# Let $G$ be a group. Why is $\operatorname{Ext}_{\mathbb{Z}G}^1(\mathbb{Z},\mathbb{Z})\cong \operatorname{Hom}_{Grp}(G,\mathbb{Z})$?

Let $G$ be a Group and $\mathbb{Z}G$ is the ring of the formal sums $$\sum_{g\in G}n_gg$$ with multiplication $$(\sum_{g\in G}n_gg)(\sum_{h\in G}m_hh)=\sum_{g\in G}(\sum_{h\in G}n_{gh^{-1}}m_h)g.$$Consider $\mathbb{Z}$ as trivial $\mathbb{Z}G$-module, and consider $\mathbb{Z}G^{\otimes (n+1)}$ as a $\mathbb{Z}G$-module by $g\cdot (g_0\otimes ...\otimes g_n)=gg_0\otimes ...\otimes g_n$.

I want to prove: $\operatorname{Ext}_{\mathbb{Z}G}^1(\mathbb{Z},\mathbb{Z})\cong \operatorname{Hom}_{Grp}(G,\mathbb{Z})$.

By definition, it is $\operatorname{Ext}_{\mathbb{Z}G}^1(\mathbb{Z},\mathbb{Z})=H^1(\operatorname{Hom}_{\mathbb{Z}G}(P_*,\mathbb{Z}))$, where $P_*$ is a projective resolution of $\mathbb{Z}$ and $H^1$ is the first cohomology group. Therefore I started with the Bar resolution of $\mathbb{Z}$:

$$\mathbb{Z}\xleftarrow{\epsilon} \mathbb{Z}G \xleftarrow{d_1} \mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G \xleftarrow{d_2} \mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G\leftarrow ...,$$where $\epsilon(g)=1$ and $d_n:\mathbb{Z}G^{\otimes(n+1)}\to \mathbb{Z}G^{\otimes(n)}$, $$d_n(g_0\otimes g_1...\otimes g_n)=\sum_{i=0}^{n-1}(-1)^i g_0\otimes ...\otimes g_ig_{i+1}\otimes ..\otimes g_n +(-1)^ng_0\otimes ..\otimes g_{n-1}.$$ For example it is $d_1(g_0\otimes g_1)=g_0g_1-g_0$ and $d_2(g_0\otimes g_1\otimes g_2)=g_0g_1\otimes g_2 -g_0\otimes g_1g_2+g_0\otimes g_1$. I already know that this is a projective resolution of $\mathbb{Z}$ such that I can use this resolution for $P_*$.

Next step is to apply the covariant $\operatorname{Hom}_{\mathbb{Z}G}(-,\mathbb{Z})$-functor on the exact sequence above to obtain the cochain-complex:

$$\operatorname{Hom}_{\mathbb{Z}G}(\mathbb{Z},\mathbb{Z})\xrightarrow{\epsilon^*} \operatorname{Hom}_{\mathbb{Z}G}(\mathbb{Z}G,\mathbb{Z}) \xrightarrow{d^1} \operatorname{Hom}_{\mathbb{Z}G}(\mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G,\mathbb{Z}) \xrightarrow{d^2} \operatorname{Hom}_{\mathbb{Z}G}(\mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G \otimes_\mathbb{Z} \mathbb{Z}G,\mathbb{Z}) \rightarrow ..$$ By definition, it is $H^1(\operatorname{Hom}_{\mathbb{Z}G}(P_*,\mathbb{Z}))=\frac{\operatorname{ker}(d^2)}{\operatorname{im}(d^1)}$, and $d^2(\alpha)=\alpha\circ d_2$, $d^1(\alpha)=\alpha\circ d_1$ (for $d_2, d_1$ see above). I.e. I have to find out, what the quotient $\frac{\operatorname{ker}(d^2)}{\operatorname{im}(d^1)}$ is, but here I'm stuck. I have no idea what this image and kernel should be.. If $d^2(\alpha)=\alpha\circ d_2=0$, I only see that $\operatorname{im}(d^2)\subseteq \ker(\alpha)$ and else I dont know how to continue. I hope my question is not a duplicate and I appreciate your help.

Regards

• Find The kernel – Mariano Suárez-Álvarez Jan 21 '16 at 1:43
• ok, now I know how it works – user197416 Jan 21 '16 at 21:02

Now I have time to write down my solution, I forgot that $\mathbb{Z}$ is a trivial $\mathbb{Z}G$-module which is important for the next calculations and therefore I got stuck. The Kernel of $d^2$ is $\operatorname{Hom}(G,\mathbb{Z})$, because: $$d^2(\alpha)(g_0\otimes g_1\otimes g_2)=\alpha (g_0g_1\otimes g_2-g_0\otimes g_1g_2+g_0\otimes g_1)=g_0\alpha(1\otimes g_2)-g_0\alpha (1\otimes g_1g_2)+g_0\alpha (1\otimes g_1),$$because $\alpha$ is $\mathbb{Z}G$-linear. It follows: $$g_0\alpha (1\otimes g_1g_2)=g_0\alpha (1\otimes g_1)+g_0\alpha(1\otimes g_2).$$ $\mathbb{Z}$ is a trivial $\mathbb{Z}G$-module, therefore we get $$\alpha (1\otimes g_1g_2)=\alpha (1\otimes g_1)+\alpha(1\otimes g_2),$$i.e. $G\mapsto \mathbb{Z},\; g\mapsto \alpha(1\otimes g)$ is a group homomorphism. With $\hat{\alpha}:\mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G\to \mathbb{Z}, \; \hat{\alpha}(g_0\otimes g_1):=g_0\alpha (g_1)$, you can understand the group homomorphism $G\mapsto \mathbb{Z},\; g\mapsto \alpha(1\otimes g)$ as a $\mathbb{Z}G$-linear map $\mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G \to \mathbb{Z},\; 1\otimes g_1\mapsto \alpha(g_1)$.
Hence: $\operatorname{ker}(d^2)=\operatorname{Hom}(G,\mathbb{Z})$.
Next we claim that $\operatorname{Im}(d^1)=0$. For arbitray $g_0\otimes g_1\in \mathbb{Z}G\otimes_\mathbb{Z} \mathbb{Z}G,\; \beta\in Hom_{\mathbb{Z}G}(\mathbb{Z}G,G)$ It is $$d^1(\beta)(g_0\otimes g_1)=\beta(g_0g_1-g_0)=g_0\beta(g_1)-\beta(g_0)=\beta(g_1)-\beta(g_0)=\beta(1)-\beta(1)=0,$$ therefore $\operatorname{Im}(d^1)=0.$
Finely: $Ext_{\mathbb{Z}G}^1(\mathbb{Z},\mathbb{Z})=H^1(\operatorname{Hom}_{\mathbb{Z}G}(P_*,\mathbb{Z}))=\frac{\operatorname{ker}(d^2)}{\operatorname{Im}(d^1)}=\operatorname{Hom}(G,\mathbb{Z})$.