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Math Help - Question about presentation of cyclic group

  1. #1
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    Question about presentation of cyclic group

    I want to list the elements of T=<a,b : a^6=1, b^2=a^3,bab^{-1}=a^{-1} >

    So from my understanding, a^ia^j=a^{i+j} \ \ \ \forall 0 \leq i,j \leq 5 with a^{i+j}=a^{i+j-6} \ \ i+j \geq 6

    But I'm having trouble trying to make senses of the two other conditions, and how to properly translate them into the table.

    Thank you!!!
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  2. #2
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    Notice first that b^4 = (b^2)^2 = (a^3)^2 = a^6 = 1.

    From the relation bab^{-1} = a^{-1} it follows that ba = a^{-1}b = a^5b. Using that, in any product of a's and b's you can always push the a's to the left of the b's. For example aba^2b = a(ba)ab = a(a^5b)ab = bab = a^5b^2. In that way, you can express any element of the group as a^ib^j, with 0\leqslant i\leqslant5 and 0\leqslant j\leqslant3. That gives you 24 elements of the group, and you still have a fair amount of work to do if you want to write down the whole multiplication table.
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  3. #3
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    and to complete Opalg's argument : also note that since b^2=a^3 and b^3=a^3b, every element of T can be written as a^ib^j with 0 \leq i \leq 5 and 0 \leq j \leq 1. it's easy to show that this

    presentation is unique. so T is a group of order 12 with elements 1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b. the group T is called dicyclic group.
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