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Groups, g-sets and stabilizer

  1. Apr 23, 2012 #1
    hey,

    if you have the group S3 and treat it also as a g-set under the action
    x*g = g-1xg

    is it correct that the stabilizer of the g-set is just the identity element?

    thanks
     
  2. jcsd
  3. Apr 23, 2012 #2

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    Hi Jesssa!

    A stabilizer of an action G on a set X is defined for a specific element x in X.
    Each stabilizer Gx contains at the very least the identity element of G.

    Now I'm unclear on what exactly your action is.
    Shouldn't it be g*x=g-1xg, with g in S3 and x in {1,2,3}?

    In that case any stabilizer consists of all elements of S3.
    Since if we pick for instance g=(123) and x=2, we get: (123)-12(123)=2.
    In other words, for any g in S3 and any x in {1,2,3}, we have g*x=x.
     
    Last edited: Apr 23, 2012
  4. Apr 23, 2012 #3

    micromass

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    That would imply that it is a left action. But it is not as [itex]h*(g*x)=h^{-1}g^{-1}xgh\neq (hg)*x=g^{-1}h^{-1}xhg[/itex].
     
  5. Apr 23, 2012 #4

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    Ah, so it is a right action!
    (Just looked up the definition on wikipedia. I was not aware yet of the requirement for associativity nor of the definition of a right action. I'll have to check my books why I didn't know that yet.)

    I think the rest of my argument remains intact?
     
  6. Apr 23, 2012 #5

    micromass

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    I don't think the rest of the argument is correct, but the OP would have to say something about it.
    You seem to imply that [itex]S_3[/itex] acts on {1,2,3}. But I rather think that it acts on [itex]S_3[/itex] itself. That is, I think that both g and x should be in [itex]S_3[/itex].
     
  7. Apr 23, 2012 #6
    hey guys,

    thanks for your replies!

    I though of it like this,

    for x*g = g-1xg = x

    then x must = g to give

    g-1gg=g=x

    So that means every element is its own stabilizer under this action right?

    Which means that there mustn't be a stabilizer for the g-set besides the identity right?

    When i first looked at the question I had a hunch that the stabilizer of the set was the identity but didn't know how to explain it, so i started doing them one by one until I thought of that,

    Do you think its right?
     
  8. Apr 23, 2012 #7

    micromass

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    I don't really get what you're doing here.

    First of all, what is your definition of "stabilizer for the g-set"?? Are you talking about "stabilizer of an element" or "stabilizer of a set". In the latter case, how is that defined?
     
  9. Apr 24, 2012 #8

    micromass

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    OK, then in general the stabilizer will depend on the element x.

    So, given an x fixed, we must find all the elements g such that

    [tex]g^{-1}xg=x[/tex]

    Or in other terms: [itex]xg=gx[/itex]. So the elements in the stabilizer are exactly the elements which commute with x.

    Now, I propose you calculate the stabilizer for each element x in S3. There are 6 cases:

    1) x=e
    2) x=(1 2)
    3) x=(2 3)
    4) x=(1 3)
    5) x=(1 2 3)
    6) x=(1 3 2)

    So I propose that in each case, you find all the elements g which commute with the given x.

    The first one is easy: you'll just need to find all the g in S3 which commute with e. But ALL elements commute with e (by definition). So the stabilizer of e is S3.

    The other 5 cases are a bit harder, but you should be able to do it.
     
  10. Apr 24, 2012 #9
    The definition we were given is,

    If X is a G-set then Stab(X) = { g in G| x*g = e for all x in X},

    Arn't permutations non-communtitative?

    So xg≠gx unless x = e.

    So e can be the only element in the stabilizer of the set

    Stab(X) = {e}
     
  11. Apr 24, 2012 #10

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    Should we assume that X is also S3?

    Anyway, some permutations are non-commutative, but not all.
    For instance, any permutation commutes with itself.
     
  12. Apr 24, 2012 #11
    yea I just introduced X to explain the stabilizer of the G-set without making confusion between the group and the g-set,

    But in S3 there is no 1 permutation which commutes with every element of the g-set,

    such that the stabilizer condition of the set Stab(X) where X = G is satisfied

    so there is no fixed g in G such that xg=gx for all x in X

    Besides the identity

    so the only element in the stabilizer must be the identity right?
     
  13. Apr 24, 2012 #12

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  14. Apr 24, 2012 #13
    cool thanks I Like Serena!

    hey could I ask something else related to this set up which I have been trying to practice doing,

    If you act on the set X by an element of S3 all that you do is rearrange the ordering of the elements within the set,

    So its just a permutation of the g-set, say its a function β(g) = σ = x*g

    for example for (1 2 3) σ = (2 4 3)(6)(5)(1)

    ....S3={e, (1 2), (13), (23), (123),(132)}
    labeling 1....2.......3.....4.......5.......6

    which you can get by doing

    β((1 2 3)) for each x in the g-set,

    for example x=(12)

    (321)(12)(123) = (23), so 2 goes to 4

    and so on for each elements

    would the imagine of this β(g) just be every element of S3 since all β does is rearrange the g-set?
     
    Last edited: Apr 24, 2012
  15. Apr 24, 2012 #14

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    Not quite.
    It's called a conjugation.
    It means that the "type" of the permutation remains the same, but the elements are shuffled.
    In your example, you'll always get a cycle of length 2, just with different elements.
     
  16. Apr 24, 2012 #15
    I think I worked it out,

    The image of β is just all the possible permutations β(x) for each x in the set,

    So it'll be Im(β)={(34)(56),(24)(56),(23)(56),(243),(234)}

    the first three correspond to x= (12),(13),(23) respectively and the last two (123) and (132) respectively
     
    Last edited: Apr 24, 2012
  17. Apr 24, 2012 #16

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    That's just as well, because you've lost me.

    I don't understand how you've moved from β(g) to β(x), from S3 to S6, nor what your Im(β) represents.
     
  18. Apr 24, 2012 #17
    β(x) = β(g) since x is in X and g is in G and G=X, sorry this makes it confusing

    when an element of g acts on the g-set the elements in the set are re-ordered which can be expressed as a permutation on the set.

    β(g) outputs the permutation of the set,

    so if you label the elements as 1 to 6 and act (123) on the set you will find that
    the set is rearranged according to (2 4 3), so the element in position 2, (12), moves to position 4, the element in position 4, (23), moves to position 3, and (13) goes to position 2. the elements in position, 1, 5 and 6 are fixed, (321)(123)(123) = (123), (321)()(123)=(), (321)(132)(123) = (132).

    So one of the elements in the image of β must be (2 4 3),

    Then similarly when you act on the set with the order 2 permutations (12) and so on you get the (34)(56), since now () and (12) are fixed, (12)(12)(12)=(12), (12)()(12)=(), and when you apply it to the rest of the set you find that (13) goes to (23), (123) goes to (132),
    since (12)(13)(12)=(23) and so on, giving (34)(56) which is another element of the Image of β

    β:G -> S(X)

    Im(β) = {s in S(X)|s = β(g) for g in G}

    I actually forgot () in my Image in the previous post as well,
     
  19. Apr 25, 2012 #18

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    Ah okay.
    It makes more sense now. :)

    Btw, how about naming the elements for instance e,a,b,c,d,d2?
    That would help eliminate the ambiguity of the numbers.
     
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