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I want to compute the first cohomology group $H^1(G,\mathbb{Z})$ for $G$ finite.

Here is what I have got so far:

If $G$ has odd order, $G$ has to act on $\mathbb{Z}$ trivially. Then $H^1(G,\mathbb{Z})=\operatorname{Hom}(G,\mathbb{Z})$. And $\operatorname{Hom}(G,\mathbb{Z})$ is trivial (right?).

If $G$ has even order, then $G$ can either act trivially on $\mathbb{Z}$ or $G$ acting on $\mathbb{Z}$ by switching the generators $1$ and $-1$. If $G$ acts trivially on $\mathbb Z$, then again, $H^1(G,\mathbb{Z})=\operatorname{Hom}(G,\mathbb{Z})$ trivial.

How to compute $H^1(G,\mathbb{Z})$ if $G$ acts on $\mathbb{Z}$ nontrivially?

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  • $\begingroup$ @PseudoNeo : Hm. I must say I haven't done any cohomology yet and I thought that $H^1(G,\mathbb Z)$ was always equal to $\mathrm{Hom}(G,\mathbb Z)$. Of course I deleted my answer immediately. Let me think about this. $\endgroup$
    – Patrick Da Silva
    Oct 14, 2014 at 12:52

1 Answer 1

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$f\in {\rm Hom}\ (G,{\bf Z})$ : $$ f(gh)=f(g)+f(h) $$

That is, $$ f(1)=2f(1)\Rightarrow f(1)=0$$

If $g$ has order $n$ then $$ 0=f(1)=f(g^n)=nf(g) \Rightarrow f(g)=0$$

Hence ${\rm Hom}\ (G,{\bf Z})=0$

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