Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation? I'm reading Serre's Local Fields,  and I'm trying to understand the proof of  Prop. 9 in $\S$5 of Chap. 8 (p.136). First, the setup:


*

*$p$ is a prime number

*$G$ is a cyclic group of order $p$

*$A$ is a $G$-module

*$h(A)$ is the Herbrand quotient of $A$

*$\varphi(A)$ is the Herbrand quotient of $A$ if it were acted on by $G$ trivially


Prop. 9 states that

When $\varphi(A)$ is defined, the quantities $\varphi(A^G)$, $\varphi(A_G)$, and $h(A)$ are all defined, and 
  $$h(A)^{p-1}=\varphi(A^G)^p/\varphi(A)=\varphi(A_G)^p/\varphi(A).$$

In the proof, one of the cases that we end up reducing to is when $A$ is a $G$-module that has the following properties:


*

*$A$ is $p$-divisible (for every $y\in A$, there is an $x\in A$ such that $px=y$)

*every element of $A$ is $p$-power torsion (for every $y\in A$, there is an $n\in\mathbb{N}$ such that $p^ny=0$)

*the $p$-torsion subgroup $\{y\in A\mid py=0\}$ of $A$ is finite


Serre says at this point that:

... it is quicker to use Pontryagin duality: it transforms $A$ into a compact group $\hat{A}$, which is a free module of finite type over the ring $\mathbb{Z}_p$ of $p$-adic integers, on which $G$ acts. It can be immediately verified that
  $$h(A)=h(\hat{A})^{-1},\qquad\varphi(A)=\varphi(\hat{A})^{-1},\qquad\varphi(A^G)=\varphi(\hat{A}_G)^{-1}.$$

Not having any more than a Wikipedia-level knowledge of Pontryagin duality, this passage is unfortunately rather opaque to me. From this section of the Wikipedia page, I deduce that we have implicitly put the discrete topology on $A$. Okay, that's reasonable. But what is the justification for the displayed formulas? Noting the multiplicativity of the Herbrand quotient on exact sequences, I assume the first equation is true because there is an exact sequence involving $A$, $\hat{A}$, and $G$-module $B$ with $h(B)=1$, and similarly with the second equation, but I can't seem to figure out what $B$ ought to be, nor why the third equation is true.
If you are feeling generous, I would also appreciate a hint/explanation as to why $\hat{A}$ is a finitely generated free $\mathbb{Z}_p$-module.
 A: I don't know the $p$-version of this, but I can tell you how things work in
the "non-$p$"-case. It should not be hard to tweak this into what you want.
Write $\hat H^\bullet(G,\mathord-)$ for Tate cohomology of a finite group
$G$. For each abelian group $A$ let $A^\wedge=\hom(A,\mathbb R/\mathbb Z)$
be the abelian group of all homomorphisms of abelian groups; if $A$ happens to
be a $G$-module, then $A^\wedge$ is also a $G$-module in the usual way.
Then there is a canonical isomorphism $$\hat
H^{p-1}(G,A^\wedge)\cong\left(\hat H^{-p}(G,A)\right)^{\wedge}.$$ In other
words, $(\mathord-)^\wedge$ flips the cohomological degree with a twist,
and dualizes. This can be proved from general nonsense using the fact that
$(\mathord-)^\wedge$ is a duality---it is done in detail in
Cartan-Eileberg.
Now, for example, suppose $A$ is a $G$-module for which $h(A)$ is defined.
We have (using periodicity: I am assuming from now one that here $G$ is cyclic, or, to be silly, that $G$ has period $2$) $$\hat H^0(G,A^\wedge)=\hat H^{-1}(G,A)^\wedge=\hat
H^1(G,A)^\wedge$$ and $$\hat H^1(G,A^\wedge)=\hat H^{-2}(G,A)^\wedge=\hat
H^0(G,A)^\wedge.$$ Since the groups $\hat H^p(G,A)$ are finite, they are
(non-canonically) isomorphic to their duals, so $h_0(A^\wedge)=h_1(A)$ and
$h_1(A^\wedge)=h_0(A)$. This gives the first formula you want.
