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Let A be a G-group, i.e. a set on which G acts on, has a group structure and satisfies $^s(xy)=^s x ^s y$ for all $x,y \in A \ , s \in G$. A homogeneous principal space P is a non-empty G-set on which A acts on the right, in a manner competible with G, and satisfy the following: $\forall x,y \in P : \exists ! a \in A : \ y = x \bullet a$ (i.e. there is a unique such $a \in A$).

A cocycle in $Z^1(G,A)$ is a map from G to A , $ s \mapsto a_s $ which is continuous and such that $a_{st} = a_s \ ^s a_t$ for all $s,t \in G$. Cocycle $a,b$ are cohomologous if there is $c \in A$ such that $b=c^{-1} a_s \ ^s c$. The set of these classes of cocycles is $H^1(G,A)$.

After these prequisties, I present the question. A theorem says that:

Let A be a G-group. There is a bijection between the set of classes of principal homogeneous over A (which we denote $P(A)$ and the set $H^1(G,S)$.

Serre's proof is as follows, he defines a map $\lambda : P(A) \to H^1(G,A)$ in the following way: For $P \in P(A)$ choose $x \in P$. If $s \in G$ then $^sx \in P$ and therefore there exists $a_s \in A$ such that $^sx = x \bullet a_s$. The map $s \mapsto a_s$ defined by choosing this x defines a cocycle, and one can check that by substituing $x \bullet b$ one get a cohomologeous cycle and thus $\lambda (P)$ is defined as the class of $a_s$.

The following isn't clear to me: Now, we want to do the opposite, define a map $\mu : H^1(G,A) \to P(A)$. If $a_s \in Z^1(G,A)$ one denotes by $P_a$ the group A on which G acts by the following "twisted" formula: $x \star s =?= \ ^{s'}x = a_s \cdot ^sx$. A is given, how then we can build another group and call it A? I think that the formula defines a new action of G on $P_a$ (which I don't understand what is it and where it lies, I think but not sure that $P_a \subset A$). Then he says that if we let A act on the right on $P_a$ by translation - I don't to what action he means - then one obtain a principal homogeneous space (need to prove that). Then $\mu$ will map $a_s$ to this space and cohomologeous cocycles to isomorphic spaces.

The final part of the proof is to check that $\lambda \circ \mu = id$ and $\mu \circ \lambda = id$.

Thanks in advance, Zachi Evenor

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up vote 2 down vote accepted

As far as I can see, $P_a$ should be the set $A$ (i.e., the underlying set of the given group $A$), considered not as a group but as a set with an action of $G$, namely the action in the "twisted" formula. In addition to this action of $G$, there is also an action of the group $A$ on the set $P_a$, namely right translation. In other words, view the multiplication operation $m:A\times A\to A$ of $A$ as a function $P_a\times A\to P_a$ (since $P_a$ and $A$ are the same as sets). The claim is then (if I understand correctly) that $P_a$ with these actions of $G$ and of $A$ is the desired element of $P(A)$.

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