Exercise 4, § 16

Let G be a group and H a subgroup of G. For any $\mathbf{Z}$[H]-module N, denote by $\widetilde{N}$ the group Hom$_{\mathbf{Z}[H]}(\mathbf{Z}[G],N)$ endowed with a $\mathbf{Z}$[G]-module structure deduced from the right $\mathbf{Z}$[G]-module structure on $\mathbf{Z}[G]$.

a) Prove that $\widetilde{N}$ can be identified with the $\mathbf{Z}$[G]-submodule of $\mathscr{F}(G,N)$ (Exercise 3) consisting of the mappings $f$ such that $f(hg) =hf(g)$ for every $h\in H$ and $g\in G$. If A is an abelian group, then the $\mathbf{Z}$[G]-module $\mathscr{F}\widetilde{(H},A)$ is canonically isomorphic to $\mathscr{F}(G,A)$.

b) Let $0\rightarrow N'\rightarrow^iN\overset{p}{\rightarrow}N''\rightarrow 0$ be an exact sequence of $\mathbf{Z}$[H]-modules; the sequence

$'$ Hom(1$,i$) Hom(1$,p$)$''$

is exact (observe that the $\mathbf{Z}$[H]-module $\mathbf{Z}[G]$ is free).

c) Let $u_N:\widetilde{N}\rightarrow N$ be the mapping $f\mapsto f(1)$; it is compatible with the canonical injection $i: H\rightarrow G$ (Exercise 2). Prove that the homomorphism

is bijective for every $q\geqslant 0$ (first treat the case $q= 0$ and then reason by induction on $q$ using the above; Exercise 3, c); and Exercise 2, c)).

d) Suppose that H has finite index $n$ in G. Let M be a $\mathbf{Z}$[G]-module, which we view as a $\mathbf{Z}$[H]-module by restriction of scalars. Let $f\in \widetilde{M}$; the element $g^{-1}f(g)$ depends only on the right class $Hg$ of $g$. Set $\pi (f) =\sum_{Hg\in H\backslash G}g^{-1}f(g)$. Prove that $\pi$ is a $\mathbf{Z}$[G]-linear homomorphism from $\widetilde{M}$ to M. For any integer $q\geqslant 0$, denote by Cor$^q: H^q(H,M)\rightarrow H^q(G,M)$ the homomorphism $H^q(\pi )\circ H^q(i, u_M)^{-1}$ (“corestriction homomorphism”).

e) Prove that the endomorphism Cor$^q\circ$ Res$^q$ of $H^q(G,M)$ is multiplication by $n$ (first treat the case $q= 0$ and then treat the general case as in the proof of c)). f ) Suppose that G is finite; prove that for every $\mathbf{Z}$[G]-module M and every integer $q\geqslant 1$, the group $H^q(G,M)$ is annihilated by the order of G.