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I When is D_{n} abelian? What's wrong with the proof?

  1. Feb 14, 2017 #1
    I agree that this could have been done more simply(i'm not looking for an alternative proof), but I don't understand how it is wrong, any insight?

    Since Dn is an dihedral group, we know its elements are symmetries, Dn = (R1,R2,R3...Ri) and since R is a symmetry, we know it's a permutation, so, each Ri can be written as a product of disjoint cycles (w1,w2...wk), now since each element in w represents the vertices's of Dn, which has n vertices's, it follows that w has a total length of n, therefore by problem 4,which states, two non equal cycles of length 2 commute if and only if they are disjoint and also that this isn't the case when the length is larger than 2, we know that Dn is commutative when n <= 2 and therefore abelian when n <= 2.

    I'm not sure if this site has latex, thanks.
  2. jcsd
  3. Feb 15, 2017 #2


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    Staff: Mentor

    So far so good.
    What do you mean by this? Why shouldn't, e.g. ##(1,2,3)(4,5,6,7)(8,9,10,11,12)## don't be commutative?

    The dihedral groups contain rotations and reflections. Wouldn't it be a lot easier to use the reflections as an argument, or if you will two-cycles?
  4. Feb 18, 2017 #3
    ##e,\ (ab),\ (cd)## and ##(ab)(cd)## gives an Abelian group that contains two-cycles from more than two elements and that are disjoint.
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