Suppose that $G$ is a group, and $\mathbb{Z}[G]$ the group ring. Then $\mathbb Z$ can be considered as a $\mathbb{Z}[G]$-module if every $g (\in G)$ acts trivally, and every $n (\in \mathbb Z)$ acts by multiplication on $\mathbb Z$.
In order to find a projective resolution of the module $\mathbb Z$, thus finding the cohomology module, a book says (translated):
Consider the elements of $G$ as vertexes, then we can define a simplicial complex $K = K(G)$. Suppose that there is a well order in $G$, and $e$ is the first element in this order. So $K_n = \{ (g_0, g_1, \cdots, g_n) | g_i \in G, g_0
In order to prove the exactness of this sequence, the author defines a group homomorphism $s_n: C_n \rightarrow C_{n+1} , (g_0, \cdots, g_n) \mapsto (e,g_0, \cdots, g_n)$ for $n =0,1,2, \cdots$, and use this to prove that the identity map in $C_n$ is null-homotopic. But this definition is unclear to me. I thought one vertex could appear only once in a simplex. So, what will happen if one of the $g_i$ in $\{g_0, \cdots, g_n \}$ is already $e$? Going back to the well-ordering assumption on $G$, is this well-order strict or not? If it is strict, then I think no duplication of vertexes in a simplex would be allowed; otherwise, the chain would be endless even if the group is finite.
Thanks to everyone.