The first cohomology group $H^1(G,\mathbb{Z})$ for $G$ finite

Question

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?

Answer 1

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

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