2

Suppose G is cyclic of finite order n with generator t. Then we have the resolution: $$\cdots \overset{N}{\rightarrow} \mathbb{Z}G \overset{t-1}{\rightarrow} \mathbb{Z}G \overset{N}{\rightarrow} \mathbb{Z}G \overset{t-1}{\rightarrow} \mathbb{Z}G \rightarrow \mathbb{Z} \rightarrow 0 $$

where $N = 1 + t + t^2 + \cdots + t^{i} + \cdots + t^{n-1}$.

And we have that $N$ induces the normed map $\bar{N}: M_{G} \rightarrow M^{G}$.

I am trying to understand why $H_{i}(G,M) = H^{i+1}(G,M) = coker \bar{N}$ for $i$ odd $(i \geq 1)$ and $H_{i}(G,M) = H^{i-1}(G,M) = ker \bar{N}$ for $i$ even $(i \geq 2)$.

I know it shouldn't be hard but... I don't know if I'm looking to the right morphisms, for example, I tried for $i = 1$, then

$H_{1} (G,M) = \dfrac{ker (t-1)}{im (t)} = \dfrac{\mathbb{Z}G \otimes M}{im (t)} = \dfrac{M}{im (t)}$ $(?)$

And for the other ($i$ even, or at least $2$) one I don't even know how to see the kernel.

The example is at Kenneth Brown books, Cohomology of Groups, page 58.

Luiza Sorice
  • 65

1 Answers1

2

To compute, say the groups $H_i(G,M)$ you take this truncated sequence $$\cdots\rightarrow \mathbb{Z}G \overset{t-1}{\rightarrow} \mathbb{Z}G \overset{N}{\rightarrow} \mathbb{Z}G \overset{t-1}{\rightarrow} \mathbb{Z}G \rightarrow 0$$ then apply the functor $\cdots\otimes_G M$ and take the homology of the new sequence. This new sequence is $$\cdots\rightarrow M\overset{t-1}{\rightarrow}M \overset{N}{\to}M \overset{t-1}{\rightarrow}M\rightarrow 0.$$

At an odd $i$, then $H_i(G,M)$ is then the kernel of $t-1:M\to M$. The kernel of $t-1$ is $M^G$, the fixed points of $M$ on $G$, so for odd $i$ $$H_i(G,M)\cong\frac{M^G}{N(M)}.$$ The image of $N:M\to M$ is contained in $M^G$, so one can think of $N$ as a map from $M$ to $M^G$. Likewise its kernel contains $(t-1)M$ so one can think of $M$ as a map from $M_G=M/(t-1)M$ to $M^G$. Thus $$H_i(G,M)\cong\frac{M^G}{N(M_G)}=\textrm{coker}(N:M_G\to M^G).$$

One can do similar with even $i>0$ and also with cohomology $H^i(G,M)$.

Angina Seng
  • 158,341