General Cayley's theorem: Let $G$ be a group and $H$ a subgroup of $G$ such that $[G:H]=n$. QUESTION:  General Cayley's theorem: Let $G$ be a group and $H$ a subgroup of $G$ such that $[G:H]=n$. Then the quotient $G/H_G$ is isomorphic to a subgroup of $S_n$.
ANSWER: In the answer I found, there is this isomorphism $\pi: G/H_G \rightarrow \operatorname{Sym}(\{yH: y \in G\})\cong S_n$. But I really didn't understand why is this a isomorphism, I didn't understand even less why this is isomorphic to $S_n$.
$H_G:=\operatorname{kernel}(\pi)$ - where $H_G$ is called normal core of the isomorphism $\pi$. Here for example, I didn't understand why $H=H_G$. Actually, I didn't undestand how to find the kernel of this isomorphism equal to $\displaystyle\bigcap_{g \in G}gHg^{-1}$.

Since $H \leq G$ is a subgroup of index $[G:H]=n$, the set $G/H$ has size $n$ and you can identify $\text{Sym}(G/H)$ with $S_n$.
Now $G$ acts on the set $G/H$ by left multiplication. This gives a homomorphism $\phi: G \to \text{Sym}(G/H)$ where $\phi(g): yH \mapsto gyH$. The kernel of this map is $\ker \phi = \{g \in G \ | \ yH = gyH \ \forall y \in G\} = \{g \in G \ | \ g \in yHy^{-1} \ \forall y \in G \} = H_G$ so this is where the normal core comes from. By the first isomorphism theorem now $G/H_G \cong \text{Im} \phi \leq \text{Sym}(G/H) \cong S_n$. As mentioned in another answer, the map is not surjective in general and $G/H_G$ is merely a subgroup of $S_n$.
