# Order of $\mathrm{Aut}(G)$

I am not sure how to approach the following problem:

Show that if $|G|=n$, then $|\mathrm{Aut}(G)|$ divides $(n-1)!$

All I can think of so far is that clearly $|\mathrm{Aut}(G)| \le |\mathrm{Sym}(G \setminus \{e\})| = (n-1)!$, but this doesn't give any suggestion as to why $|\mathrm{Aut}(G)|$ divides $(n-1)!$

Thank you.

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The embedding $\mathrm{Aut}(G)\to \mathrm{Sym}(G \setminus \{e\})$ is a homomorphism, right? So there is some subgroup of $\mathrm{Sym}(G \setminus \{e\})$ isomorphic to $\mathrm{Aut}(G)$... so ... –  martini Oct 7 '12 at 22:10