How do I prove the following statement about the kernel?

Question

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?

I'd say your definition of $\rho(g)$ is a bit off. Is $x\in G$? Or $x\in G/H$? Maybe $\rho(g)(xH) = gxH$ — podiki, Nov 06 '21 at 14:51
Sorry I edited it, now it should be okei right? Could you still help me? — user123234, Nov 06 '21 at 15:02

podiki · Accepted Answer · 2021-11-06T15:25:26.053

1

$\ker(\rho) \subset \bigcap xH x^{-1}$:

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 $\rho(g)=Id_{X}$.

edited Nov 06 '21 at 15:25

answered Nov 06 '21 at 15:04

podiki

2,265

ah wow perfect thanks – user123234 Nov 06 '21 at 15:14
but I think in the second direction it should be for every $x\in G$ – user123234 Nov 06 '21 at 15:20
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Yes thank you, edited. – podiki Nov 06 '21 at 15:25

How do I prove the following statement about the kernel?

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