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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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$\,S_g\,$ is a rather unusual notation. I'd rather go with $\,\operatorname{Sym}_G\,$ , or at least $\,S_G\,$ –  DonAntonio Nov 22 '12 at 15:50
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When you say, "I have a group $G=\{1,2,3\}$." That is not a group, it is a sets of three elements. A group is a set with an operation. –  Thomas Andrews Nov 22 '12 at 16:32
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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 $|G| > 1$. –  Mikko Korhonen Nov 22 '12 at 17:15
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2 Answers 2

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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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$0 \to 1$, $1 \to 0$, $2 \to 2$. It maps each elements of $G$ to an element of $G$. So why is it not an automorphism? –  sonicboom Nov 22 '12 at 16:01
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Because an automorphism of a group must first be a homomorphism of groups and, as such, it must map the group's unit (or neutral element) to itself. Here, it must be that $\,0\to 0\,$, otherwise it is not a homomorphism and, thus, not an automorphism. –  DonAntonio Nov 22 '12 at 16:28
    
Cheers, I get it now. Just wondering, what is $Aut(G)$ in the case of my example in the original post...is it the elements $()$ and $(2\, 3)$? These are the only ones that fix $1$, which is the identity element in my group $G$? –  sonicboom Nov 22 '12 at 16:51
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Well, in your example you have a set with three elements, what'd be a cyclic group of order three...which is exactly what is $\,G\,$ in my answer! Here, $\,|Aut(G)|=2\,$ , and the only two automorphisms are: $$0\to 0\,\,,\,\,1\to 1\,\,,\,\,2\to 2$$and $$0\to 0\,\,,\,\,1\to 2\,\,,\,\,2\to 1$$ –  DonAntonio Nov 22 '12 at 17:04
    
Note however, that mapping $0$ to itself is not all that is required for a permutation to be an automorphism in general. It just happens that all counterexamples are for groups with at least 4 elements. –  Henning Makholm Nov 22 '12 at 17:08
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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. –  sonicboom Nov 22 '12 at 15:58
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