Interpretations of the first cohomology group

Question

I've been revisiting group cohomology, and I realized that there is something I never quite understood. Let $G$ be a finite group, and let $A$ be a $G$-module (i.e. $\mathbb{Z}[G]$-module). Then the second cohomology $H^2(G,A)$ classifies group extensions $1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$ such that the action of $G$ on $A$ by left conjugation jibes with the given action of $G$ on $A$.

The interpretation of the first cohomology group, however, I still find somewhat elusive. On the one hand, there is the following interpretation: $H^1(G,A)$ is isomorphic to $\operatorname{Aut} ( 1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1) / \sim$ (where $\sim$ is conjugation by an element of $A$, and where $\operatorname{Aut}$ is taken in the category of group extensions of $G$ by $A$) by $\langle \rangle \mapsto (e \mapsto \langle f(e)\rangle e)$, where $f$ is the map $E\rightarrow G$, and where $\langle \rangle$ is a $1$-cocycle. This is indeed an isomorphism.

However, there is another interpretation! Some places say that there is a bijection between $H^1(G,A)$ and sections of $f:E\rightarrow G$ up to conjugation by an element of $A$! Here the problem is that I can't quite decipher if this should mean set-theoretic sections or group-theoretic sections.

Here's how I'm thinking about this: you can let $\operatorname{Aut}(E)/\sim$ act on the set of set-theoretic sections of $f$ by $\phi$ acts on $\lambda$ taking it to the section $\phi\circ \lambda$. It seems like this action is free. Furthermore, it seems like it is transitive. Indeed, if you want to take $\lambda$ to $\lambda'$ you can use the automorphism $e\mapsto \lambda'(f(e))\lambda^{-1}(f(e))e$ ($e\mapsto \lambda'(f(e))\lambda^{-1}(f(e))$ is a $1$-cocycle). So it seems like this gives a (non-canonical) bijection between $H^1(G,A)$ and the set-theoretic sections. But something is wrong -- this action takes the group-theoretic sections to group-theoretic sections! So it can't act freely and transitively on the set of set-theoretic sections...

I'm not quite sure what the resolution of this is since most sources don't distinguish between group-theoretic sections and set-theoretic sections.

Also, if you can inform me a way to think about this that relates these two interpretations with the interpretation that the first cohomology classifies torsors, I would be very happy indeed!

Answer 1

My standard way to think about $H^1(G,A)$ is as functions $G \to A$ such that $f(gh)= g \cdot f(h) + f(g)$ ("cocycles"), modulo functions of the form $g \mapsto (g-1)\cdot a$ for some $a \in A$ ("coboundaries"). This is what you get by working with the bar resolution.

In answer to your question about sections, it's group homomorphisms that you need. You can choose the semidirect product $A \rtimes G$ as $E$. Then $H^1(G,A)$ as defined above is in bijection with the group homomorphisms $G \to A \rtimes G $ that split the projection $A\rtimes G\to G$, up to the action by conjugation of $A \subset A \rtimes G$. Any such map is of the form $g \mapsto (\alpha(g), g)$ where $\alpha: G \to A$; you can see that $\alpha$ must be a cocycle if the map is to be a group homomorphism, and that conjugation by $A$ affects $\alpha$ by a coboundary $g \mapsto (g-1)\cdot a$.

Returning to the sequence definition, an automorphism of your sequence (again with $E = A \rtimes G$) is really a group homomorphism $E \to E$ that induces the identity on $A$ and on $A \rtimes G / A \cong G$. Thus you can write it as $(a,g) \mapsto (a \alpha(g), g)$, where $\alpha : G \to A$. This map being a group homomorphism forces $\alpha$ to be a cocycle (compute the image of $(1,gh)$, compare to the product of the images of $(1,g)$ and $(1,h)$), and that conjugation by elements of $A$ changes $\alpha$ by a coboundary. So the sequence definition agrees with the `standard' one. This shows that your two definitions are equivalent.

I don't know the definition of a torsor I'm afraid so I can't help with your last question. Could you say what it means?