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Math Help - Burnsides Theorem

  1. #1
    Junior Member maths900's Avatar
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    Burnsides Theorem

    Using Burnsides Theorem

    How many distinguishable regular octagons can be painted if each edge of a octagon is painted one of eleven colours and different edges can be the same colour?
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    Quote Originally Posted by maths900 View Post
    Using Burnsides Theorem

    How many distinguishable regular octagons can be painted if each edge of a octagon is painted one of eleven colours and different edges can be the same colour?
    the symmetric group of a regular octagon is D_8, the dihedral group of order 16, which is generated by two permutations \sigma = (1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8) and \tau=(2 \ 8)(3 \ 7)(4 \ 6).

    that is D_8 = \{\tau^i \sigma^j: \ \ 0 \leq i \leq 1, \ 0 \leq j \leq 7 \}. for each \tau^i \sigma^j suppose n(\tau \sigma^j) is the number of cycles in the complete decomposition of \tau^i \sigma^j. then Burnside's lemma says

    that the number of coloring for a regular octagon using 11 colors is \frac{1}{|D_8|}\sum_{i,j}11^{n(\tau^i \sigma^j)}=\frac{1}{16} \sum_{i,j} 11^{n(\tau^i \sigma^j)}. as an example, there are 8 cycles (of length 1) in the complete

    decomposition of \text{id}, the identity element of D_8, and so n(\text{id})=8. more examples: \sigma and \tau have one cycle and 5 cycles respectively and thus n(\sigma)=1, \ n(\tau)=5, etc.
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    Junior Member maths900's Avatar
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    So would the answer be 11^(8*1) + 11^(8*5) all divided by 16?
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    Quote Originally Posted by maths900 View Post

    So would the answer be 11^(8*1) + 11^(8*5) all divided by 16?
    no! you need to read my solution again.
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    Junior Member maths900's Avatar
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    Sorry, i'm lost towards the end. I'm with you up until where you've written more examples.
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    Junior Member maths900's Avatar
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    Is the answer 13442286?
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