I have the following problem:
Given a finite group $G$ and $p$ the smallest prime dividing $card(G)$. Let $H$ be a subgroup s.t. $card(G\setminus H)=p$ Let $X=G\setminus H$ and consider the action $$G\times X\rightarrow X; \,\,(g,xH)\mapsto gxH$$ Let $$\rho:G\rightarrow Bij(X,X);\,\,g\mapsto \rho(g):X\rightarrow X\,\,\text{given by}\,\,\rho(g)(xH)=gxH$$the associated morpism. I have just shown that $Stab_G(xH)=xHx^{-1}$ is the stabilizer. Now I need to deduce that $ker(\rho)=\bigcap_{x\in G} xHx^{-1}$.
I'm somehow unsure since they wrote deduce I think we should use that $Stab_G(xH)=xHx^{-1}$. I wanted to take $g\in ker(\rho)$ and then show that for all $x\in G$ $gxH=xH$ But i somehow struggle. Could someone help me please?
$\ker(\rho) \subset \bigcap xH x^{-1}$:
If $g\in\ker(\rho)$ then $\rho(g) = id_{X}$ and we have $gxH = xH$ for all $x \in G$. Thus $g\in\text{Stab}_G(xH)=xHx^{-1}$ for every $x\in G$.
$\ker(\rho)\supset \bigcap xH x^{-1}$:
Let $g\in xHx^{-1}$ for every $x\in G$. Then let $y\in G$ and consider $\rho(g)(yH)= gyH$. Since $g\in yHy^{-1}$, we have an $h\in H$ such that $g=yhy^{-1}$ and can compute
$$gyH = yhy^{-1}yH= yhH= yH.$$ So $\rho(g)(yH)=yH$. But $y$ was arbitrary hence $\rho(g)=Id_{X}$.