The paper "Cohomology of p-adic analytic groups" by Symonds and Weigel is considered one of the main references for continuous cohomology of profinite groups. There is a passage I do not understand: the context is the following.

We consider $G$ to be a pro-$p$ group Poincaré dual: this is defined in §4.4 to be a group of finite cohomological dimension $n$ such that $\mathbb{Z}_p$ (considered as a trivial module) admits a projective resolution of finite length. Moreover we have $H^k(G; \mathbb{Z}_p [\![G]\!] )=0$ for $k \neq n$ and $H^k(G; \mathbb{Z}_p [\![G]\!] )=\mathbb{Z}_p$.

The first two conditions are only for finiteness and well behavior of the associated cohomology groups, the key property is the third which ensures $H^*(G;-)$ and $H_{n-*}(G;{-})$ are related by an appropriate duality.

In fact we can show the existence of a right $\mathbb{Z}_p [\![G]\!] $-module $D_p(G)$ whose underlying abelian group is $\mathbb{Z}_p$ and such that there are natural bjiections \begin{equation} H^*(G;M)\cong H_{n-*}(G; D_p(G)\otimes M) \end{equation} for $M$ in an appropriate category of left $\mathbb{Z}_p [\![G]\!] $-modules.

My question regards Corollary 4.5.2 of this paper: it states that for such $G$ and any module $M$ which is isomorphic to $\mathbb{Z}_p$ as abelian group then the following two conditions are equivalent

- $H^n(G;M)\neq 0$
- $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M \cong \mathbb{Z}_p$ as $\mathbb{Z}_p [\![G]\!]$-modules.

I do not see why 1 should imply 2: the duality implies $H^n(G;M)\cong H_{0}(G; D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)\neq 0$ and we know $H_0(G;N)\cong N_G=N/_{ \overline{\langle g.n-n \mathrel: g \in G, \ n \in N \rangle}}$. It seems to state that $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G \neq 0$ implies $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G=\mathbb{Z}_p$ and by duality the top cohomolgy $H^n(G;M)$ can only be $0$ or $\mathbb{Z}_p$.

But this should be false: a priori we could have that $\overline{\langle g.x-x \mathrel: g \in G, \ x \in D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M}\rangle$ is a proper subgroup of $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M$ (which is just $\mathbb{Z}_p$ as abelian group).

Also it is easy to provide a counterexample to the statement that $H^*(G;M)$ is either $0$ or $\mathbb{Z}_p$. Take $G=1+p\mathbb{Z}_p$ and $M=\mathbb{Z}_p$ with the action by multiplication, we can compute $H^1(G;M)=\mathbb{Z}/p$ and $H^k(G;M)=0$ for $k \neq 1$.