1. ## 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?

2. 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.

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

4. 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.

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