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Symmetric and Alternating Groups disjoint cycles

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Let a = (a1a2..ak) and b = (c1c2..ck) be disjoint cycles in Sn. Prove that ab = ba.

    3. The attempt at a solution
    Sn consists of the permutations of the elements of T where T = {1,2,3,...,n}
    so assume we take an i from T. Then either i is in a, i is in b, or i is in neither a or b

    1. assume i is in a.
    then a o b(i) = a(i) as b(i) = i since b maps it to itself
    then b o a(i) = b(a(i)) = a(i) as b maps this a(i) to itself

    2. assume i is in b.

    then a o b (i) = a(b(i)) = b(i) as a maps this b(i) to itself
    then b o a(i) = b(i)

    3. assume it is in neither

    then a o b (i) = a(i) = i as both a and b map i to itself
    then b o a (i) = b(i) = i

    so regardless of which permutation i is in, we see ab = ba for disjoint cycles a and b.
     
  2. jcsd
  3. May 25, 2012 #2
    Your proof is fine. Is there a question about this?
     
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