Kernel of group action on set of cosets by left side multiplication

Question

I was reading this article that explains some ways you can show a certain group is not simple. One of the methods described is by observing the action of the group $G$ on the set of cosets $G/H$ (for some subgroup $H$) defined by left side multiplication - for any $g∈G$ and $xH∈G/H$ let $g⋅xH:=gxH$. We can then define a homomorphism $φ$ from $G$ to $Sym(G/H)≅Sn$ by $φ(g)=σ_{g}$ where $σ_{g}$ is the permutation given by $σ_{g}(xH)=g⋅xH=gxH$.

In the article, it is said that the kernel of the homomorphism satisfies - $Ker(\phi)=\bigcap_{x\in G} xHx^{-1}$, but I'm afraid I don't quite understand why that is. Why does the intersection act trivially on all cosets? And how do we know that there aren't other group members that satisfy this? I know that the kernel is the intersection of all stabilizers, so I guess this implies that $ xHx^{-1}$ is necessarily a stabilizer?

I'd appreciate it if anyone could explain - thanks in advance.

Answer 1

I assume there's a typo and you meant $\ker(\varphi)$.

The following sequence of equivalent statements should answer your questions:

1. $g\in\ker(\varphi)$
2. $\varphi(g)=\text{id}_{G/H}$
3. $gxH=xH$ for all $x\in G$
4. $x^{-1}gx\in H$ for all $x\in G$
5. $g\in xHx^{-1}$ for all $x\in G$
6. $g\in\bigcap_{x\in G} xHx^{-1}$.

I guess this implies that $ xHx^{-1}$ is necessarily a stabilizer?

Yes, indeed.