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I'm learning some group cohomology from the third section in Serre's Local Fields, and I'm up to the section on change of group. If $f:G'\rightarrow G$ is a homomorphism of groups and $A$ is a $G$-module, there is an induced $G'$-module structure given by $s'\cdot a=f(s')\cdot a$, for $s'\in G'$ and $a\in A$ (this is Serre's notation, I would reserve $G'$ for the commutator subgroup, but oh well). This induces both maps on cohomology $H^q(G,A)\rightarrow H^q(G',A)$ and homology $H_q(G',A)\rightarrow H_q(G,A)$.

My question is: If $A$ is a projective/injective/relatively projective/relatively injective $G$-module, must $A$-as-a-$G'$-module be such a $G'$-module? If not, are there any assumptions we could make about $f:G'\rightarrow G$ so that this would be true (e.g., $f$ surjective)?

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If $f:G' \to G$ is injective, this is true in all cases: (relatively) projective/injective. Identify the $G$-modules with $\mathbb{Z}G$-modules and observe that $\mathbb{Z}G$ is a free (hence also projective and flat) $\mathbb{Z}G'$-module (it is free as a $\mathbb{Z}G'$-module because $\mathbb{Z}G \cong \bigoplus_{G/G'} \mathbb{Z}G'$ as a $\mathbb{Z}G'$-module). The functors $\mathbb{Z}G \otimes_{G'} {-}$ (induction) and $\operatorname{Hom}_{\mathbb{Z}G'}{(\mathbb{Z}G,{-})}$ (co-induction) are therefore exact and left/right adjoint to restriction.

By checking the lifting/extension conditions, it is easy to see that a left/right adjoint to an exact functor preserves projectives/injectives. Here "exact" of course means "sends $\mathbb{Z}$-split exact sequences to $\mathbb{Z}$-split exact sequences" in the relative case. For instance, if $L$ has an exact right adjoint $R$ and $P$ is projective then $LP$ is projective because a lifting problem of a map $LP \to N$ over an epimorphism $M \twoheadrightarrow N$ corresponds to a lifting problem of a map $P \to RN$ over the morphism $RM \twoheadrightarrow RN$ which is epi because $R$ is exact.

However, if $f$ is not injective, this is wrong in general. Just consider the case where $G = \{e\}$ is the trivial group. It is certainly not the case that $\mathbb{Z}$ is (rel.) projective or $\mathbb{Q}$ is (rel.) injective as a $\mathbb{Z}G'$-module (otherwise group homology with $\mathbb{Z}$-coefficients or group cohomology with $\mathbb{Q}$-coefficients would always have to vanish).

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