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**Giraffro** If $\displaystyle G$ is a finite simple group and $\displaystyle H$ is a proper, non-trivial subgroup, then $\displaystyle |G|$ divides $\displaystyle \frac{[G:H]!}{2}$.

__What I have so far:__

Define $\displaystyle \forall g,h \in G, (Hh)(g)\phi := Hhg$. Then $\displaystyle \phi : G \rightarrow \text{Sym}(\cos(G:H))$ is a well-defined group homomorphism with kernel $\displaystyle \text{core}_G(H) = \{e\}$. Therefore by the First Isomorphism Theorem, $\displaystyle |G| \, \mid \, |\text{Sym}(\cos(G:H))| = |\cos(G:H)|! = [G:H]!$. However I'm stuck at this point. I think I need to prove that the image of $\displaystyle \phi$ consists of even permutations on $\displaystyle \cos(G:H)$, but I can't seem to find a way to prove that. The image must be simple as $\displaystyle \phi$ is injective, but I can't see how that helps.

Any help would be greatly appreciated. Thanks!