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More automorphisms

  1. Nov 20, 2004 #1
    Okay, so i'm having trouble understanding the following question:
    Determine the group of automorphisms of [tex]S_3[/tex].

    I understand that the automorphisms must match orders of the same element, and since there are three permutations of order 2 and two of order 3, there are 6 "possible" permutations. But I don't know where to go from here. I'm pretty sure there's a better way than to tediously go through all six possible automorphisms, and explicitly check whether each work or not. Am i missing something here? :confused:
    (I mean, how do you know when you've determined the group?)
     
  2. jcsd
  3. Nov 21, 2004 #2

    matt grime

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    You only need to look at the elements of order two since you know where (123) goes once you know where (12) and (23) go - that is the point of automorphisms - and this doesn't involve much work.

    Of course, you could just prove it has no outer automorphisms...

    In this case, every permutation of the 3 elements of order two is an automorphism, obviously, and this determines every automorphism, equally clearly, and hence the group Aut(S_3) = S_3.
     
  4. Nov 22, 2004 #3

    matt grime

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    I was thinking that this may well need more explanation.

    Every relabelling of the numbers 1,2,3 gives an automorphism of S_3 (and a relabelling of 1,..,n would give an automorphism of S_n). These correspond to the inner automorphisms - the ones where the group acts by conjugation. If you've done linear algebra it's a lot like a change of basis.

    This tells us that there is a copy of S_3 inside Aut(S_3). Now all we need to do is show that Aut(S_3) has at most 6 elements and we are done.

    Because S_3 has exactly 3 elements of order 2, and they generate S_3, then any permutation of them *might* be an automorphism, and any two automorphisms permute them in distinct ways, so there are at most 6 possible automorphisms, as we needed to show.

    Hence S_3 <= Aut(S_3) <=S_3

    so they are equal.

    The reason I thought I needed to clarify this was that I suspected it wasn't clear why this didn't show that the Aut(S_n) was something that it wasn't.

    It's important that n=3 here, so that all the elements of order 2 are of the same cycle type and generate the group.

    In general there is always a map from G to Aut(G) given by [tex] x \to f_x[/tex] where [tex]f_x(y)=x^{-1}yx[/tex] is how the aut f_x acts on G.

    Exercises:
    1. Show this map from G to Aut(G) is a homomorphism
    2. What is the kernel?
    3. Hence conclude that when G=S_n it is an injection unless n=2


    FACT it is an isomorphism on S_n except for n=2 and n=6.
    Proof n=2 is easy since S_2 is abelian. n=6 is too hard for me to recall but essentially seems to be because 2*3=2+3+1. Have a look for a proof somewhere - there are some nice geometric ones available.
     
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