Herbrand quotient - definition and multiplicativity We defined the Herbrand quotient as follows:
Let $G$ be a cyclic group and let $A$ be a $G$-module. Then the Herbrand quotient is defined as
$$h(A)= \frac{|H^0(G,A)|}{|H^1(G,A)|}.$$
We then had the proposition: If the cyclic group $G$ of order $n$ acts trivially on $A$, then $$h(A)=q_{0,n}(A),$$ where, for $nA:= \{ a \in A | na=0\}$, we defined 
$q_{0,n}(A):= \frac{ (A:nA)}{|nA|}.$
First question: Why do I have this equality $h(A)=q_{0,n}(A)$? I tried to work with the definition that $H^0(G,A)=A^G$ and $H^1(G,A)=Hom(G,A)$, but I don't know how to proceed?
In the course of the proposition above, we then stated as a consequence: If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence of abelian groups, then $q_{0,n}(B)=q_{0,n}(A) \cdot q_{0,n}(C)$.
Second question: Again, why does this hold? I tried to reformulate this statement by using the def of $q_{0,n}(.)$, but I don't see how I can use the exactness of this sequence.
Thank you for your ideas!
 A: Remember that this is defined for the Tate cohomology groups, which differ from the "regular" cohomology groups in degree $0$: $\hat{H}^i(G,A) = H^i(G,A)$ if $i > 0$ and $\hat{H}^0(G,A) = H^0(G,A)/\text{Im}(N)$ where $N: A \to A$ is the norm map induced by the action of $G$. The reason we use the modified groups is because this allows them to fit into a long exact sequence with the homology groups $H_i(G,A)$, again with some modification of $H_0$.
Now for your first question, just remember that $\text{Hom}(G,A)$ is the same thing as picking out an element of $A$ of order dividing $n$, which you write as "$nA$", but that's confusing because you also use that notation to mean $\{na~|~a \in A\}$ which is the standard meaning of $nA$. The more standard notation for the $n$-torsion in $A$ is $A[n]$.
So show that $\text{Hom(G,A)} = A[n]$, and for $\hat{H}^0(G,A) = A^G/\text{Im}(N)$, recall that if the action of $G$ on $A$ is trivial then $A^G = A$, and $\text{Im}(N) = nA$ (in my notation). That should finish part $1$ for you.
For the second part, use the long exact sequence in cohomology induced from the short exact sequence $0 \to A \to B \to C \to 0$ and recall that when $G$ is cyclic, its Tate cohomology groups are periodic: $\hat{H}^i(G,A) \cong \hat{H}^{i-2}(G,A)$. Use this to get an "exact hexagon" and you can read the multiplicativity from that.
