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Homework Help: Dic12 element orders

  1. Dec 23, 2017 #1
    1. The problem statement, all variables and given/known data
    The dicyclic group of order 12 is generated by 2 generators x and y such that: ##y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1} ## where the element of Dic 12 can be written in the form ##x^{k}y^{l}, 0 \leq x < 6, y = 0,1##. Write the product between two group elements in the form ##(x^{k}y^{l})(x^{m}y^{n})## and show that ##a^3## is the only order 2 element and ##a^2## is the only order 3 element in Dic12.

    2. Relevant equations
    ##y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1}##

    3. The attempt at a solution
    The answer I was able to obtain is ##(x^{k}y^{l})(x^{m}y^{n}) = x^{k-m}y^{l+n}## for ##l = -1, 1## and ##(x^{k}y^{l})(x^{m}y^{n}) = x^{k+m}y^{n}## for ##l = 0##. Is this correct? Also, Isn't it that ##a^4## also an order 3 element? So now, I'm confused T_T
     
  2. jcsd
  3. Dec 25, 2017 #2

    Dick

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    For one thing you seem to be using ##x## and ##a## to denote the same symbol. That's confusing. I do think your product relation looks like it works out. Though I'd choose to write it differently (I'm not sure what ##l=-1## is for). But sure, order 3 elements always come in pairs.
     
  4. Dec 25, 2017 #3
    Sorry, a is supposed to be ##x##. I'm confused by the problem that ##x^2## is the only order 3 element, but I also think that ##x^4## is also an order 3. I thought that maybe my product is wrong or something.
     
  5. Dec 25, 2017 #4

    fresh_42

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    Staff: Mentor

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