kernel of action.

Question

If $\phi : G \to Perm(G/H)$ where $\phi$ is the group action on $G/H$ by $G$.

$\phi := g(g'H) = (gg')H$

Why is the kernel of $\phi$ equal to $\cap_{x\in G} xHx^{-1}$

I thought the kernel is $\ker \phi = \{g | gx =x \}$ (so ker is a stabilizer).So in our case it is $g(g'H) = g'H$ . So all forms of $g$ are $\{ e_G, g'hg'^{-1} \}$

I don't understand why we have intersection and why is it not $g'$ but $x$

I use notes from http://www.math.unl.edu/~s-wmoore3/notes/groupThry.pdf (page 3/25)

Answer 1

$\ker \phi$ = {g | gx =x } = {g | gxH=xH} = {g | $x^{-1}$gx in H } = {g | g in xH$x^{-1}$ } = {g | g in $\cap_{x\in G} xHx^{-1}$ = $\cap_{x\in G} xHx^{-1}$

Answer 2

$g(g'H)=g'H$ iff there exists an $h\in H$ such that $gg'=g'h$, i.e., iff $g\in g'Hg'^{-1}$. The condition that this be true for all $g'\in G$ defines the kernel of $\phi.$