I tried to understand the answer to the question here, which is about why $H^n(G,U(1))=H^{n+1}(G,\mathbb{Z})$. I don't quite understand what the notation "$b_1(G)=k$" means and why we will "get a factor $U(1)^k$". Can someone help me understand this?
Probably it is also related to the fact that $H^n(G,\mathbb{R})=0$ for $n>0$ with trivial group action. Since if it vanishes, then by the long exact sequence $$...\rightarrow H^n(G,\mathbb{R})\rightarrow H^n(G,U(1))\rightarrow H^n(G,\mathbb{Z})\rightarrow H^n(G,\mathbb{R})\rightarrow ...$$we can get the desired isomorphism. However, I have no idea how to prove $H^n(G,\mathbb{R})=0$ with $n>0$ and $G$ a finite group. I tried to use the universal coefficient theorem but in vain.
Thanks in advance!
If $|G|=m$ and $n>0$ then $H^n(G,A)$ is $m$-torsion for all $G$-modules $A$, that is $m H^n(G,A)=0$. But $\Bbb R$ is divisible, so $mH^n(G,\Bbb R)=H^n(G,m\Bbb R)=H^n(G,\Bbb R)$. Therefore $H^n(G,\Bbb R)=0$.