weibel's exercise 6.2.4 
I think we should use the corollary to solve this problem.
For the first question, I think we can use the property that  $J$ is a free $\mathbb{Z}[G]$ module with a basis ${s-1,t-1}$. Hence $J\otimes \mathbb{Z^{'}}= 2\mathbb{Z}[G]\otimes \mathbb{Z^{'}}$.
But I don't know how to solve the second question.
Thank you for your help.
 A: Since $J$ is a free $\mathbb Z[G]$-module of rank two (the two generators being $(s-1)$ and $(t-1)$), the tensor product $J \otimes_{\mathbb Z[G]} \mathbb Z'$ is the direct sum of two copies of $\mathbb Z [G] \otimes_{\mathbb Z[G]} \mathbb Z' \cong \mathbb Z$. So the tensor product of the resolution of the trivial representation $0 \rightarrow J \rightarrow \mathbb Z[G] \rightarrow 0$ with $\mathbb Z'$ is:
$$0 \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0.$$
Then you need to understand the middle map. The calculations that you have already done in your question (which are in fact just a reformulation of the definition of $\mathbb Z'$) give you that the two elements $(s-1) \otimes 1$ and $(t-1) \otimes 1$, which form a basis of the source, are both sent to $1 \otimes 2 = 2 (1 \otimes 1)$ in the target, which means that this map identifies with $(m,n) \mapsto 2(m+n)$ (it sends both $(1,0)$ and $(0,1)$ to $2$). Its cokernel is thus $\mathbb Z/2$, and its kernel is $\mathbb Z$, as expected (and you can write these generators as the classes of explicit cycles if you like).
For computing the homology $H_*(T, \mathbb Z')$, you have the same resolution (since $T$ is free on one generator), and you can use exactly the same reasoning.
A: I think the hardest part of the question is how to prove $H_1(G ; \mathbb{Z}^{'}) = \mathbb{Z}$.
We just need to prove that the kernel of $J\otimes \mathbb{Z^{'}} \longrightarrow  \mathbb{Z}[G]\otimes \mathbb{Z^{'}}$ is $\{(s-t) \otimes n\mid n\in \mathbb{Z}\}$.
First it is easy to check that $(s^2-1)\otimes 1=(s-1)(s+1)\otimes 1 =0 \in J \otimes \mathbb{Z^{'}}$.
For every $g(s,t)\otimes 1\in  J\otimes \mathbb{Z^{'}}$ which is zero in $ \mathbb{Z}[G]\otimes \mathbb{Z^{'}}$, we can assume that g(s,t) dones't have constant.
Hence,$g(s,t)=sf_1(s,t)-tf_2(s,t)=t(f_1(s,t)-f_2(s,t))+(s-t)(f_1(s,t))$ with $g(1,1)=$$g(-1,-1)=0$.
We can continue this process for $(f_1(s,t)-f_2(s,t))$.
Ultimately, $\{(s-t) \otimes n\mid n\in \mathbb{Z}\}$ is the kernel.
