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 apologize if this is a duplicate. I don't know enough about group cohomology to know if this is just a special case of an earlier post with the same title.

Let $G=\langle\sigma\rangle$ where $\sigma^m=1$. Let $N=1+\sigma+\sigma^2+\cdots+\sigma^{m-1}$. Then it is claimed in Dummit and Foote that $$\cdots\mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \cdots \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;\text{aug}\;} \mathbb{Z} \longrightarrow 0$$ is a free resolution of the trivial $G$-module $\mathbb{Z}$. Here $\mathbb{Z} G$ is the group ring and $\text{aug}$ is the augmentation map which sums coefficients. It's clear that $N( \sigma -1) = 0$ so that the composition of consecutive maps is zero. But I can't see why the kernel of a map should be contained in the image of the previous map. any suggestions would be greatly appreciated. Thanks for your time.

share|cite|improve this question
for some reason my computer isn't loading the type face so I apologize if there are egregious typos. – Rory Pulvino May 22 '12 at 21:36
This follows from a direct computation, really: you pick an element of the kernel, and look at it enough time until you see that it must have a certain form which implies it is in the image. – Mariano Suárez-Alvarez May 22 '12 at 23:01
yeah yeah... I've already slapped my forehead :) – Rory Pulvino May 22 '12 at 23:30
up vote 4 down vote accepted

As $(\sigma-1)(c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1})=(c_n-c_0)+(c_0-c_1)\sigma+\dots (c_{n-2}-c_{n-1})\sigma^{n-1}$, the element $a=c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1}$ is in the kernel of $\sigma-1$ iff all $c_i$'s are equal, i.e. iff $a=Nc$ for some $c\in\mathbb{Z}$. Similarly, $Na=(\sum c_i)N$, so here the kernel is given by the condition $\sum c_i=0$, but this means $a=(\sigma-1)(-c_0-(c_0+c_1)\sigma-(c_0+c_1+c_2)\sigma^2-\cdots)$.

share|cite|improve this answer
Fantastic! Thanks so much. – Rory Pulvino May 22 '12 at 22:24

I just wanted to elaborate a bit on minu's excellent answer.

Suppose that $\alpha=\sum_{i=0}^{m-1}a_i\sigma^i$ is in the kernel of $\sigma-1$. Then $$\textstyle(\sigma-1)(\alpha)=\sum\limits_{i=0}^{m-1}a_{i}(\sigma^{i+1}-\sigma^i)=(a_{m-1}-a_0)+(a_0-a_1)\sigma+\cdots+(a_{m-2}-a_{m-1})\sigma^{m-1}=0$$ so that $a_{i-1}-a_i=0$ for all $i$, and so all the $a_i$ are equal. Therefore, $$\alpha=a\left(\textstyle\sum\limits_{i=0}^{m-1}\sigma^i\right)=N(a)$$ where $a=a_0=a_1=\cdots=a_{m-1}$. Thus, $\ker(\sigma-1)\subseteq\operatorname{im}(N)$.

Now suppose that $\alpha=\sum_{i=0}^{m-1}a_i\sigma^i$ is in the kernel of $N$. Then $$N(\alpha)=\textstyle\sum\limits_{i=0}^{m-1}a_i\left(\sum\limits_{j=0}^{m-1}\sigma^{i+j}\right)=\sum\limits_{i=0}^{m-1}a_i\left(\sum\limits_{j=0}^{m-1}\sigma^{j}\right)=\left(\sum\limits_{i=0}^{m-1}a_i\right)\left(\sum\limits_{j=0}^{m-1}\sigma^{j}\right)=0$$ which implies that $\sum_{i=0}^{m-1}a_i=0$, and therefore $$\textstyle(\sigma-1)\left(\sum\limits_{i=0}^{m-1}\left(-\sum\limits_{j=0}^ia_j\right)\sigma^i\right)=\left(a_0-\sum\limits_{j=0}^{m-1}a_j\right)+\sum\limits_{i=1}^{m-1}\left(\sum\limits_{j=0}^{i}a_j-\sum\limits_{j=0}^{i-1}a_j\right)\sigma^i$$ $$=(a_0-0)+\textstyle\sum\limits_{i=1}^{m-1}a_i\sigma^i=\alpha.$$ Thus, $\ker(N)\subseteq\operatorname{im}(\sigma-1)$.

share|cite|improve this answer
Thanks for taking the time to spell it out, Zev – Rory Pulvino May 22 '12 at 23:34

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.