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!
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.