$G$ simple group and there exists a subgroup of index $n$. Show that: $|G|$ divides $n!/2$

Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question here.

There was written: "The proof is really simple. Let $X$ be the set of left cosets of $H$. Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$. Then $\phi$ is a homomorphism. Consider now $K=\ker \phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. Finally, $G/K$ is isomorphic to a subgroup of $\text{Sym}(X)$, which has order $n!$, where $n=[G:H]$. Thus, $[G:K]$ is finite and divides $n!$."

In fact, I don't understand the last sentence. "Thus, $[G:K]$ is finite and divides $n!$." Also did I understand it right that $ker(\phi)=H$? Can you also give me an advice how to proof it for $n \ge 3$. (maybe induction)

Thank you for taking your time.

Solution 1:

Sym(X), is the group of permutations of the collections of cosets, $G/H$. As $H$ has index $n$, this group Sym($X$) is of order $n!$. As $K$ is the kernal of $\phi$ we see$\phi$ leads to an injective homomorphism $G/K\to Sym(X)$ and by Lagrange's theorem its order should divide $n!$. As there is anormal subgroup of index 2 (consisting of even permutations) in Sym($X$), the image of the simple group $G$ should be contained in that normal subgroup otherwise its intersection will lead to a proper normal subgroup in $G$, contradicting its simplicity. So $|G|$ divides $n!/2$

EDIT: I should explore if it is possible for image of $\phi$ to have trivial intersection with Alt(X). In that case the sign homomorphism $S_n\to\{\pm1\}$ restricts to a non-trivial homomorphism on Image of $G$, whose kernel gives a proper nontrivial normal subgroup in image of $G$ and hence on $G$ contradicting the simplicity of $G$.