Why is the first cohomology group of a group $G$ with coefficients in $\mathbb Q/\mathbb Z$ the group $Hom(G,\mathbb Q/\mathbb Z)$?

Question

I've been reading some notes on group cohomology and there are repeated mentions of the following:

For a group $G$, $H^1(G,\mathbb Q/\mathbb Z)=Hom(G,\mathbb Q/\mathbb Z)$. It may be very easy, but can anyone please point out to me why that is true? I also noted that the notes deals with finite groups $G$, but I think this may be true for all groups. The interpretation of first cohomology group I'm considering is the one discusses here: Interpretations of the first cohomology group. But I'm still missing the point why this is true.

Answer 1

In general: If $A$ is a $G$- module then $$Z^1(G,A)=\{f:G\to A; f(xy)=xf(y)+f(x) \}$$ $$B^1(G,A)=\{f:G\to A; f(x)=xa-a; a\in A \}$$

Let $A$ be a trivial $G$- module(i.e: $ax=a;\forall a\in A, x \in G$), then $B^1(G,A)=\{f:G\to A; f(x)=0; a\in A \}=0$ and $Z^1(G,A)=Hom(G,A)$, because $xf(y)+f(x)=xya+xa=(xy+x)a=a=(xy)a=x(ya)=(xa)(ya)=f(x)f(y)\Rightarrow f(xy)=f(x)f(y)$

Answer 2

Write down the definition of a $1$-cocycle: $$f(\sigma \tau) = \sigma f(\tau) + f(\sigma).$$ As the action of $G$ on $\mathbb Q/ \mathbb Z$ is trivial, the above means that a $1$-cocycle is exactly a homomorphism: $$f(\sigma \tau) = f(\tau) + f(\sigma).$$ Similarly, co-boundaries are identically equal to $0$.

(This is anomaly's comment above, of course.)