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.



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