Projective resolution induced by subgroup $H\subset G$ in group cohomology I'm studying group homology and cohomology following the book An introduction to homological algebra from Weibel and I have difficulties in understanding a statement regarding projective resolutions.
From now on, let $G$ a group, $H\subset G$ a subgroup, and $\mathbb{Z}G$ the integral group ring.
At the beginning of the proof of Shapiro's lemma (6.3.2) its said the following:

Note that $\mathbb{Z}G$ is a free $\mathbb{Z}H$-module. Hence any projective right $\mathbb{Z}G$-module resolution $P\rightarrow\mathbb{Z}$ is also a projective $\mathbb{Z}H$-module resolution.

I have no problems in proving that $\mathbb{Z}G$ is free as $\mathbb{Z}H$-module.
However, I am not able to undestand why any projective right $\mathbb{Z}G$-module resolution $P\rightarrow\mathbb{Z}$ has also to be a projective $\mathbb{Z}H$-module resolution.
I would be very grateful if someone could help me with this.
Thanks in advance!
 A: We have a subring $S$ ($= ℤ[H]$) of a ring $R$ ($= ℤ[G]$) such that $R$ is free as an $S$-module, and we have for some $R$-module $M$ a projective resolution
$$
  \dotsb
  \xrightarrow{\enspace d_2 \enspace} P_2
  \xrightarrow{\enspace d_1 \enspace} P_1
  \xrightarrow{\enspace d_0 \enspace} P_0
  \xrightarrow{\enspace ε \enspace} M
  \longrightarrow 0 \,.
  \tag{$\ast$}
$$
By restriction, both $M$ and each $P_i$ become $S$-modules.
To show that $(\ast)$ is also a projective resolution of $M$ as an $S$-module, we need to show that

*

*the sequence $(\ast)$ is also exact as $S$-modules, and


*each $P_i$ is projective as an $S$-module.
Part 1 is true because exactness can be checked on the underlying level of abelian groups.
Part 2 can be seen in three steps:

*

*We know that $R$ is free as an $S$-module.


*Consequently, every free $R$-module $F$ is also free as an $S$-module.
Indeed, we have $R ≅ S^{⊕ I}$ as $S$-modules for some set $I$, as well as $F ≅ R^{⊕ J}$ as $R$-modules for some set $J$.
Therefore $F ≅ R^{⊕ J} ≅ ((S^{⊕ I})^{⊕ J} = S^{⊕ (I × J)}$ as $S$-modules.


*Let $P$ be a projective $R$-module.
This means that there exists another $R$-module $P'$ such that $P ⊕ P'$ is free as an $R$-module.
As seen above, this means that $P ⊕ P'$ is also free as an $S$-module.
The existence of $P'$ thus shows that $P$ is also projective as an $S$-module.
