4
$\begingroup$

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)?

1 Answers 1

3

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).