Let
F n = Z G ⊗ Z Z G ⊗ Z ⋯ ⊗ Z Z F_{n}=\mathbb{Z} G \otimes_{\mathbb{Z}} \mathbb{Z} G \otimes_{\mathbb{Z}} \cdots \otimes_{\mathbb{Z}} \mathbb{Z}
F n = Z G ⊗ Z Z G ⊗ Z ⋯ ⊗ Z Z
G(n + 1 factors for n ≥ \geq ≥ 0 with G-action defined on simple tensors by
g ⋅ ( g 0 ⊗ g 1 ⊗ ⋯ ⊗ g n ) = ( g g 0 ) ⊗ g 1 ⊗ ⋯ ⊗ g n g \cdot\left(g_{0} \otimes g_{1} \otimes \cdots \otimes g_{n}\right)=\left(g g_{0}\right) \otimes g_{1} \otimes \cdots \otimes g_{n}
g ⋅ ( g 0 ⊗ g 1 ⊗ ⋯ ⊗ g n ) = ( g g 0 ) ⊗ g 1 ⊗ ⋯ ⊗ g n
(a) Prove that F n F_{n} F n is a free Z G \mathbb{Z} G Z G -module of rank ∣ G ∣ n \operatorname{rank}|G|^{n} rank ∣ G ∣ n with Z G \mathbb{Z} G Z G basis
1 ⊗ g 1 ⊗ g 2 ⊗ ⋯ ⊗ g n 1 \otimes g_{1} \otimes g_{2} \otimes \cdots \otimes g_{n}
1 ⊗ g 1 ⊗ g 2 ⊗ ⋯ ⊗ g n
with g i ∈ G g_{i} \in G g i ∈ G .
Denote the basis element
1 ⊗ g 1 ⊗ g 2 ⊗ ⋯ ⊗ g n 1 \otimes g_{1} \otimes g_{2} \otimes \cdots \otimes g_{n}
1 ⊗ g 1 ⊗ g 2 ⊗ ⋯ ⊗ g n
in (a) by ( g 1 , g 2 , … , g n ) \left(g_{1}, g_{2}, \ldots, g_{n}\right) ( g 1 , g 2 , … , g n ) and define the G-module homomorphisms d n d_{n} d n for n ≥ \geq ≥ 1 on these basis elements by d 1 ( g 1 ) = g 1 − 1 d_{1}\left(g_{1}\right)=g_{1}-1 d 1 ( g 1 ) = g 1 − 1 and
d n ( g 1 , … , g n ) = g 1 ⋅ ( g 2 , … , g n ) + ∑ i = 1 n − 1 ( − 1 ) i ( g 1 , … , g i − 1 , g i g i + 1 , g i + 2 , … , g n ) + ( − 1 ) n ( g 1 , … , g n − 1 ) \begin{array}{c}
d_{n}\left(g_{1}, \ldots, g_{n}\right)=g_{1} \cdot\left(g_{2}, \ldots, g_{n}\right)+\sum_{i=1}^{n-1}(-1)^{i}\left(g_{1}, \ldots, g_{i-1}, g_{i} g_{i+1}, g_{i+2}, \ldots, g_{n}\right) \\
+(-1)^{n}\left(g_{1}, \ldots, g_{n-1}\right)
\end{array}
d n ( g 1 , … , g n ) = g 1 ⋅ ( g 2 , … , g n ) + ∑ i = 1 n − 1 ( − 1 ) i ( g 1 , … , g i − 1 , g i g i + 1 , g i + 2 , … , g n ) + ( − 1 ) n ( g 1 , … , g n − 1 )
for n ≥ 2 \geq 2 ≥ 2 . Define the Z \mathbb{Z} Z -module contracting homomorphisms
Z → s − 1 F 0 ⟶ s 0 F 1 ⟶ s 1 F 2 ⟶ s 2 ⋯ \mathbb{Z} \stackrel{s_{-1}}{\rightarrow} F_{0} \stackrel{s_{0}}{\longrightarrow} F_{1} \stackrel{s_{1}}{\longrightarrow} F_{2} \stackrel{s_{2}}{\longrightarrow} \cdots
Z → s − 1 F 0 ⟶ s 0 F 1 ⟶ s 1 F 2 ⟶ s 2 ⋯
on a Z \mathbb{Z} Z basis by s − 1 ( 1 ) = 1 s_{-1}(1)=1 s − 1 ( 1 ) = 1 and
s n ( g 0 ⊗ ⋯ ⊗ g n ) = 1 ⊗ g 0 ⊗ … ⊗ g n . s_{n}\left(g_{0} \otimes \cdots \otimes g_{n}\right)=1 \otimes g_{0} \otimes \ldots \otimes g_{n}.
s n ( g 0 ⊗ ⋯ ⊗ g n ) = 1 ⊗ g 0 ⊗ … ⊗ g n .
(b) Prove that
ϵ s − 1 = 1 , d 1 s 0 + s − 1 ϵ = 1 , d n + 1 s n + s n − 1 d n = 1 , for all n ≥ 1 \epsilon s_{-1}=1, \quad d_{1} s_{0}+s_{-1} \epsilon=1, \quad d_{n+1} s_{n}+s_{n-1} d_{n}=1, \text { for all } n \geq 1
ϵ s − 1 = 1 , d 1 s 0 + s − 1 ϵ = 1 , d n + 1 s n + s n − 1 d n = 1 , for all n ≥ 1
where the map aug : F 0 → Z F_{0} \rightarrow \mathbb{Z} F 0 → Z is the augmentation
map aug ( ∑ g ∈ G α g g ) = ∑ g ∈ G α g . \operatorname{map} \operatorname{aug}\left(\sum_{g \in G} \alpha_{g} g\right)=\sum_{g \in G} \alpha_{g}.
map aug ⎝ ⎛ g ∈ G ∑ α g g ⎠ ⎞ = g ∈ G ∑ α g .
(c) Prove that the maps s n s_{n} s n are a chain homotopy between the identity (chain) map and the zero (chain) map from the chain
⋯ ⟶ F n ⟶ d n F n − 1 ⟶ d n − 1 ⋯ ⟶ d 1 F 0 ⟶ aug Z ⟶ 0 ( ∗ ) \cdots \longrightarrow F_{n} \stackrel{d_{n}}{\longrightarrow} F_{n-1} \stackrel{d_{n-1}}{\longrightarrow} \cdots \stackrel{d_{1}}{\longrightarrow} F_{0} \stackrel{\text { aug }}{\longrightarrow} \mathbb{Z} \longrightarrow 0 \quad {(\ast)}
⋯ ⟶ F n ⟶ d n F n − 1 ⟶ d n − 1 ⋯ ⟶ d 1 F 0 ⟶ aug Z ⟶ 0 ( ∗ )
of Z \mathbb{Z} Z -modules to itself.
(d) Deduce from (c) that all Z \mathbb{Z} Z -module homology groups of ( ∗ ) (\ast) ( ∗ ) are zero, i.e., ( ∗ ) (\ast) ( ∗ ) is an exact sequence of Z \mathbb{Z} Z -modules. Conclude that ( ∗ ) (\ast) ( ∗ ) is a projective G-module resolution of Z \mathbb{Z} Z .