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What is the number of automorphisms (including identity) for permutation group $S_3$ on 3 letters?

I believe the answer for this is 6. As we can write the group elements as below

  1. (a)(b)(C)
  2. (ab)(c)
  3. (ac)(b)
  4. (bc)(a)
  5. (abc)
  6. (bac)

Can we generalize that for any $S_n$ onto $n$ the number of automorphisms will be $n!$ Also I cannot find this anywhere in my text, can we have a permutation $S_n$ onto $m$ where $n\neq m$? (for eg: is it possible to have a permutation group say $S_5$ on say 2 elements?)

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    Are you asking about the number of *elements*, or the number of *automorphisms*? An automorphism of $S_3$ is a group homomorphism $f\colon S_3\to S_3$ that is one-to-one and onto. (It just so happens that there is a bijection between this group and $S_3$, but that is non-trivial, certainly not a matter of just listing elements). I don't understand your last question.2012-03-25
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    By definition, $S_n$ means the group of all permutations of $\{1,2,\ldots,n\}$. "A group $S_5$ on 2 elements" is unclear at best. You are still not making much sense, and you are still not clarifying whether you are trying to count elements or automorphisms. This suggests that you are very confused, regardless of anything else.2012-03-25
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    I updated the question. Yes I am talking of automorphisms. So this number is the number of bijections we can have on $S_n$ ?2012-03-25
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    If you are talking about "automorphisms", then what you wrote is useless. Your list is giving the **elements** of $S_3$; how are you relating an element to an automorphism, and how you are you showing that what you wrote are *all* the automorphisms? It is a theorem of Otto Holder that $\mathrm{Aut}(S_n)\cong S_n$ for all $n\neq 2,6$.2012-03-25
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    @ArturoMagidin I thought it was, couldn't find anywhere , It is confusing that all text say $S_n$ on n when $S_n$ itself is self explanatory as you said2012-03-25
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    $S_n$ can **act** on other sets; for example, you can interpret $S_n$ as a group of permutations of $m$ elements, with $m\gt n$ (by fixing all elements greater than $n$); you can also let, for instance, $S_3$ act on a set with $6$ elements in a different way (say, take $\{a,b,c,d,e,f\}$, and then let $S_3$ act on the subsets $\{a,b\}$, $\{c,d\}$, and $\{e,f\}$ by identifying them with "1", "2" and "3", and always mapping lexicographically. But those are group *actions*.2012-03-25
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    Ok I got confused between group elements and automorphisms :( By Aut($S_n$)≅$S_n$ mean number of automorphisms is equivalent to $S_n$ ?2012-03-25
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    By $\mathrm{Aut}(S_n)\cong S_n$ we mean that the two groups are *isomorphic*. In *particular*, they have the same number of elements, but the statement is much stronger than merely "the same number of elements". You don't need so much to solve this particular problem.2012-03-25
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    So after listing the elements I write down all possible mappings of $S_n$ to $n$ elements. The number of the these mappings is the number of automorphisms. A shortcut for $n$!=2,6 is that the number of bijective functions on $S$ which is given by $n!$2012-03-25
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    "All possible mappings of $S_n$ to $n$ elements" is nonsense, and it does not help you in finding the automorphism. An automorphism is a **group homomorphism** from $S_n$ to itself. **Not** a "map from $S_n$ to $n$ elements" (which would mean a *set theoretic* function $S_n\to \{1,2,\ldots,n\}$). Nothing you wrote in the last comment makes any sense in the context of this question.2012-03-25

1 Answers 1

6

What you did doesn't really work: you are listing elements of $S_3$, instead of automorphisms of $S_3$. While there is a natural isomorphism between the two, I suspect that if you are confusing elements and automorphisms, you are not expected to know this yet.

Try this:

  1. If $f\colon S_3\to S_3$ is a group homomorphism, and you know what $f(12)$ and $f(123)$ are, then you know $f(\sigma)$ for all $\sigma\in S_3$.

  2. If $f,g\colon S_3\to S_3$ are group homomorphisms and have the same values at $(12)$ and at $(123)$, then $f=g$.

  3. $f(12)$ must be an element of order $2$. So there are, at most, three possibilities.

  4. $f(123)$ must be an element of order $3$, so there are at most two possibilities.

  5. Conclude that there are at most six automorphisms of $S_3$.

  6. Exhibit six different automorphisms of $S_3$ (Hint: Conjugation...)

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    Plus one for this comment.2012-03-25