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Abstract Algebra Concept-based Question

  1. Jan 6, 2010 #1
    I have no abstract algebra background (only matrices and calculus and stats) but this problem came up in one of my classes and this time I'm completely clueless:

    1. The problem statement, all variables and given/known data

    A group is cyclic if an element, g, of the group generates the entire group in the sense that if h is any other element of the group, then h = gk = g * g * g * ... ¤ g for some k. Show that the symmetric group on a finite set, S, is not cyclic if the set has more than 2 elements.

    2. Relevant equations

    no equations, just definitions I think.

    3. The attempt at a solution

    I'm sorry, I really have no idea on this (never seen "groups" before) but I'll take a try on it. (not even sure how to show this properly):

    if the group has 2 elements g and h, then they are related by g * h --> group S
    for some other element x, then gk is not possible since there are only 2 parameters, it cannot accept 3 parameters.
    so g * h * x --> not group S
     
  2. jcsd
  3. Jan 6, 2010 #2

    Dick

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    That 'proof' really doesn't say much. You probably know that. Try taking this approach: i) show a cyclic group is abelian, and ii) show S(n) for n>2 is not abelian.
     
  4. Jan 7, 2010 #3
    Sorry, I don't understand what you mean by an abelian group. Are there certain properties that an abelian group has that a regular group doesn't have?
     
  5. Jan 7, 2010 #4

    Dick

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    In an abelian group every two elements commute. I.e. f*g=g*f. It's true for cyclic groups (prove that). It's not true for Sn with n>2. So Sn can't be cyclic.
     
  6. Jan 11, 2010 #5
    i) so if I have 2 elements a and b of a cyclic group:
    then let a=gx and b=gy
    So... a*b=(gx)*(gy)=g^x+y=g^y+x
    So the cyclic group is Abelian, great.

    ii) Ok then, so if I were to say elements h and x and y are in G, then h*x*y = ga*gb*gc = ga+b+c = gc+b+a = y*x*h

    I think this shows that it is Abelian, but does this show that it is not cyclic (since it is impossible to generate all of the group g with only just one element?)
     
  7. Jan 11, 2010 #6

    Dick

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    You don't really need part ii). Showing a*b=b*a is sufficient. This shows you that any group where you can find two elements that don't commute cannot be cyclic. Can you show Sn for n>=3 has two elements that don't commute?
     
  8. Jan 11, 2010 #7
    Not sure how, and if I could, that would've been the answer to my original problem :)
     
  9. Jan 11, 2010 #8

    Dick

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    Sn is the set of permutation of {1,2,3,...n}, right? Take 'a' to be the permutation that interchanges 1 and 2 and take 'b' to be the permutation that interchanges 2 and 3. Is ab=ba?
     
  10. Jan 11, 2010 #9
    in that case, then no, ab is not equal to ba; ooh, so then those two symmetric sets (simply by switching two (not same) elements in both sets!?!) will show the set used is not commutative, which shows it is not abelian, which finally shows it's not cyclic! Yay! But, how does your set of {1,2,3,...n} qualify as symmetric? Why not say, { n, n+1, n+2, ... etc,} ?
     
  11. Jan 11, 2010 #10

    Dick

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    I'm not sure what you are saying there. Sn is a group of permutations, not a group of sets. I think you might want to review the definition of Sn again. It's seems a little hazy in your mind.
     
  12. Jan 11, 2010 #11
    I'll look it up again, might be confusing permutations with sets again. So that was enough to show that the finite sets were not symmetric?
     
  13. Jan 11, 2010 #12

    Dick

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    Well, no. It's enough to show symmetric groups aren't commutative, hence not cyclic. Your words seem to be all scrambled up.
     
  14. Jan 11, 2010 #13
    sorry, meant cyclic, not symmetric, not sure how that came out (was thinking group theory).
    Thanks so much for your help. :D
     
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