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Two questions about group

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data

    1, If Card G=2n, which is even, prove that there exists a g in G, such that g^2=e (e=identity)

    2, show that Sn=< (1 2), (1 2 3 4....n) >


    2. Relevant equations



    3. The attempt at a solution

    1, Well, I cannot come up with a practical idea....I've no idea how to find it directly so I assume that there is no g such that g^(-1)=g, then I came to a contradiction, well, in a particular group G with Card G=4....I've no idea how to generalize this procedure to an arbitrary group.

    2, I have proved that Sn=< (1 2) (1 3) (1 4)....(1 n) > So I want to show (1 i) can be represented by (1 2) and (1 2 ...n), which will be sufficient. But I checked in S4, and found that the product might be quite difficult to find. (e.g. Let a=(1 2 3 4), b=(1 2) in S4, and (1 3) = abaabaa, (1 4)= babaa )

    I'm just start learning group, any suggestion would be greatly appreciated!
     
  2. jcsd
  3. Oct 3, 2008 #2

    morphism

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    Hints:

    1. Pair off elements with their inverses.

    2. What happens when you conjugate an element by a cycle (where conjugating h by g means looking at g^-1 h g)?
     
  4. Oct 4, 2008 #3
    1. It turns out to be quite easy.
    2. Sorry but I've no idea of what happens really... I checked some different conjugations and was confused, since their results seem quite different and have nothing in common. Could you give me more hints about this one? Thanks
     
  5. Oct 4, 2008 #4

    morphism

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    Here's an example (I'm going to use 'conjugate' to mean look at ghg^-1 instead of g^-1hg). Let's conjugate (1 2)(3 4) by (1 2 3) in S_4:
    (1 2 3)(1 2)(3 4)(3 2 1) = (3 2)(1 4)

    Now, the 3-cycle (1 2 3) is the permutation that sends 1 to 2, 2 to 3, 3 to 1, and 4 to 4. What happens when we take the element (1 2)(3 4) and replace each number in it by its image under the permutation (1 2 3)? 1 goes to 2 and 2 goes to 3, so (1 2) becomes (2 3). Similarly, (3 4) becomes (1 4). Thus, (1 2)(3 4) becomes (2 3)(1 4) = (1 2 3)(1 2)(3 4)(3 2 1).

    This sort of behavior is true in general: if [itex]\sigma[/itex] and [itex]\tau[/itex] are arbitrary elements of S_n, then [itex]\tau\sigma\tau^{-1}[/itex] has the cycle structure of [itex]\sigma[/itex], where every entry i in [itex]\sigma[/itex] is replaced by [itex]\tau(i)[/itex].
     
  6. Oct 4, 2008 #5
    Yea, I get it now. This is pretty cool.

    If a(i) = j, then bab^-1( b(i) ) = b(j)
    hence they have same structure.

    so if I have a=(1 2) and b=(1 2 3 ...n) in hand, then bab^-1 = (2 3) = (1 2)(1 3)(1 2), then we get (1 3). By the same procedure we get (3 4) (4 5).....(n-1 n) = (1 n-1)(1 n)(1 n-1), we get (1 4) (1 5)....(1 n).
    Hence <(1 2) (1 2 3....n)>=<(1 2) (1 3)...( 1 n)>=Sn

    Thanks!
     
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