Burnsides Theorem

**maths900** · Oct 15th 2009, 05:32 AM

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?

**NonCommAlg** · Oct 15th 2009, 07:47 AM

Originally Posted by maths900

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.

**maths900** · Oct 15th 2009, 08:49 AM

So would the answer be 11^(8*1) + 11^(8*5) all divided by 16?

**NonCommAlg** · Oct 15th 2009, 08:57 AM

Originally Posted by maths900

So would the answer be 11^(8*1) + 11^(8*5) all divided by 16?

no! you need to read my solution again.

**maths900** · Oct 15th 2009, 09:02 AM

Sorry, i'm lost towards the end. I'm with you up until where you've written more examples.

**maths900** · Oct 15th 2009, 12:36 PM

Is the answer 13442286?

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