I have been reading about cohomology of groups. For example, consider the cohomology group $H^{1}(Z/2,Z_-)$ of Wikipedia article https://en.wikipedia.org/wiki/Group_cohomology. Here $Z/2= \{1,-1\}$ acts on $Z=\{n\}$ as, $-1.n = -n$. Wikipedia computes $H^{1}(Z/2,Z_-)$ as the group of crossed homomorphisms $\{f: Z/2\to Z\}$ satisfying $f(ab)= f(a)+ af(b)$ modulo the group of principal crossed homomorphisms $f:Z/2 \to Z; f(a)= an-n$ for some integer $n$. It is obvious from this definition that $H^{1}(Z/2,Z_-)=Z/2$.

However, the same article says that the cohomology group $H^1(G,A)$ can also be computed in the following way. One considers an exact sequence of Abelian groups $0\to A\to B \to C\to 0$ with some specified action of $G$ on $A,B$ and $C$. Then one sends each group to the group of invariants (under $G$) $A^G, B^G, C^G$ obtaining the long exact sequence $0\to A^G \to B^G\to C^G \to H^1(G,A) \to H^1(G,B)\to H^1(G,C)\to \cdots$.

I want to apply the same procedure for computing $H^{1}(Z/2,Z_-)$. In order to do that I consider the exact sequence $0\to Z \to^{d_1} Z\times Z\to^{d_2} Z\to 0$ with $d_1(n)=(n,0)$ and $d_2(n_1,n_2)=(0,n_2)$. Here $Z/2=\{1,-1\}$ acts on the left most $Z$ as $-1.n=-n$, on $Z \times Z$ as $-1.(n_1,n_2)=(-n_1,n_2)$ and trivially on the right most $Z$. Now we send each group to the group of invariants giving the exact sequence $0\to \{0\} \to \{0,n\} \to Z\to 0$. So $H^{1}(Z/2,Z_-)=0$ instead of $Z/2$.

It seems that I have misunderstood something here. It will be helpful if someone can clarify this.


1 Answer 1


What you're missing is that the exactness of your sequence does not imply that $H^1(\mathbb Z/2, \mathbb Z_-) = 0$ : indeed in the long exact sequence you have other groups on the right.

In fact, your sequence in some sense cannot be useful to compute any cohomology, since it is split (of the form $0\to A\to A\oplus B\to B\to 0$ with the canonical maps), therefore the long exact sequence will also split into a bunch of short exact sequences of the form $0\to H^i(G,A)\to H^i(G,A\oplus B)\to H^i(G,B)\to 0$; connected by the zero morphism.

So what's happening here is that you have a long exact sequence $0\to 0\to \mathbb Z\to \mathbb Z \overset{0}\to H^1(\mathbb Z/2, \mathbb Z_-) \to H^1(\mathbb Z/2, \mathbb {Z_-\oplus Z})\to ...$ which can be decomposed as your short exact sequence, followed by $0\to H^1(\mathbb Z/2,\mathbb Z_-)\to ...$

  • $\begingroup$ Maybe the exact sequence $0 \to Z\to R \to R/Z\to 0$, where $Z/2$ acts on $R$ as $r \to -r$, will help since it does not split. $\endgroup$ Oct 27, 2019 at 3:32
  • $\begingroup$ Since 'sending' each group to the group of invariants gives the exact sequence $0\to 0 \to 0\to Z/2 \to H^{1}(Z/2,Z) \to \cdots$. It is obvious that $H^{1}(Z/2,Z)$ must contain $Z/2$ as a subgroup. What tells us that it has to be $Z/2$? $\endgroup$ Oct 27, 2019 at 4:12
  • 1
    $\begingroup$ Well for that you would have to compute $H^1(\mathbb Z/2, \mathbb R)$. There's a general argument that shows that it's $0$ : that's because it is killed by $2=|\mathbb Z/2|$, but at the same time $2$ is invertible because it is so in $\mathbb R$ $\endgroup$ Oct 27, 2019 at 10:26

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