How is this an group homomorphism?

Question

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.

Answer 1

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

1. $\tau(x,gh)=\tau(\tau(x,g),h)$, for all $x\in X$ and for all $g,h\in G$;
2. $\tau(x,1)=x$ for all $x\in X$.

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

1. $x(gh)=(xg)h$, for all $x\in X$ and for all $g,h\in G$;
2. $x1=x$, 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.