Herbrand Quotient I am trying to solve exercises from Lang's Algebra, and I am stuck on a problem about Herbrand quotients. 
Let $G$ be a finite cyclic group of order $n$ generated by an element $\sigma$. Assume that $G$ operates on an abelian group $A$, and let $f, g : A \rightarrow A$ be the endomorphisms of $A$ given by $$f(x) = \sigma x - x$$ and $$g(x) = x + \sigma x + \ldots + \sigma ^ {n -1} x.$$
Define the Herbrand quotient by the expression $q(A) = (A_f : A ^ g) / (A_g : A ^ f)$, provided both indices are finite. Assume now that $B$ is a subgroup of $A$ such that $G B \subseteq B$. 
(a) Define in a natural way an operation of $G$ on $A / B$. 
(b) Prove that $$q(A) = q(B) q(A / B)$$
in the sense that if two of these quotients are finite, so is the third, and the stated equality holds.
(c) If $A$ is finite, show that $q(A) = 1$. 
The $A_f, A ^ f$ in the question denotes the image and kernel of $f$ as a map from $A$ to $A$. Similarly for $B_g$ and $B ^ g$. 
I am currently stuck on $(b)$ and also I do not understand why $f, g$ are endomorphisms of groups to start with. If $G$ is acting on $A$, it is not necessary that each $g \in G$ induces an automorphism on $A$ right? Thanks in advance!
 A: Statement (b) can be proved by applying long homology sequence.
For consider the complexes of abelian groups:
\begin{align}
&A:\ldots\to A\xrightarrow f A\xrightarrow g A\xrightarrow f\ldots\\
&B:\ldots\to B\xrightarrow f B\xrightarrow g B\xrightarrow f\ldots\\
&A/B:\ldots\to A/B\xrightarrow f A/B\xrightarrow g A/B\xrightarrow f\ldots\\
\end{align}
Then we get an exact sequence of complexes
$$\{0\}\to B\to A\to A/B\to\{0\}$$
giving rise to the long homology sequence
$$H_1(A/B)\xrightarrow{\delta_1}H_0(B)\xrightarrow{\kappa_0}H_0(A)\xrightarrow{\pi_0}H_0(A/B)\xrightarrow{\delta_0}H_1(B)\xrightarrow{\kappa_1}H_1(A)\xrightarrow{\pi_1}H_1(A/B)\tag 1$$
where we noted that
$$H_n(A)=\begin{cases}A_g/A^f&n\equiv 0\pmod 2\\A_f/A^g&n\equiv 1\pmod 2\end{cases}$$
so that $q(A)=|H_1(A)|/|H_0(A)|$.
The equation $q(A)=q(B)q(A/B)$ then follows from the exactness of $(1)$ for:$\newcommand\Ker{\operatorname{Ker}}\renewcommand\Im{\operatorname{Im}}$
\begin{align}
q(B)(A/B)
&=\frac{|H_1(B)|}{|H_0(B)|}\frac{|H_1(A/B)|}{|H_0(A/B)|}\\
&=\frac{|\Ker\kappa_1||\Im\kappa_1|}{|\Ker\kappa_0||\Im\kappa_0|}
\frac{|\Ker\delta_1||\Im\delta_1|}{|\Ker\delta_0||\Im\delta_0|}\\
&=\frac{|\Im\kappa_1|}{|\Im\kappa_0|}
\frac{|\Ker\delta_1|}{|\Ker\delta_0|}\\
&=\frac{|\Ker\pi_1|}{|\Ker\pi_0|}
\frac{|\Im\pi_1|}{|\Im\pi_0|}\\
&=\frac{|H_1(A)|}{|H_0(A)|}\\
&=q(A)
\end{align}
A: I will use the result in this page. Let $\pi:A\rightarrow A/B$ be the canonical surjection.
It is sufficient to prove:
$$[A_f:A^g][B_g:B^f][(A/B)_g:(A/B)^f]=[A_g:A^f][B_f:B^g][(A/B)_f:(A/B)^g].$$
But:
$$[A_f:A^g]=[(A_f)^\pi:(A^g)^\pi][(A_f)_\pi:(A^g)_\pi]$$
$$[B_g:B^f]=[(A_\pi)_g:(A_\pi)^f]=[(A_g)_\pi:(A^f)_\pi][(A^f)_\pi:(A_\pi)^f]$$
$$[(A/B)_g:(A/B)^f]=[(A^\pi)_g:(A^\pi)^f]=[(A^\pi)_g:(A_g)^\pi][(A_g)^\pi:(A^f)^\pi]$$
Analogously:
$$[A_g:A^f]=[(A_g)^\pi:(A^f)^\pi][(A_g)_\pi:(A^f)_\pi]$$
$$[B_f:B^g]=[(A_f)_\pi:(A^g)_\pi][(A^g)_\pi:(A_\pi)^g]$$
$$[(A/B)_f:(A/B)^g]=[(A^\pi)_f:(A_f)^\pi][(A_f)^\pi:(A^g)^\pi]$$
So the problem is equivalent to prove that:
$$[(A_f)^\pi:(A^g)^\pi][(A_f)_\pi:(A^g)_\pi][(A_g)_\pi:(A^f)_\pi][(A^f)_\pi:(A_\pi)^f][(A^\pi)_g:(A_g)^\pi][(A_g)^\pi:(A^f)^\pi]$$
is equal to:
$$[(A_g)^\pi:(A^f)^\pi][(A_g)_\pi:(A^f)_\pi][(A_f)_\pi:(A^g)_\pi][(A^g)_\pi:(A_\pi)^g][(A^\pi)_f:(A_f)^\pi][(A_f)^\pi:(A^g)^\pi].$$
So it suffices to prove that:
$$[(A^f)_\pi:(A_\pi)^f][(A^\pi)_g:(A_g)^\pi]=[(A^g)_\pi:(A_\pi)^g][(A^\pi)_f:(A_f)^\pi].$$
We will prove that $[(A^f)_\pi:(A_\pi)^f]=[(A^\pi)_f:(A_f)^\pi]$. Let $P$ be a representant system of cosets of $(A_\pi)^f$ in $(A^f)_\pi$. For $p\in P$ choose an element $a_p\in A$ such that $f(a_p)=p$. We will prove that the family $(\pi(a_p))_{p\in P}$ is a representant system of cosets of $(A_f)^\pi$ in $(A^\pi)_f$.
For $y\in (A^\pi)_f$, then there is $x\in A$ such that $y=\pi(x)$, so $0=f(y)=f(\pi(x))=\pi(f(x))$, so $f(x)\in(A^f)_\pi$, so there is $p\in P$ such that $f(x)\in p+(A_\pi)^f$, so there is $b\in A_\pi$ such that $f(x)=f(a_p)+f(b)$, so $x-a_p-b\in A_f$, so $\pi(x)-\pi(a_p)-\pi(b)\in (A_f)^\pi$, so $y\in\pi(a_p)+(A_f)^\pi$.
If $\pi(a_p)+(A_f)^\pi=\pi(a_q)+(A_f)^\pi$, then $\pi(a_p)-\pi(a_q)\in(A_f)^\pi$, so there is $x\in A_f$ such that $\pi(a_p)-\pi(a_q)=\pi(x)$, so $a_p-a_q-x\in A_\pi$, so $f(a_p)-f(a_q)-f(x)\in(A_\pi)^f$, so $p-q\in(A_\pi)^f$, so $p+(A_\pi)^f=q+(A_\pi)^f$, so $p=q$.
