This is an exercise from Field and Galois theory by Patrick Morandi. This determines the second cohomology group of a cyclic group $G$.

Let $M$ be a Abelian group and $G$ be cyclic group of order $n$. Let $M$ be a $G$-module. Suppose $G=\langle\sigma\rangle.$ Let $M^{G} =\{\ m \in M\mid\sigma m=m \}$. Define the norm map $N: M\rightarrow M^{G}$ by $N(m)=\sum_{i=0}^{n-1} \sigma^{i}m$. We want to prove $H^{2}(G,M) \cong M^{G}/im(N)$. It asks to prove this result in the following steps:

a) If $m\in M^{G}$ Let $f_{m}$ be the cochain given by $$f_{m}(\sigma^{i}, \sigma^{j})=\left\{\begin{array}{cc} 1, & \quad \text{ if $i+j < n$}\\ m, & \quad \text{ if $i+j \geq n$} \end{array} \right. $$ Prove that $f_{m}$ is a cocycle.

b) Suppose $f_m$ and $f_n$ are cocycles that are cohomologous. Then there are $c_{i} \in M$ with $f_{m}(\sigma^{i}, \sigma^{j})=f_{n}(\sigma^{i}, \sigma^{j}).c_{i}\sigma^{i}(c_{j})c_{i+j}^{-1}$, where we are writing $c_{i}$ for $c_{\sigma^{i}}$. Show that $m-n=N(c_{1})$.

c) Prove that a cocycle $f \in Z^{2}(G,M)$ is cohomologous to $f_m$, where $m=\sum_{i=0}^{n-1}f(\sigma^{i}, \sigma)$.

d) Conclude from these steps that the map $m \rightarrow f_{m}$ induces an isomorphism $M^{G}/im(N) \cong H^{2}(G,M)$.

My attempt: I have tried to use the cocycle condition for the first one , that is $f_m$ is 2-cocycle if for all $\sigma^{i}, \sigma^{j}, \sigma^{k}$ we have $\sigma^{i}f_m(\sigma^{j},\sigma^{k})- f_{m}(\sigma^{i+j},\sigma^{k})+f_{m}(\sigma^{i}, \sigma^{j+k})-f_{m}(\sigma^{i}, \sigma^{j})=0$. But I have tried to do it by considering certain cases like when

1) $i+j+k\leq n$ in which case the relation is satisfied.

2) $i+j+k \geq n$ but $i+j \leq n$ and $j+k \leq n$ in which case also the relation is satisfied.

3) but for the cases that $i+j+k \geq n$ but $i+j \leq n$ and $j+k \geq n$ or, $i+j+k \geq n$ but $i+j \geq n$ and $j+k \leq n$ it doesn't seem to work.

Also assuming the a) I have done the second part about finding $c_{i}$'s ( which is just by definition) but I haven't been able to show that $m-n=N(c_{1})$. Also I haven't been able to proceed with c) and d).Any kind of help will be appreciated.

Thanks in advance!

  • $\begingroup$ Alternatively, use abstract nonsense and the fact that $K(\mathbb{Z}/n, 1)$ is an infinite-dimensional lens space. $\endgroup$
    – anomaly
    Apr 21, 2017 at 17:02
  • $\begingroup$ How to prove part (c)? $\endgroup$
    – m-agag2016
    Mar 8, 2018 at 8:47

1 Answer 1


The cocycle $f_m$ is essentially the celebrated "carry digit" cocycle.

Let's analyse this in detail. As $m$ is fixed under $G$ the cocycle condition you need to check is just $$f_m(\sigma^j,\sigma^k)-f_m(\sigma^{i+j},\sigma^k) +f_m(\sigma^i,\sigma^{j+k})-f_m(\sigma^i,\sigma^j)=0.$$ Write $f_m(\sigma^i,\sigma^j)=F(i,j)m$ for $0\le i,j<n$. Then the cocycle condition is now $$F(j,k)-F(i+_n j,k)+F(i,j+_nk)-F(i,j)=0.$$ We then have $F(i,j)=0$ if $i+j<n$ and $F(i,j)\ge n$ if $i+j\ge n$ (recall $i,j\in\{0,1,\ldots,n-1\}$). Here I write $i+_nj$ to emphasise that this addition is modulo $n$. Thus (as we are assuming $i$, $j\in\{0,1,\ldots,n-1\}$) $$i+_nj=i+j-nF(i,j).$$ Now $$i+(j+k)=i+(j+_nk)+nF(j,k) =i+_n(j+_nk)+nF(j,k)+nF(i,j+_nk)$$ and $$(i+j)+k)=(i+_nj)+_k+nF(i,j) =(i+_nj)+_nk+nF(i,j)+nF(i+_nj,k).$$ As both $+$ and $+_n$ are associative, the subtracting these identities gives the cocycle condition.

  • $\begingroup$ what is the meaning of $"i+_{n}j=i+j-nF(i,j)$?? I didn't get it $\endgroup$
    – Riju
    Apr 21, 2017 at 16:05
  • $\begingroup$ @Riju Let's take an example, $n=10$ say. Then for instance $6+_{10}8=4$ and $F(6,8)=1$ so $6+_{10}8=6+8-10F(6,8)$ etc, $\endgroup$ Apr 21, 2017 at 16:08
  • $\begingroup$ $F(6,8)=1$ meaning??? M is an abelian group so let us say just say that 0 is the identity(although in the question 1 is taken to be the identity). Anyway that doesn't matter. but by your notation $F(i,j)$ must be $f_{0}(\sigma^{i},\sigma^{j})$ or $f_{1}(\sigma^{i},\sigma^{j})$ (whatever you take the identity notation to be in M). In the example that you have given it will mean it is 1 (or 0), that is the identity of M but how are you multiplying n with 1(or 0)? $\endgroup$
    – Riju
    Apr 21, 2017 at 16:25
  • $\begingroup$ @Riju I have added a bit extra: I hop it is more clear now. $\endgroup$ Apr 21, 2017 at 16:28
  • $\begingroup$ what I don't understand is F(i,j) where from this map comes?? $\endgroup$
    – Riju
    Apr 21, 2017 at 16:38

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