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I have read that $Aut(G)$ is a subset of $S_g$.

So say I have a group $G = \{1, 2, 3\}$ for example. Then $S_G = S_3$ is the group of all permutation of the three elements of $G$.

But I don't see why $Aut(G)$ is a subset of $S_G$ as opposed to $Aut(G) = S_G$.

Each element of $S_3$ maps each element of $G$ to an element of $G$. I.e. each element is an automorphism. So why is $Aut(G) \subset S_3$ instead of $Aut(G) = S_3$?

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    Not every permutation of $G$ is an automorphism. For example, automorphisms always fix the identity, but there are definitely permutations of $G$ that do not fix the identity if |G| > 1.2012-11-22

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There are elements in $\,\operatorname{Sym}_G\,$ which are not automorphisms of the group $\,G\,$, say the permutation $\,(01)\,$ in $\,S_3\,$ is not an automorphism of cyclic group $\,\Bbb Z_3:=\Bbb Z/3\Bbb Z:=\{0,1,2\}\,$, with operation modulo $\,3\,$

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    Of course. I wrote about a necessary condition for a map to be "automorphism of a group", which in general is very remote from being sufficient.2012-11-22
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(I'm assuming that $G$ here is a group, but I now notice that you're not saying so explicitly. If $G$ is not considered to be a group, then you have to ask what does $\operatorname{Aut}(G)$ mean at all?)

Merely "mapping each element of $G$ to an element of $G$" is not enough to be an automorphism. An automorphism is a bijection $G\to G$, which is also a homomorphism. Most elements of $S_3$ will not correspond to homomorphisms $G\to G$.

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    Sorry, I'ved edited in that $G$ is a group.2012-11-22