Prove that ${\rm Ker}f=\bigcap_{x\in G} xHx^{-1}$, where $H$ is a subgroup of the group $G$, and $f:G\to \text{Sym}(G/H)$ is the left-coset action.

Question

I am trying to prove that ${\rm Ker}f=\bigcap_{x\in G} xHx^{-1}$, where $H$ is a subgroup of the group $G$, and $f:G\to \text{Sym}(G/H)$ is the left-coset action.

Here is what I have

$$\begin{align} k\in K=\text{Ker} f &\iff k\cdot xH=xH, \forall xH\in G/H\\ &\iff k\in G_{xH}, \forall xH\\ &\iff k\in \bigcap_{xH\in G/H} G_{xH}\tag{$*$} \end{align}$$

Now we want to show that $G_{xH}=xHx^{-1}=\{ xhx^{-1} \mid x\in G\}$.

$$\begin{align} g\in G_{xH} &\iff g\cdot xH=xH\\ &\iff gxh_1=xh_2, \ \ \ h_1,h_2\in H\\ &\iff g=xh_2 h_1^{-1} x^{-1}\\ &\iff g \in xHx^{-1} \end{align}$$

Hence $G_{xH}=xHx^{-1}=\{ xhx^{-1} \mid x\in G\}$.

Back to $(*)$,

$$\begin{align} k\in \bigcap_{xH\in G/H} G_{xH} &\iff k\in \bigcap_{x\in G} xHx^{-1}, \end{align}$$

which is true iff

$$K=\bigcap_{x\in G} xHx^{-1}.$$

Is it correct? I appreciate any comments. Thanks.

Answer 1

By definition: \begin{alignat}{1} \operatorname{Stab}(xH) &=\{g\in G\mid gxH=xH\} \\ &=\{g\in G\mid x^{-1}gxH=H\} \\ \tag1 \end{alignat} Now, call $g':=x^{-1}gx$, whence $g=xg'x^{-1}$. From $(1)$: \begin{alignat}{1} \operatorname{Stab}(xH) &=\{xg'x^{-1}\in G\mid g'H=H\} \\ &=x\{g'\in G\mid g'H=H\}x^{-1} \\ &=x\{g'\in G\mid g'\in H\}x^{-1} \\ &=xHx^{-1} \\ \end{alignat} Therefore: \begin{alignat}{1} \operatorname{ker}f &=\bigcap_{xH\in G/H}\operatorname{Stab}(xH) \\ &=\bigcap_{x\in G}\operatorname{Stab}(xH) \\ &=\bigcap_{x\in G}xHx^{-1} \\ \end{alignat}