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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

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

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.

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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.

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