Proof verification: Every group of order n! contains non trivial normal subgroup (without Sylow)

Question

Suppose $ |G| = n!$ then either

$ G \simeq S_{n} $ $=>$ the preimage of $A_{n}$ is a normal subgroup which means that G contains a non trivial normal subgroup

or if $G$ is not isomorph there is a Grouphomomorphism to $S_{n}$ because the order of $G$ is finite but since they are not isomorph the homomorphism is not injective therefore we can choose a kernel and are done

is this correct or am i missing something?

Answer 1

As observed in the comments, the proof proposed in the OP is incorrect because Cayley's theorem does not guarantee a group homomorphism to $S_n$.

Since $|G| = n!$, what it guarantees is an embedding into the much bigger group $S_{n!}$. Since the map constructed this way is injective, it does not hand you the kernel you want.

Your claimed result seems likely to me (that $n!$ is not the order of a finite simple group - I can't think of a finite simple group with factorial order), but it is hard for me to imagine a proof that doesn't use Sylow theory or something more advanced (like the classification theorem).