How is this an group homomorphism? Let $G$ be a group and $H \leq G$. If $[G:H]=n$, then exists $\phi: G \to S_n$ homomorphism with $\operatorname{Ker}(\phi) = \bigcap_{x \in G} x^{-1}Hx$.
My idea is to show that, for the set $S=\{Hg_1,Hg_2,...,Hg_n\}$ of the right cosets of $H$ in $G$, we can take the action of $G$ in $H$ by right multiplication, and that will induce an homomorphism from $G$ to $S_n$ defined as:
$$\phi(x) = \left(\begin{matrix}
    Hg_1 & Hg_2 & \cdots & Hg_{n-1} & Hg_n \\
    Hg_1x & Hg_2x & \cdots & Hg_{n-1}x & Hg_nx
  \end{matrix}\right)$$
But my question is: how is that an homomorphism?
Because for $x,y \in G$, we should have:
\begin{align*}
\phi(xy) &= \left(\begin{matrix}
    Hg_1 & \cdots & Hg_n \\
    Hg_1xy & \cdots & Hg_nxy\end{matrix}\right)\\
 &= 
\left(\begin{matrix}
    Hg_1 & \cdots & Hg_n \\
    Hg_1x & \cdots & Hg_nx
  \end{matrix}\right)\left(\begin{matrix}
    Hg_1 & \cdots & Hg_n \\
    Hg_1y & \cdots & Hg_ny 
  \end{matrix}\right)\\ &= \phi(x)\phi(y)
\end{align*}
But I can't see how is that true. The kernel part is easy to see, but that homomorphism I can't see at all.
 A: It depends on how you compose permutations. From your approach I guess that you compose them from left to right. So a right action of $G$ on the set $A$ is a mapping
$$
\tau\colon X\times G\to X
$$
such that

*

*$\tau(x,gh)=\tau(\tau(x,g),h)$, for all $x\in X$ and for all $g,h\in G$;

*$\tau(x,1)=x$ for all $x\in X$.

However, $\tau(x,g)$ is more commonly written as $xg$, so the properties are

*

*$x(gh)=(xg)h$, for all $x\in X$ and for all $g,h\in G$;

*$x1=x$, for all $x\in X$.

In this case you can define a map $\varphi\colon G\to S_X$ (the codomain is the group of permutations of the set $X$) by
$$
\varphi(g)=\hat{g},\qquad (x)\hat{g}=xg
$$
(maps are written on the right, to comply with the order of composition). This is a group homomorphism for every action. The proof is just doing the verification:
$$
(x)(\hat{g}\hat{h})=((x)\hat{g})\hat{h}=(xg)\hat{h}=(xg)h=x(gh)=(x)\widehat{gh}
$$
What's the kernel of this homomorphism? It is the set of $g\in G$ such that $xg=x$, for every $x\in X$, that is, $\hat{g}$ is the identity map.
In your case $X=G/H$, the set of right cosets and, if you want, you can identify $S_{G/H}$ with $S_n$, where $n=[G:H]$, by enumerating the cosets (in whatever order you prefer).
The action is $(Hx)g=H(xg)$ and the axioms are easily verified. The kernel is the set
$$
K=\{g\in G:Hxg=Hx,\text{ for all }x\in X\}
$$
Now, $Hxg=Hx$ if and only if $x\in Hxg$, that is, $x=hxg$, for some $h\in H$. In particular $g=x^{-1}h^{-1}x\in x^{-1}Hx$. So if $g\in K$, then $g\in x^{-1}Hx$, for all $x\in G$.
Conversely, if this happens, then $Hxg=Hxx^{-1}hx=Hhx=Hx$, for every $x\in G$. Therefore
$$
K=\bigcap_{x\in G}x^{-1}Hx
$$
as required.
Note that using the abstract approach frees you from considering the actual cosets until really needed.
