# To find image of map $(C_{1}(H)_H, C_{2}(G)_G)\otimes_{\mathbb{Z}} \mathbb{Z}[G] \to C_1(G)$, where $G$ is a group and $H$ its subgroup.

Let $$G$$ be a group. For any $$G$$-set $$Y$$ consider a chain complex $$(C_*(Y)_G : = C_*(Y) \otimes_{\mathbb{Z[G]}} \mathbb{Z}, \tilde\partial=\partial \otimes id_{\mathbb{Z}})$$, where $$C_n(Y)$$ is the free $$\mathbb{Z}$$-module generated by $$(n+1)$$-tuples $$(y_0, y_1, \ldots,y_n)$$ of elements of $$Y$$, $$\partial = \sum_i (-1)^i \partial_i$$ and $$\partial_i(y_0, y_1, \ldots,y_n)=(y_0, \ldots,y_{i-1}, y_{i+1}, \ldots, y_n)$$.

Now let $$H$$ be a subgroup of $$G$$. Consider a chain complex $$\big( D_*= (C_{*-1}(H)_H, C_{*}(G)_G), \delta \big)$$, where $$\delta=\begin{bmatrix} \tilde\partial &0\\i_*&-\tilde\partial\end{bmatrix}$$.

Now consider $$F_i=D_i \otimes_\mathbb{Z} \mathbb{Z}[G]$$ for $$i \geq 2$$, and $$F_1= \ker(D_1 \to D_0) \otimes_\mathbb{Z} \mathbb{Z}[G]=(0,C_1(G)\otimes_{\mathbb{Z}[G]} \mathbb{Z})\otimes_{\mathbb{Z}} \mathbb{Z}[G]\cong C_1(G).$$

I have to find the image of the map $$F_2 \to F_1$$.

The following are the calculations I have done.

\begin{align} &\delta\big(((h,1) \otimes 1, (g_1,g_2,1) \otimes 1) \otimes g \big)\\ &=\delta\big(((h,1) \otimes 1, (g_1,g_2,1) \otimes 1)\big) \otimes g\\ &=\big((h-1) \otimes 1, (h,1) \otimes 1 -(g_2,1) \otimes 1 + (g_1,1) \otimes 1 -(g_1,g_2) \otimes 1)\big) \otimes g\\ &=(0, (h,1) \otimes 1 -(g_2,1) \otimes 1 + (g_1,1) \otimes 1 -(g_1g_2^{-1},1) \otimes 1) \otimes g. \end{align}

Under the isomorphism $$F_1 \cong C_1(G)$$, we have $$\delta\big(((h,1) \otimes 1, (g_1,g_2,1) \otimes 1) \otimes g \big)=(hg,g) -(g_2g,g)+(g_1g,g) -(g_1g_2^{-1}g,g).$$

Thus the image of the map $$F_2 \to F_1 \to C_1(G)$$ is generated by the set $$\{ n(hg,g) -m(g_2g,g)+m(g_1g,g) -m(g_1g_2^{-1}g,g)~|~h \in H, g_1,g_2,g \in G, n,m\in \mathbb{Z}\}$$.

I request you to please check whether my calculations are correct or not.

The above clarification is needed to understand this question.