved that:

• $$\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_0 \simeq \mathbb{Z}$$ simply because $$C_0 \simeq \mathbb{Z}[G]$$ as a $$\mathbb{Z}[G]$$-module, with basis $$(1)$$.

• We have $$1 \otimes d_1 = 0$$. In effect: as any element of $$C_1$$ is a $$\mathbb{Z}[G]$$-linear combination of elements of the form $$(1,g)$$, it suffices to note that \begin{align} 1 \otimes d_1 (n \otimes (1,g)) &= n \otimes [(g)-(1)] = n \otimes (1) - n \otimes (g) = n \otimes (1) - n \otimes g(1) \\ &= n \otimes (e) - gn \otimes (1) = 0. \end{align} because the action of $$G$$ on $$\mathbb{Z}$$ is trivial.

From here one can see that $$H_0(G) = \frac{\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_0}{\operatorname{im}1 \otimes d_1} = \frac{\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_0}{0} = \mathbb{Z} \otimes_{\mathbb{Z}[G]} C_0 \simeq \mathbb{Z}.$$

I also know by a direct calculation that

\begin{align} 1 \otimes d_2(n \otimes (1,g,h)) &= n \otimes [(g,h) - (1,h) + (1,g)]\\ &= n \otimes [(1,g^{-1}h) - (1,h) + (1,g)]. \end{align}

and so

$$H_1(G) = \frac{\ker 1 \otimes d_1}{\operatorname{im} 1 \otimes d_2} = \frac{\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_1}{\operatorname{im} 1 \otimes d_2} = \frac{\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_1}{_{n \in \mathbb{Z}, \ g,h \in G}}$$

So far, I've failed to come up with a good description for $$\mathbb{Z} \otimes_{\mathbb{Z}[G]} C_1$$ and $$\operatorname{im} 1 \otimes d_2$$.

Any hints on how to continue would be greatly appreciated! Thanks in advance.

As you have pointed out, in $$\Bbb Z\otimes_{\Bbb Z[G]}C_1$$ you have the relation $$1\otimes(g,h)=1\otimes(1,g^{-1}h)$$. Then $$\Bbb Z\otimes_{\Bbb Z[G]}C_1$$ is the free Abelian group on the symbols $$1\otimes(1,g)$$, which I'll denote by convenience $$[g]$$. Then $$H_1(G)$$ is the quotient of $$\Bbb Z\otimes_{\Bbb Z[G]}C_1$$ by the symbols $$[g^{-1}h]-[h]+[g]$$, as you point out. That is $$[g]+[k]-[gk]$$ where $$k=g^{-1}h$$. So $$H_1(G)$$ is the quotient of the free Abelian group on $$G$$ by the relations $$[gk]=[g]+[k]$$. That's just the Abelianisation of $$G$$: $$G/[G,G]$$. In more detail, $$G\mapsto H_1(G)$$ via $$g\mapsto [g]$$ is a homomorphism, and as the target is Abelian, it factors through $$G/[G,G]$$, so in effect is a map $$G/[G,G]\to H_1(G)$$. Considering the map going the other way taking $$[g]$$ in the free Abelian group over $$G$$ to $$g[G,G]\in G/[G,G]$$ yields the inverse map.