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Math Help - Symmetric Groups Questions

  1. #1
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    Symmetric Groups Questions

    1. Let f = (x1, x2, ...., xr) in Sn. Show that o(f) = r.
    (I have to do this using cycles or transpositions or something, but I don't know how it would work.)

    2. Prove that Sn is non-abelian if n is greater than or equal to 3.
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  2. #2
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    Quote Originally Posted by Janu42 View Post
    1. Let f = (x1, x2, ...., xr) in Sn. Show that o(f) = r.
    (I have to do this using cycles or transpositions or something, but I don't know how it would work.)
    To show that |f|=r you need to show that f^r = \text{id} (identity permuation) and that f^k \not = \text{id} if k<r. Also it would be helpful to write f = (x_0,x_1,...,x_{r-1}) (you will see why) instead.

    First we show that f^r is the identity permuation. This means we need to show f^r(x) = x for all x\in S_n. We can assume that x\in \{x_0,...,x_{r-1}\} because otherwise f(x) = x and so there is no need to check that case. Remember that f(x_j) = x_{j+1\bmod n} (this is why we rewrote the indicies differently). Therefore, f^k (x_j) = x_{j+k\bmod n}. If we want f^k(x_j) = x_j it must mean that j\equiv j+k(\bmod n) \implies n|k. Therefore, the smallest such k is n and so |f|=n.

    2. Prove that Sn is non-abelian if n is greater than or equal to 3.
    Compare (12)(23)\text{ and }(23)(12).
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    And for 2,
    (12)(24) and (24)(12) is different as well, and so on... correct?

    Like to prove it, I could take an x greater than or equal to 3, and do
    (12)(2x) does not equal (2x)(12)?
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  4. #4
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    Quote Originally Posted by Janu42 View Post
    And for 2,
    (12)(24) and (24)(12) is different as well, and so on... correct?
    You need to be careful, the problem said S_n for n\geq 3 so we can find (12),(23). However, we do not necessarily have (24) because the group is not necessary S_4 it can be S_3.
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