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Center of Symmetric Group

  1. Jun 2, 2008 #1
    [SOLVED] Center of Symmetric Group

    The problem statement, all variables and given/known data
    Show that for n ≥ 3, Z(Sn) = {e} where e is the identity element/permutation.

    The attempt at a solution
    It is obvious that e is in Z(Sn). If there is another element a ≠ e in Z(Sn), then... There must be some sort of contradiction and it has to do with the fact that n ≥ 3 but I can't figure it out. Any tips?
     
  2. jcsd
  3. Jun 2, 2008 #2

    Dick

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    Can't you find two transpositions that don't commute?
     
  4. Jun 3, 2008 #3
    I know that (12)(23) ≠ (23)(12). But how does this show that Z(Sn) is trivial?
     
  5. Jun 3, 2008 #4

    Dick

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    You want to show that for any element g, except for e, there exists an element h such that gh!=hg. You've just done it if g is a transposition. I was thinking you could decompose g into disjoint cycles and construct an element that it doesn't commute with. It should work. Can you try it?
     
  6. Jun 3, 2008 #5
    I've only done it for a particular h and g. Not for every g.

    I'm thinking of the following: say the disjoint cycle decomposition of g is (ab...)(cd...). Let h = (ac). Then gh = (ad...) and hg = (ab...), which are not equal. Hmm...If this is to work in general, I will have to deal with the various ways in which g is decomposed.
     
  7. Jun 3, 2008 #6

    Dick

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    You've done the case where g consists of a single transposition. Just generalize your point labelling. Now suppose g contains a cycle of three or more elements, say (123...). Hint: use only 1,2 and 3, that way your other permutation h will commute with all of the other cycles. That leaves only the case where g contains only transpositions, like (12)(34)... Can you find something that doesn't commute with that? Actually I think you are already on the right track. You don't need to know EVERYTHING about the cycle decomposition, you only need to know enough to i) find an h and ii) make sure your cases cover all permutations.
     
    Last edited: Jun 3, 2008
  8. Jun 3, 2008 #7
    What do you mean by "point labelling"?

    I have no idea what you mean by that hint. Anyways, if g has a cycle (abc...) then by letting h = (bc), h will not commute with g since gh(b) = x where x is not b or c and hg(b) = b. Is that what you meant?

    This is similar to what I wrote in my previous post. h = (13) will not commute with g = (12)(34)...
     
  9. Jun 3, 2008 #8

    Dick

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    By 'point labelling', I just mean do what you are doing, call the elements a,b,c... instead of 1,2,3.... Not that it really matters. I think you are doing all the right things. Can you just pull them altogether into a proof?
     
  10. Jun 3, 2008 #9
    OK. Let g ≠ e be an arbitrary permutation in Sn, n ≥ 3, in disjoint cycle form. Now either (i) g has a 2-cycle, i.e. g = (ab)..., or (ii) g has a m-cycle where m ≥ 3, i.e. g = (abc...)...

    Let h = (bc), c ≠ a. Obviously h is in Sn. Given (i), gh(b) = g(c) ≠ a since only g(b) = a, and hg(b) = h(a) = a. Given (ii) gh(b) = g(c) = x where x is neither b nor c (since only g(a) = b and g(b) = c) and hg(b) = h(c) = c. In either case, gh ≠ hg. Since g is arbitrary, then for every g in Sn, n ≥ 3, there is a permutation h such that gh ≠ hg. Ergo, there can't be a permutation other than e that commutes with all the permutations in Sn and so Z(Sn) = {e}.

    How is that?
     
  11. Jun 3, 2008 #10

    Dick

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    It looks fine. I might have stated it in a more awkward way with even more cases. But that's my problem.
     
  12. Jun 4, 2008 #11
    Thanks a lot Dick.
     
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