Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!!

share|cite|improve this question
up vote 2 down vote accepted

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..


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

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

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

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

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

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


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

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

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

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

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

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


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

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

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


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


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

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


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

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

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

share|cite|improve this answer
Firstly, thank you! I was hoping that maybe you could explain a little more in the steps. I am just learning about complexes and chains and, while I have yet to digest all of this, I am afraid I may not be able to understand it... For instance, could you explain a little bit more about the use of the complex E in the hint, especially with respect to the kernels and images of the maps? Thank you again! – nate Sep 14 '12 at 3:50
@nate Do you understand that there exits the following exact sequence of complexes? $0\rightarrow K(B) \rightarrow K(A) \rightarrow K(A/B) \rightarrow 0$ – Makoto Kato Sep 14 '12 at 4:00
Hi, I read about complex chains and found a link onsite,‌​x?rq=1, and even though I'm learning dealing with cochains, I was able to follow it and do it myself. Now to try your question.... – nate Sep 14 '12 at 5:27
@nate Chapter 20 of Lang's algebra treats chain complexes and their homologies. I think it gives you enough knowledge to understand my proof. – Makoto Kato Sep 14 '12 at 5:31
Okay, well thanks a lot for your help - I'll give the Lang book another try (it is sort of terse for me!) :) – nate Sep 14 '12 at 6:29

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.

share|cite|improve this answer
Okay, I understand the action definition - was pretty standard looking. Would anyone happen to have class notes / links of something to study? Not a tome or in-depth book at first ;) Thanks! – nate Sep 14 '12 at 1:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.