2
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\begin{bmatrix}1&2&3&4&5&6&7&8\\2&3&4&5&1&7&8&6\end{bmatrix}

I have already written this permutation as disjoint cycles: (12345)(678)

My attempt at a product of two cycles: (12)(23)(34)(45)(67)(78), but I don't really think this is right, and if it is right, I'm not sure why. It's just an honest guess.

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  • $\begingroup$ Your answer looks correct to me. $\endgroup$ – jwc845 Sep 22 '18 at 23:57
  • $\begingroup$ Why don't you just work out the product of those six $2$-cycles and see what you get? Hint: it's correct. $\endgroup$ – bof Sep 22 '18 at 23:57
3
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Indeed, your product of disjoint cycles correctly represent the given permutation. And also your transpositions from the disjoint cycles are spot on.

See also this answer for a more thorough discussion of ways to write the product of transpositions for disjoint permutations.

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  • $\begingroup$ I've already seen that answer, but as I can't comment yet due to not having enough reputation, I figured it might be better to ask my own question. In that answer, I have no idea how they were able to get their answer in method 1. $\endgroup$ – Alexia Paskevicius Sep 23 '18 at 0:02
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You are correct. In general we have $(x_1, x_2, x_3,...,x_k)=(x_1, x_2)(x_2, x_3)(x_3, x_4)...(x_{k-1}, x_k)$. Very easy to see why this is correct, just check where does each of the two permutations send each element.

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