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Let $G$ be a group.

1- What is the kerner of (non-faithful) transitive group action (since the stabilizer is a subgroup of G and the kernel is normal subgroup of G and there exist only one stabilizer because of transitive action and from wikipedia " the kerner is the intersection of all stabilizers")?

2- if a group action is faithful transitive action then the kernel is {e} and the stabilizer is a non trivial subgroup ? .we know that the kerner is the intersection of all stabilizers and the stabilizer ( the only one ) is a non trivial subgroup how could that the kerner be trivial , and if we say that the stabilizer = kernel , then the stabilizer is a normal subgroup of G ? Example : SO(3) faithful transitive action on 2-sphere with stabilizer SO(2) ( not normal subgroup of SO(2) )

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Let $G$ be a group acting on $X$ we have a general way to express the kernel of the action in terms of stabilizers:

1st fact $$Ker(action)=\bigcap_{x\in X}Stab(x) $$

If the action is transitive for any $x,y$, there is $g$ such that $g\cdot y=x$.

2nd fact, in this case for $x\in X$, $$Ker(action)=\bigcap_{g\in G}Stab(g\cdot x) $$

Now we can express $Stab(g\cdot x)$ differently:

3rd fact, $$Stab(g\cdot x)=g Stab(x) g^{-1}$$

Using these facts, we have a nice expression for the kernel of the action. A few remarks about what you are writing. You are wrong to say that there is only one stabilizer. There is only one stabilizer $\textbf{up to conjugation}$ if the action is transitive (this is the 3rd fact).

If I were you I would not talk about 'the' stabilizer. Remark also that to avoid doing this mistake you should take the habits to always talk about the stabilizer of some $x\in X$.

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  • $\begingroup$ Thank you very much . You are right ( one stabilizer up to conjugation, since stabilizers of all points g.x in the orbit of x form what's called the "the class of conjugate subgroups " or "conjugate class of subgroup H") $\endgroup$ – Mike Alex Jul 3 '18 at 14:08

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