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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.

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  • $\begingroup$ If $G$ acts trivially on $M$, then $H^1(G, M) = \operatorname{Hom}(G, M)$. The proof is just unwinding the construction of $H^*$. $\endgroup$
    – anomaly
    Feb 12, 2017 at 20:06
  • $\begingroup$ I deleted my answer ($H^1=$ crossed homomorphisms / boundaries ), as it probably was not what you were looking for... $\endgroup$
    – peter a g
    Feb 12, 2017 at 20:25
  • $\begingroup$ Sorry, I didn't see your answer, but I'd like to see an interpretation in terms of quotients of crossed homomorphisms. Why did you delete it? $\endgroup$
    – adrija
    Feb 12, 2017 at 20:32

2 Answers 2

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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 $xf(y)+f(x)=xya+xa=(xy+x)a=a=(xy)a=x(ya)=(xa)(ya)=f(x)f(y)\Rightarrow f(xy)=f(x)f(y)$

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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.)

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  • $\begingroup$ I had deleted the answer, as it pretty much was "it is because it is," and not why the cocycle definition matches up with other definitions or interpretations. Still, here it is back.. $\endgroup$
    – peter a g
    Feb 12, 2017 at 20:39

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