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?