Lang's Algebra: Herbrand quotient

Question

I've looked around a lot and couldn't find much help (at least that I could understand) on this question - it is 1.45 in Lang's Algebra book:

Let $G$ be a cyclic group of order $n$, generated by $\sigma$. Assume $G$ acts on an abelian group $A$ as groups s.t. $\sigma(x+y) = \sigma(x)+\sigma(y)$ for $x,y \in A$, and let $f,g: A \to A$ be the homomorphisms defined as:

$$f(x) = \sigma\,x - x $$ and $$g(x) = x + \sigma\,x + \cdots + \sigma^{n-1}\,x$$

Herbrand quotient is given as $q(A) = (A_f: A^g)/(A_g:A^f)$, provided both indices are finite. And, $A_f$ and $A^f$ are the kernel and image of map $f$. Assume $B$ is a subgroup of $A$ s.t. $GB \subset B$. Then

a.) Define in a natural way an operation of $G$ on $A/B$

b.) Prove that $q(A) = q(B)\,q(A/B)$ Hint: consider complex: $E: 0 \to A_g \to A \overset{g}{\to} A \overset{f}{\to} A \overset{g}{\to} A^g \to 0$

Hint: $K(A): \cdots A_i \overset{d_i}{\to} A_{i+1} \overset{d_{i+1}}{\to} \cdots$ where $A_i = A$ for all $i$ and $d^i = f$ if $i$ is even and $d^i = g$ if $i$ is odd. Similarily consider $K(B)$ and $K(A/B)$. Examine long exact sequence on cohomology associated to the exact sequence of complexes $0 \to K(B) \to K(A) \to K(A/B) \to 0$. Keep in mind complexes $K$ are periodic so the long exact sequence will also be periodic, of the form $H^0(K(B)) \to H^0(K(A)) \to H^0(K(A/B)) \to H^1(K(B)) \to H^1(K(A)) \to H^1(K(A/B))$

c.)If $A$ is finite, then $q(A) = 1$.

So I fiddled around with $C_n$ groups and their subgroups to see the homomorphism work, and I can think of the isomorphism theorems - making me think of the quotient group - but I am stuck on the first part of the question.

I would appreciate any help with this!!

Answer 1

a) is trivial.

b) Let $K(A)$ be the following complex, where $A_i = A$ for all $i$ and $d^i = f$ if $i$ is even and $d^i = g$ if $i$ is odd.

$$\cdots\rightarrow A_i\rightarrow A_{i+1} \rightarrow\cdots$$

Similarly we define $K(B)$ and $K(A/B)$. Then there exits the following exact sequence of complexes.

$$0\rightarrow K(B) \rightarrow K(A) \rightarrow  K(A/B) \rightarrow 0$$

Let $H_i(A)$(resp. $H_i(B)$, $H_i(A/B)$) be the $i$-th homology group of $K(A)$(resp. $K(B)$, $K(A/B)$).

Then we get the following exact sequence.

$\cdots \rightarrow H_1(A/B) \rightarrow H_0(B) \rightarrow H_0(A) \rightarrow H_0(A/B) \rightarrow H_1(B) \rightarrow H_1(A) \rightarrow H_1(A/B) \rightarrow H_0(B) \rightarrow\cdots$

We denote $|H_0(A)|$ by $h_0(A)$. Similarly we define $h_1(A)$, $h_0(B)$, etc..

We denote by $m_0(A)$ the order of image of $H_0(B) \rightarrow H_0(A)$. Similarly we define $m_1(A)$, $m_0(B)$, etc..

Then

$h_0(B)/m_0(B) = m_0(A)$

$h_0(A)/m_0(A) = m_0(A/B)$

$h_0(A/B)/m_0(A/B) = m_1(B)$

$h_1(B)/m_1(B) = m_1(A)$

$h_1(A)/m_1(A) = m_1(A/B)$

$h_1(A/B)/m_1(A/B) = m_0(B)$

Hence

$h_0(A) = m_0(A)m_0(A/B)$

$h_1(A) = m_1(A)m_1(A/B)$

$h_0(B) = m_0(B)m_0(A)$

$h_1(B) = m_1(B)m_1(A)$

$h_0(A/B) = m_0(A/B)m_1(B)$

$h_1(A/B) = m_1(A/B)m_0(B)$

Hence

$$q(A) = \frac{h_0(A)}{h_1(A)} = \frac{m_0(A)m_0(A/B)}{m_1(A)m_1(A/B)}$$

$$q(B) = \frac{h_0(B)}{h_1(B)} = \frac{m_0(B)m_0(A)}{m_1(B)m_1(A)}$$

$$q(A/B) = \frac{h_0(A/B)}{h_1(A/B)} = \frac{m_0(A/B)m_1(B)}{m_1(A/B)m_0(B)}$$

Hence

$$q(A) = q(B)q(A/B)$$

c)

$$0 \subset A^g \subset A_f \subset A$$

$$0 \subset A^f \subset A_g \subset A$$

Hence

$|A| = [A:A_f][A_f : A^g]|A^g|$

$|A| = [A:A_g][A_g : A^f]|A^f|$

Since $[A : A_f] = |A^f|$ and $[A : A_g] = |A^g|$, $[A_f : A^g] = [A_g : A^f]$. Hence $q(A) = 1$.

Answer 2

This should be very standard: To define an action of $G$ on $A/B$, you need to specify $g(aB)$ for $g\in G$ and cosets $a+B\in A/B$. The obvious way is to set $g(a+B)=g(a)+B$, but you need to check that this is well-defined: If $a+B=a'+B$, then $g(a)-g(a')=g(a-a')\in g(B)\subseteq B$, hence $g(a)+B=g(a')+B$ as desired.