Exercise 17, § 6

Let G be a finite group. We denote by N the element $\sum_{g \in G} e_g$ of $\mathbf{Z}^{(G)}$. For every $\mathbf{Z}^{(G)}$-module M, multiplication by N defines by passing to the quotient a homomorphism $N_0 : H_0(G, M) \to H^0(G, M)$. We put $\hat{H}^0(G, M) = \mathrm{Coker}\,(N_0)$, $\hat{H}^{-1}(G, M) = \mathrm{Ker}\,(N_0)$, $\hat{H}^q(G, M) = H^q(G, M)$ for $q \geq 1$, and

a) Let $\mathscr{E} : 0 \to M' \to M \to M'' \to 0$ be an exact sequence of $\mathbf{Z}^{(G)}$-modules; define an exact sequence

b) If M is projective relative to $\mathbf{Z}$, show that one has $\hat{H}^q(G, M) = 0$ for every $q \in \mathbf{Z}$ (cf. exercise 12). c) Let H be a subgroup of G, $i : H \to G$ the canonical injection. For every $\mathbf{Z}^{(G)}$-module M and every rational integer q, define homomorphisms $\hat{i}_M^q : \hat{H}^q(G, M) \to \hat{H}^q(H, M)$ and $\hat{i}_M^q : \hat{H}^q(H, M) \to \hat{H}^q(G, M)$ (denoted simply by $\hat{i}^q$ and $\hat{i}^q$ if there is no ambiguity about M) in the following way: for $q \geq 1$, put $\hat{i}^q = \hat{i}^q$ and $\hat{i}^q = t^q$ (exercises 10 and 14); for $q \leq -2$, put $\hat{i}^q = \theta_{-q-1}$ and $\hat{i}^q = i_{-q-1}$; finally, define $\hat{i}^0$ and $\hat{i}^0$ (resp. $\hat{i}^{-1}$ and $\hat{i}^{-1}$) by passing to the quotients (resp. to the subsets) from $i^0$ and $t^0$ (resp. $t_0$ and $i_0$). Prove that one has $\hat{i}^q \circ \hat{i}^q = n.\mathrm{Id}$ for every $q \in \mathbf{Z}$, where n is the index of H in G. Deduce that the groups $\hat{H}^q(G, M)$ are annihilated by the order of G (cf. exercise 6, p. 189). d) With the notations of b), show that one has $\hat{i}_{M'}^{q+1} \circ \partial^q(G, \mathscr{E}) = \partial^q(H, \mathscr{E}) \circ \hat{i}_{M''}^q$ and

for every $q \in \mathbf{Z}$. e) Let p be a prime number and H a p-Sylow subgroup (I, p. 74) of G. Show that the homomorphism $\hat{i}^q : \hat{H}^q(G, M) \to \hat{H}^q(H, M)$ has for kernel the sum of the l-primary components of the $\mathbf{Z}$-module $\hat{H}^q(G, M)$ for $l \neq p$. In particular, if for every prime number p, one chooses a p-Sylow subgroup $H_p$ of G, the mapping $\hat{H}^q(G, M) \to \prod_p \hat{H}^q(H_p, M)$ deduced from the homomorphisms $\hat{i}^q$ is injective.