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Math Help - D4

  1. #1
    Super Member Bernhard's Avatar
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    Orders of Elements of Dihedral Group D8 (Correction from earlier post)

    Dummit and Foote (D&F) Section 1.2 Dihedral Groups, exercise 1(b) reads as follows:

    Compute the order of each of the elements of D_8.

    D&F give the following as one presentation of D_{2n} :

    D_{2n} = < r,s | r^n = s^2 = 1, rs = sr^{-1}>

    Following this notation I proceeded to compute r, s and sr and (among other elements) was going to determine the order of sr. I proceeded as follows:

    Given the set-up mentioned for the square - see attached pages from D&F on the Dihedral Groups - i defined r and s as follows:

    r = \left(\begin{array}{cccc} 1&2&3&4 \\ 2&3&4&1 \end{array}\right)

    s = \left(\begin{array}{cccc} 1&2&3&4 \\ 1&4&3&2 \end{array}\right)

    Thus

    sr = \left(\begin{array}{cccc} 1&2&3&4 \\ 4&3&2&1 \end{array}\right)

    So from these permutations I calculated that

    (sr)^2 = \left(\begin{array}{cccc} 1&2&3&4 \\ 1&2&3&4 \end{array}\right) = 1


    BUT!! from the relations given by D&F we have s^2 = 1 and so we can proceed thus to find (sr)^2

    (sr)^2 = s^2 r^2 = 1. r^2 ????

    Can anyone help?
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    Last edited by Bernhard; October 12th 2011 at 03:17 AM.
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  2. #2
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    Re: Orders of Elements of Dihedral Group D8 (Correction from earlier post)

    in cycle notation, we have:

    r = (1 2 3 4), s = (1 4)(2 3).

    so sr = (1 3), a transposition, which is clearly of order 2.

    your mistake is in your calculation of (sr)^2 algebraically.

    it is NOT the case that (sr)^2 = s^2r^2. typically, that is only true in abelian groups.

    what you should have is this:

    (sr)^2 = (sr)(sr) = s(rs)r (by associativity)

    = s(sr^-1)r (from rs = sr^-1)

    = (ss)(r^-1r) = (1)(1) = 1.
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