If a group $G$ of order $1200$ with action of $X$ and orbit $6$ then the stabilizer contains a subgroup so $H\triangleleft G$

Question

So I was trying to solve the following question:

Let $G$ be a group of order $1200$ with action on $X$. Let $a\in X$. We know that the orbit of $a$ is $6$. Prove that the stabilizer of $a$ has subgroup $H$ so $H\triangleleft G$.

Also I saw the solution:

We know that orbit is the index of the stabilizer, so $|G\,:\,Stab_{\alpha}|=6$. By Cayley theorem we can conclude that there is a group $H$ of $G_a$ so $H\triangleleft G$ and $G/H$ is isomorphic to a subgroup of $S_6$.

I don't understand the solution. The Cayley theorem is:

Let $G$ be a group. so $G$ is isomorphic to a subgroup of a permutation group.

First of all, I do understand why $|G\,:\,Stab_{\alpha}|=6$. But why from Cayley theorem we conclude that there is a group $H$ of $G_a$ so $H\triangleleft G$ and $G/H$ is isomorphic to a subgroup of $S_6$? Also how do we know that it is $S_6$?

Answer 1

So $Stab(a)$ has order $ 200 $ and index $6$. Consider the action of $G$ on $Orb(a)$. This gives u a group homomorphism $G\rightarrow S_6$. Now we claim this map is not injective since $1200$ does not divide $6!=720$. So there must be a kernel i.e. a non trivial normal subgroup of $G$ contained in $Stab(a)$ ( since any element of the kernel acts trivially on $orb(a)$ and hence $ \in Stab(a)$) such that $G/H$ is isomorphic to a subgroup of $S_6$