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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

Hint: The permutations of a given set form a group under composition. An automorphism of a group is, in particular, a permutation of the group which fixes the identity, which is 'the same' as a permutation of the non-identity elements of the group. How many of these are there? What do you know about sizes of subgroups?

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