Group homology Künneth Formula I am trying to understand Weibel's proof of Künneth Formula for group homology (Prop. 6.1.13 from An Introduction to Homological Algebra), and I am struggling with following statement:
Let $G,H$ be groups. Let $P\rightarrow\mathbb{Z}$ and $Q\rightarrow\mathbb{Z}$ be free resolutions of $\mathbb{Z}$ respectively over $\mathbb{Z}G$ and $\mathbb{Z}H$. Then, by the Kunneth Formula for chain complexes, we get that the homology of $P\otimes_\mathbb{Z}Q$ (seen as total complex of tensor product double complex) is $0$ for non-zero degree.
How one can derive this statement from Künneth Formula for chain complexes?

We have, by Künneth formula,
$$H_n(P_{\cdot} \otimes_{\Bbb Z} Q_{\cdot}) \cong 
\left(
\bigoplus_{p+q=n} H_p(P_{\cdot}) \otimes_{\Bbb Z} H_q(Q_{\cdot})
\right) \oplus 
\bigoplus_{p+q=n-1} \mathrm{Tor}_1^{\Bbb Z}(H_p(P_{\cdot}), H_q(Q_{\cdot}))$$
When $r>0$, we know that $H_r(P_{\cdot}) = H_r(Q_{\cdot}) = 0$, since these complexes are free resolutions.
When $n>0$ and $p+q=n$, we must have $p>0$ or $q>0$, thus the first direct summand is $0$.
When $n>1$ and $p+q=n-1$, we must have $p>0$ or $q>0$, thus the second direct summand is $0$. If $n=1$, we are left with 
$\mathrm{Tor}_1^{\Bbb Z}(P_0, Q_0) = 0$, since both modules are free over $\Bbb Z[G]$, hence over $\Bbb Z$ (hence flat over $\Bbb Z$). In all cases, when $n>0$, the second direct summand also vanishes.
In other words, we've got : $H_n(P_{\cdot} \otimes_{\Bbb Z} Q_{\cdot}) = 0$ when $n>0$.
