[Math] Stabilizer, Kernel and Orbit of the right action.

abstract-algebragroup-actionsgroup-theory

If $G$ act on the set of all right cosets of a subgroup $K$ then I have the following questions:-

  1. What is the stabilizer of an element $Kx$.
  2. What is the kernel of the Action.
  3. What is the orbit of $Kx$.

My attempt ( I know its short):-

  1. $Stb_G(Kx)=\{g\in G : Kxg=Kx\}$.
  2. $Ker = \{g\in G : Kxg=Kx \, \, \mbox{for all} \, Kx \}$ also I know the Ker is intersection of all stabilizer but since the stabilizer is not nice not sure if useful to use it.
  3. Okay the action is transitive, but not sure why.

Best Answer

For 1, it follows from what you wrote that $xg=k'x$ for some $k'\in K$, so $g=x^{-1}k'x$ and the answer is that the the stabilizer is $x^{-1}Kx$.

The kernel is, as you say, the intersection of all $x^{-1}K x$ for all $x\in G$. I'm not sure if it has a name: in case of commutative groups it is all of $K$, in some cases it can even be trivial. (It is always a normal subgroup of $G$.)

For transitivity, note that, in a group, you can convert an element $kx$ to any other element by suitable right multiplication.