Question about action on groups in Bourbaki (Algebra I) In Bourbaki, Algebra I, chapter I, §5 "Groups operating on a set" paragraph 1, Bourbaki defines the operation of a group $G$ on a set $E$ as a morphism $\alpha \in G\mapsto f_\alpha \in S(E)$ ($S(E)$ is the group of permutations of $E$). Then he defines a faithfull operation asking the morphisme to be injective.
I call $f$ the morphism above, and I define $\varphi_x:G\to E$, $\alpha \mapsto \varphi_x(\alpha)=f(\alpha)(x)=f_\alpha(x)$.
In paragraph 4, Bourbaki calls orbital mapping defined by $x$ the map $\alpha\in G\mapsto \alpha x\in E$ , which is exactly, according to me, the map $\varphi_x$: correct ?
I claim that the faithfulness of the operation is equivalent to the injectivity of all $\varphi_x$ (to my mind, suffice to write the definitions wich only imply "for all"): correct ?
Then in the same paragraph, Bourbaki defines a "free" operation writing that it's an operation where all orbital maps are injective... 
$\Rightarrow$ For me, this definition is equivalent to the faithfulness of the operation, according to what I claimed above.
This is not a problem in itself, everyone can give two different names for the same thing (especially in maths...). But the thing is that in paragraph 6, he defines a "simply transitive" operation and asserts that this definition amounts to the same to say that the operation is free and transitive (NB: there is no problem about the notion of transitivity). And in the example 2) on next page, he claims something about faithfulness wich shows that this notion is different from that of a free operation.
According to other texts I read, all this would be correct if the definition of a free operation was replaced by the "traditional" one can find in many books: an operation is free when for all $(x,y)\in E^2$, there exists at most one $\alpha\in G$ such that $y=gx$.
Can anyone tell me where I am wrong?
 A: *

*Yes, the orbital mapping is your $\phi_x$

*Faithful does not mean that all $\phi_x$ are injective. For example $G$ acts naturally on itself by left multiplication and we can extand this action to an action on $G\cup\{\infty\}$ (where $\infty$ is an arbitrary object $\notin G$) by declaring $\alpha \infty=\infty$ for all $\alpha \in G$. Then this action is faithful because $f_\alpha=f_\beta$ implies $\alpha=f_\alpha(1)=f_\beta(1)=\beta$, i.e., the map $\alpha\mapsto f_\alpha$ is injective. Nevertheless, the orbit of $\infty$ isjust $\{\infty\}$, hence $\phi_\infty$ is constant. So unless $G$ is trivial, it fails to be injective.
In a faithful action, no group element (except $1$) leaves everything fix. In a free action, no group element (excep $1$) leaves anything fix.

*"All orbital maps are injective" is equivalent to: For all $x$ we  have that $g_1x= g_2x$ implies $g_1=g_2$. And this is equivalent to: For all $x$ and $y$ there exists at most one $g$ with $y=gx$.
A: In response to the question of Hoot, here is a detailed proof of: "commutative + (transitive + faithful) $\Rightarrow$ transitive + free".
Let $f$ be the morphism associated with the operation, and $(g,g')\in G^2$ such that $gx=g'x$, which is the same than $x={g'}^{-1} gx$. The idea is to show that $(*)$ $f({g'}^{-1} g)=f(e)$ and conclude that $g=g'$ using the injectivity of $f$ thanks to the faithfulness of the operation. The equality $(*)$ is equivalent to: for all $z\in E$, $f({g'}^{-1} g)(z)=f(e)(z)$, or $({g'}^{-1} g)z=ez=z$. Hence, suffice to prove the latter equality.
But by transitivity, there exists at least an $h\in G$ such that $x=hz$. Then we get, using here the commutativity of $G$: $({g'}^{-1} g)z = {g'}^{-1} g(h^{-1} h)z = {g'}^{-1} g h^{-1}(hz)=h^{-1}({g'}^{-1} gx)=h^{-1} x=z$, which permits to conclude.
